EngineeringMechanics by J L Meriam 6th Solutions by Khalid Yousaf

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587

Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

By

Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.



What is Mechanics? Mechanics is the physical science which deals with the effects of forces on objects. The subject of mechanics is logically divided into two parts: statics,which concerns the equilibrium of bodies under action of forces, and dynamics, which concerns the motion of bodies. BASIC CONCEPTS The following concepts and definitions are basic to the study of mechanics, and they should be understood at the outset. Space is the geometric region occupied by bodies whose positions are described by linear and angular measurements relative to a coordinate system. For three-dimensional problems, three independent coordinates are needed. For two-dimensional problems, only two coordinates are required. Time is the measure of the succession of events and is a basic quantity in dynamics. Time is not directly involved in the analysis of statics problems. Mass is a measure of the inertia of a body, which is its resistance to a change of velocity. Mass can also be thought of as the quantity of matter in a body. The mass of a body affects the gravitational attraction force between it and other bodies. This force appears in many applications in statics. Force is the action of one body on another. A force tends to move a body in the direction of its action. The action of a force is characterized by its magnitude, by the direction of its action, and by its point of application. Thus force is a vector quantity.. A particle is a body of negligible dimensions. In the mathematical sense, a particle is a body whose dimensions are considered to be near zero so that we may analyze it as a mass concentrated at a point. We often choose a particle as a differential element of a body. We may treat a body as a particle when its dimensions are irrelevant to the description of its position or the action of forces applied to it. Rigid body. A body is considered rigid when the change in distance between any two of its points is negligible for the purpose at hand. For instance, the calculation of the tension in the cable which supports the boom of a mobile crane under load is essentially unaffected by the small internal deformations in the structural members of the boom. For the purpose, then, of determining the external forces which act on the boom, we may treat it as a rigid body. Statics deals primarily with the calculation of external forces which act on rigid bodies in equilibrium. Determination of the internal deformations belongs to the study of the mechanics of deformable bodies, which normally follows statics in the curriculum.







PROBLEMS Introductory Problems 2/1 The force F has a magnitude of 800 N. Express F as a vector in terms of the unit vectors i and j. Identify the x and y scalar components of F.

Soln. Step1: Free body Diagram

Step2:Magnitude of force 800N x component of force, Fx=‐Fsin35o

Fx=‐800sin35o Fx=‐459 N y component of force, Fy=Fcos35o

Fy=800 cos35o Fy=655 N Force vector, F=(‐459i‐655j)N

2/2 The magnitude of the force F is 600 N. Express F as a vector in terms

of the unit vectors i and j. Identify both the scalar and vector components of F.

Soln.

Step 1: Magnitude of force F=600 lb Step 2: Free body diagram

Step3: Force vector F=600cos30i ‐ 600sin30j F=520i – 300j

Step 4:Scalar components of force Along x‐axis Fx=520 lb Along y‐axis Fy= ‐300 lb Step5:Vector components of force Along x‐axis Fx=520i lb Along y‐axis Fy= ‐300j lb



2/3 The slope of the 4.8-kN force F is specified as shown in the figure. Express F as a vector in terms of the unit vectors i and j.

Soln. Step1:Free body diagram

Step2: Magnitude of force, F=4.8 kN

Unit vector of force, n= - i- j Force vector F= F n

F=4.8(- i- j) =(-2.88i-3.84j)kN 2/4 The line of action of the 9.6-kN force F runs through the points A and B as shown in the figure. Determine the x and y scalar components of F.

Soln. Step1: Magnitude of force F=4800 lb Position of point A= -15i-20j Position of point B=30i+10j Step2: Free body diagram

Stpe3: Position vector of AB, AB=OB – OA AB=45i+30j Magnitude of AB, AB =√45 30 = 54.08 in

Unit vector of AB, n=

= .

2/5 A cable stretched between the fixed supports A and B is under a tension T of 900 lb. Express the tension as a vector using the unit vectors i and j, first, as a forceTA acting on A and second, as a force TB acting on B.




Step2: Magnitude of tension in cable AB, T =

900 lb

Unit vector of AB=nAB=√

-

√j

nAB = 0.832i - 0.55j TA= TA nAB TA=900(0.832i - 0.55j) TA=(749i-499j)lb But TA= - TB =-(749i-499j)lb =(-749i+499j)lb 2/6 The 1800-N force F is applied to the end of the I beam. Express F as a vector using the unit vectors i and j.

Soln. Stpe1: Magnitude of force F=1800 N

Unit vector of force, n= - i- j

Force vector F= F n

F=1800(- i- j) =-1080iN-1440jN 2/7 The two structural members, one of which is in tension and the other in compression, exert the indicated forces on joint O. Determine the magnitude of the resultant R of the two forces and the angle θ which R makes with the positive x-axis.


x-components of resultant force Rx=∑Fx

Rx= -3cos60o-2cos30o Rx= -3.23 kN y-components of resultant force Ry=∑Fy

Ry= 3sin60o+2 sin30o Ry= -1.598 kN Magnitude of the resultant force

R= R R

R= 3.23 1.598

R=3.6 kN



Step3: Angle θ made by ‘R’

θ=tan-1( )

=tan-1(.

.)

=26o Angle θ made by ‘R’ with positive x-axis θ=180o+26o =206o

2/8 Two forces are applied to the construction bracket as shown. Determine the angle θ which makes the resultant of the two forces vertical. Determine the magnitude R of the resultant.


Step2:

F1=800 lb F2=425 lb Given that the resultant force is normal to x-axis.Therefore the x-component of resultant force is zero. Rx=∑Fx = 0 Rx= -425cosθ+800cos70o = 0

Cosθ=800 70425

θ=49.90

y-components of resultant force Ry=∑Fy

Ry= -425sin49.9-800 sin70o Ry= -1070 lb Magnitude of the resultant force

R= R R

R= 0 1077 R=1077 lb

Representative Problems

2/9 In the design of a control mechanism, it is determined that rod AB transmits a 260-N force P to the crank BC. Determine the x and y scalar components of P. Soln. Step1: Free body Diagram

Step2:Magnitude of force P=260 N

Unit vector of force, n= - i- j Force vector P= P n

P=260(- i- j) =-240iN-100jN Scalar component of P along x Px=-240 N Scalar component of P along y Py=-100 N 2/10 For the mechanism of Prob. 2/9, determine the scalar components Pt and Pn of P which are tangent and normal, respectively, to crank BC. Soln. Step1:Free body diagram



Step2:magnitude of force F=260 N

tan‐1( )

θ=22.6o Scalar components of force P along ‘n’ Pn=260 cos(30o‐22.6o) Pn=258 N Scalar components of force P along ‘t’ Pt=260 sin(30o‐22.6o) Pt=33. 5N 2/11 The t-component of the force F is known to be 75N. Determine the n-component and the magnitude of F.


Step2: The t-component of the force F,Ft=75 N.Let Fn be the n-component of the force F

Ft=Fcos40o (1) Fn=-Fsinθ (2) Dividing equation (2) by (1)

=-tan40o

Fn=-Ft tan40o Fn=-75 tan40o Fn=-62.9o By equation (1) Ft=Fcos40o 75=F cos40o F=97.9 N 2/12 A force F of magnitude 800 N is applied to point C of the bar AB as shown. Determine both the x-y and the n-t components of F.

Step1: Free body diagram

Step2: Components of force F along ‘x’ Fx= ‐800 cos200



Fx= ‐752 lb Components of force F along ‘y’ Fy= 800 sin200

Fy= 274 lb Step3: Components of force along ‘t’

Ft= ‐800 cos400

Ft= ‐613 lb Components of force along ‘n’

Fn= ‐800 sin400

Fn= ‐514 lb 2/13 The two forces shown act at point A of the bent bar. Determine the resultant R of the two forces.

Step1: Free body diagram

Step2: x-components of resultant force Rx=∑Fx

Rx= 7cos45o-3cos30o Rx= 2.35 kips y-components of resultant force Ry=∑Fy

Ry= -7sin60o+3 sin30o Ry= -3.45 kips The resultant of the two forces is R=Rxi+Ryj R= (2.35i-3.45j)kips 2/14 To satisfy design limitations it is necessary to determine the effect of the 2-kN tension in the cable on the shear, tension, and bending of the fixed I-beam. For this purpose replace this force by its equivalent of two forces at A, Ft parallel and Fn perpendicular to the beam. Determine Ft and Fn.


Step2: Components of force along ‘t’

Ft= 2 cos(200+300) Ft=1.286 kN Components of force along ‘n’

Fn=2 sin(200+300) Fn=1.532 kN 2/15 Determine the magnitude Fs of the tensile spring force in order that the resultant of Fs and F is a vertical force. Determine the magnitude R of this vertical resultant force.




Step2:Magnitude of force F=120 lb Tensile spring force be Fs

Since the horizontal component of ‘R’ is zero F=Fs cos60o F=120 cos60o F=60 lb The resultant force R=Fsin60o R=120sin60o R=103.9 lb 2/16 The ratio of the lift force L to the drag force D for the simple airfoil is L/D =10. If the lift force on a short section of the airfoil is 200 N, compute the magnitude of the resultant force R and the angle θ which it makes with the horizontal.

Soln. Step1: Free body diagram

Step2: Lift force L=50 lb

Ratio of the lift force to drag force =10

=10

D=5 lb Magnitude of resultant force R =√ R=√5 50 R=50.2 lb Step3: Angle made by the resultant with ‘D’

tan-1( )

tan-1( ) θ 84.3o 2/17 Determine the components of the 2-kN force along the oblique axes a and b. Determine the projections of F onto the a- and b-axes.




Step2: By the law of sine’s

=

Fa= ×2

Fa=0.598 kN

=

Fb= ×2

Fb=1.633 kN Step3:

Let Pa and Pb be the projections of force P along a and b. Pa=2 cos45o

Pa=1.414 kN

Pb=2 cos15o

Pb=1.932 kN

2/18 Determine the scalar components Ra and Rb of the force R along the non rectan -gular axes a and b. Also determine the orthogonal projection Pa of R onto axis a.


Step2: By the law of sine

30= 40 Rb= 622 N

110= 40 Ra= 1170 N Step3:Let Pa be the orthogonal projection of P onto a-axis

Step4: Pa=R cos300

=800 cos300

=693 N 2/19 Determine the resultant R of the two forces shown by (a) applying the parallelogram rule for vector addition and (b) summing scalar components.




Step2:(a) From the law of cosine’s R2=4002+6002 – 2(400)(600)cos1202 R2=760000 R2=872 N Let θ be the angle made with the vertical,then by the law of sine’s

= =36.6o

Step3:(b) x-components of resultant force Rx=∑Fx

Rx=600cos30o

Rx=520 N y-components of resultant force Ry=∑Fy Ry=-400-600sin30o Ry=-700 N Magnitude of the resultant force

R= R R

R= 520 700 R=872 N Angle θ made by ‘R’

θ=tan-1( )

=tan-1( ) =53.4o So the angle made by resultant with the vertical θ=90o-53.4o

θ=36.6o 2/20 It is desired to remove the spike from the timber by applying force along its horizontal axis. An obstruction A prevents direct access, so that two forces, one 1.6 kN and the other P, are applied by cables as shown. Compute the magnitude of P necessary to ensure a resultant T directed along the spike. Also find T.


Step2: From figure

θ1=tan-1( )

θ1=26.57o

θ2=tan-1( ) θ2=36.87o



α=180o – (θ1+θ2) α=180o – (26.57o+36.87o) α=116.56o

By the law of sine

=

P= 537 lb

=2

T= 800.541 lb 2/21 At what angle θ must the 800-N force be applied in order that the resultant R of the two forces has a magnitude of 2000 N? For this condition, determine the angle θ between R and the vertical.


Step2: Magnitude of the resultant force R=2000 lb From the law of cosine’s 20002=14002+8002 – 2(1400)(800)cos(180o-θ) But cos(180o‐θ)=‐cosθ Therefore 20002=14002+8002 + 2(1400)(800)cosθ θ=51.3o From the law of sine’s

=

=R×

β=18.19o 2/22 The unstretched length of the spring of modulus k =1.2 kN/m is 100 mm.When pin P is in the position θ=30o determine the x- and y-components of the force which the spring exerts on the pin. (The force in a spring is given by F=kx, where x is the deflection from the unstretched length.)


Step2: Givent hat spring modulus k=1.2 kN/m

Unstretched length of spring l =0.1m Position of point A=0.06i+0.04j Position of point P=0.08sin30i-0.08cos30j P=0.04i-0.0693j Position vector PA=OA-OP PA=(0.06-0.04)i-[(0.04)-(-0.0693)]j PA=0.02i+0.1093j



Magnitude of PA=√0.02 0.1093 PA=0.1111m

Unit vector nPA=

nPA= 0.02 0.10930.02 2 0.1093 2

nPA=0.18i+0.984j Magnitude of spring force F= Kx Where x is change in length.

x=PA- l x =0.1111-0.1 x=0.0111m F=1.2 kN×0.0111 F=13.32 N Step3: Spring force vector

F=F nPA

=13.32(0.18i+0.984j) =(2.4i+13.1j) N x and y components of force are Fx=2.4 N Fy=13.1 N 2/23 Refer to the statement and figure of Prob. 2/22. When pin P is in the position θ=20o,determine the n- and t-components of the force F which the spring of modulus k =1.2 kN/m exerts on the pin.

Step1:Given that Spring constant of the spring k=1.2 kN/m Unstrectched length of the spring, l=0.1m The angle made by the pin with the vertical θ=20o Let the centre of the coordinate system be positioned at 0. Step2:The arrangement of the spring system at the present instant is as shown by the free body diagram

Step3: The coordinates of points A and P are A=(0.06i,0.04j) P=(0.08 sin20oi-0.08cos20oj) P=(0.0274i-0.0752j) Hence the position vector PA is given by PA=OA-OP PA=(0.06-0.0274)i + [0.04-(-0.0752]j PA=0.0326i+0.1152j The magnitude of the vector is given by PA= 0.0326 0.1152 PA=0.1197m The magnitude of the spring force is given by F=kx Where x=PA - 0.01 x=0.1197-0.01 x=0.0197m F=1.2×0.0197 F=0.0237 kN F=23.7 kN 2/24 The cable AB prevents bar OA from rotating clockwise about the pivot O. If the cable tension is 750N.Determine the n- and t-components of this force acting on point A of the bar.




Step2: Tension in the cable T=750 N Let Tn and Tt be the components of force ‘T’ along ‘n’ and ‘t’axes respectively By the cosine law AB= OA OB 2 OA OB cos120

AB= 1.5 1.2 2 1.5 1.2 0.866

AB=2.34 m Step3: Applying sine rule for triangle BOA

.=

Sin =.

×1.2

=26.37o Step3: Components of force T along ‘n’ Tn=T Sin Tn=750 Sin26.37 Tn=333.12 N

Components of force T along ‘t’ Tt=‐T cos Tn=‐750 cos26.37 Tn=‐672 N 2/25 At what angle θ must the 400-N force be applied in order that the resultant R of the two forces have a magnitude of 1000 N? For this condition what will be the angle θ between R and the horizontal?

Step1:Free body diagram

Step2: Magnitude of the resultant force R=1000 lb Let θ be the angle between the resultant force ‘R’ and horizontal force 700 lb. From the law of cosine’s 10002=4002+7002 – 2(400)(700)cos(180o-θ) cosθ=0.625 θ=51.32o By the law of sine’s

=

=400× .

β=18.19o



2/26 In the design of the robot to insert the small cylindrical part into a close-fitting circular hole, the robot arm must exert a 90-N force P on the part parallel to the axis of the hole as shown. Determine the components of the force which the part exerts on the robot along axes (a) parallel and perpendicular to the arm AB, and (b) parallel and perpendicular to the arm BC.

Soln. Step1: Given Force exerted by robot P=90 N Let parallel force = Pt

Let perpendicular force = Pn Step2: (a) Components of force which the part exerts on the robot along parallel and perpendicular to the angle arm AB.

By resolving forces, Pt=‐90cos30 Pt=‐77.9 N (parallel force) Pn=90sin30 Pn=45 N (perpendicular force) Step3: (b) Parallel and perpendicular forces to the arm BC.

By resolving forces Pt=90sin45 Pt=63.6 N (parallel force) Pn=90cos45 Pn=63.6 N (perpendicular force) 2/27 The guy cables AB and AC are attached to the top of the transmission tower. The tension in cable AC is 8 kN. Determine the required tension T in cable AB such that the net effect of the two cable tensions is a downward force at point A. Determine the magnitude R of this downward force.


Step2:



Tension in the cable AC,Tac=8 kN Let the tension in the cable AB be Tab

Let the magnitude of the resultant force be R From the figure

θ1=tan-1( ) θ1=51.3o

θ2=tan-1( ) θ2=37.3o

θ3=180o - θ1 - θ2 θ3=95o By the law of sine

=

Tab=400×

Tab=400×.

.

Tab=5.68 kN

=

R=8×

R=8×.

R=10.21 kN 2/28 The gusset plate is subjected to the two forces shown. Replace them by two equivalent forces, Fx in the x-direction and Fa in the a-direction. Determine the magnitudes of Fx and Fa. Solve geometrically or graphically.

Soln. Step1:Geometric solution

Step2: Let Fx

and Fa be the forces along x and a-axes respectively. Let R be the resultant force. From law of cosine R2=9002+8002-2(900)(800)cos75o R2=1077300.575 R=1037.93 Step3: By the law of sine

=.

=0.744 α=48.12o.

From triangle ‘OCD’ 180o=β+45o+65o+ α 180o=β+45o+65o+48.12o β =21.88o By the law of sine

=.

Fx=547.02 N Step3: Now considering triangle ‘OCa’ By the law of sine

=

.

.=

.

.=

.

Fa=1349.963 N





PROBLEMS Introductory Problems 2/29 The 10 kN force is applied atpoint A. Determine the moment of F about point O. Determine the points on the x- and y-axes about which the moment of F is zero.


Step2: The magnitude of force F=10 kN Let B and C be the points on y and x-axes respectively. Components of force along x and y-axes.

Fx=10× Fx=8 kN

Fy=10× Fy=6 kN Taking moment about ‘O’ (CW +) Mo=Fx×5 - Fy×4 Mo=8×5 - 6×5 Mo=10 kNm

Step3:Considering similar triangles

=

4+x= x=2.67 m

=

y=2.67× y=2 m point B(0,2) C(2.67,0) 2/30 Determine the moment of the 800-N force about point A and about point O.


Step2:Magnitude of force F=200 lb



Taking moment about ‘A’ MA=200cos30o×35 MA=6062.2 lb-in (CW) Taking moment about ‘O’ MO = 200cos30o×35 – 200sin30o×35 MO = 6062.2 – 2500 MO = 3562.2 lb –in (CW) 2/31 Determine the moment of the 50N force (a) about point O by Varignon’s theorem and (b) about point C by a vector approach.


Step2: Given that Force acting along line AB,F=50 kN From figure. Position vector of A,OA=‐15i‐20j Position vector of B,OB=40i+10j Position vector of C,OC =25j Vector form of AB,AB=OB‐OA AB=(40+15)i+(10+20)j AB=55i+30j

Force vector along line AB,F=Fǀ ǀ

F=F[ √

]

F=50[ .

]

F=43.89i+23.94j Step3:(a) Taking moment about ‘O’ Mo=OA×F Mo=(‐15i‐20j)×( 43.89i+23.94j) Mo=(‐15×23.94)k+(‐20×43.89)(‐k) Mo=518.7k Magnitude of moment about ‘O’ is Mo=518.7 Nmm(CCW) Step4:(b) Moment about ‘C’ Mc=CA×F Position vector of CA,CA=OA‐OC CA=‐15i+(‐20‐25)j CA=‐15i‐45j Mc=(‐15i+(‐20‐25)j)×( 43.89i+23.94j) Mc=(‐15×23.94)k+(‐45×43.89)(‐k) Mc=1615.95k Magnitude of moment about ‘C’ is Mc=1615.95 Nmm(CCW) 2/32 The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O.


Step2: Point A(0,h) Point B(b,0) Vector AB=bi‐hj

Unit Vector AB=nAB=√



Vector OA=hj Moment about ‘O’ MO= OA × F

MO= hj ×[ √ ]

MO= ‐ √

k MO=√ (CW) 2/33 In steadily turning the water pump,a person exerts the 120N force on the handle as shown. Determine the moment of this force about point O.


Step2: Magnitude of force F=120 N Moment of force about ‘O’ (CW +) Mo=F cos(20o+15o)×0.15 Mo=120cos35o×0.15 Mo=14.74 Nm 2/34 The throttle-control sector pivots freely at O. If an internal torsional spring exerts a return moment M =1.8 on the sector when in the position shown, for design purposes determine the necessary throttle-cable tension T so that the net moment about O is zero. Note that when T is zero, the sectorrests against the idle-control adjustment screw at R.


Step2: Let the tension required be ‘T’ Taking moment about ‘O’ (Anticlock wise +) MO=0 1.8 - T×0.05=0

T=.

.

T=36 N 2/35 A force F of magnitude 60 N is applied to the gear. Determine the moment of F about point O.




Step2: Magnitude of force F=60 N Fy=F cos cos20o Fy=60 cos cos20o Fy=56.38 N Taking moment about ‘O’ (CW+) Mo=Fy×0.1 Mo=56.38×0.1 Mo=5.64 Nm 2/36 Calculate the moment of the 250-N force on the handle of the monkey wrench about the center of the bolt.


Step2: Given Magnitude of force F=250 N x and y components of force ‘F’ Fy=250 cos15o

Fy=241.48 N Fx=250 sin15o Fx=64.7 N Step3: Taking moment about ‘O’ Mo – Fy(0.2)+Fx(0.03)=0 Mo – 241.48(0.2)+64.7(0.03)=0 Mo - 48.296+1.92 =0 Mo=46.36 Nm 2/37 A mechanic pulls on the 13-mm combination wrench with the 140-N force shown. Determine the moment of this force about the bolt center O.


Step2: Magnitude of force F=140 N



Taking moment about ‘O’ (CCW+) Mo=Fcosθ×0.095 Mo=Fcos(25o-10o)×0.095 Mo=140cos10o×0.095 Mo=13.1 Nm 2/38 As a trailer is towed in the forward direction, the force F =500 N is applied as shown to the ball of the trailer hitch. Determine the moment of this force about point O.


Step2: Given that force F=120 lb Let Fx and Fy be the components of force along x and y‐axes respectively. Taking moment about ‘O’ (Anticlock wise +) MO=Fx ×1.5+Fy × 11 Mo= 120 sin30o

×1.5+120cos30o × 11 Mo=90+1143.15 Mo=1233.15 lb-in (CW) 2/39 A portion of a mechanical coin sorter works as follows:Pennies and dimes roll down the 20o incline,the last triangular portion of which pivots freely about a horizontal axis through O. Dimes are light enough (2.28 grams each) so that the triangular portion remains stationary, and the dimes roll into theright collection column. Pennies, on the other hand, are heavy enough (3.06 grams each) so that the

triangular portion pivots clockwise, and the pennies roll into the left collection column. Determine the moment about O of the weight of the penny in terms of the slant distance s in millimeters.


Step2: mgcos200 and mgsin200 are components of weight perpendicular parallel to the inclined plane respectively. Taking moment about pivot ‘O’ (CW +) Mo= mgcos200×s+ mgsin200(9.5+3.5) Mo=3.06×10‐3×9.81cos200×s+3.06×10‐3

×9.81sin200(9.5+3.5) Mo=(0.0282s+0.1335) Nmm (s in mm) 2/40 Elements of the lower arm are shown in the figure.The mass of the forearm is 2.3 kg with mass center at G. Determine the combined moment about the elbow pivot O of the weights of the forearm and the 3.6kg homogeneous sphere. What must the biceps tension force be so that the overall moment about O is zero?




Step2: Given that Weight of the fore arm W1=5 lb Weight of the sphere W2=8 lb Let ‘T’ be the tension in the bicep. Moment about ‘O’ due to the weights 5 lb and 8 lb. (Clock wise +) Mo=8×13+5×6sin55o

Mo=104+24.57 Mo=128.57 lb‐in (CW) The combined moment due to the all forces is zero. (Clock wise +)∑ Mo=0 128.57-T×2=0 T=64.29 lb 2/41 A 32 lb pull T is applied to a cord,which is wound securely around the inner hub of the drum.Determine

the moment of T about the drum center C. At what angle θ should T be applied so that the moment about the contact point P is zero?


Step2: Magnitude of force T=32 lb Taking moment about ‘C’(CW+) Mc=T×5 Mc=32×5 Mc=160 lb-in For the moment about the contact point P to be zero,the applied force should pass through point P. Let θ be the angle of T with the horizontal such that

From the triangle PCB

Cosθ=

Cosθ= θ=51.3o



2/42 A force of 200 N is applied to the end of the wrench to tighten a flange bolt which holds the wheel to the axle. Determine the moment M produced by this force about the center O of the wheel for the position of the wrench shown.


Step2: Magnitude of force F=200 N OA=450-62.5 cos20 OA=391.27 mm Moment about ‘O’ (Clock wise +) Mo=F×OA Mo=200×391.27 Mo=78253.84 Nmm 2/43 In order to raise the flagpole OC, a light frame OAB is attached to the pole and a tension of 780 lb is developed in the hoisting cable by the power winch D. Calculate the moment Mo of this tension about thehinge point O.


Step2: Magnitude of tenion T=780 lb Taking moment about ‘O’ (CCW +) Mo=780cos20o×10cos30o-780sin20o×5 Mo=6347.62-1333.88 Mo=5013.74 lb-ft 2/44 The uniform work platform, which has a mass per unit length of 28 kg/m, is simply supported by cross rods A and B. The 90-kg construction worker starts from point B and walks to the right. At what location ‘s’ will the combined moment of the weights of the man and platform about point B be zero?


Step2: Given that Mass of the platform M=28 kg/m Mass of the worker m=90 kg Weight of the platform W=28×(1+4+3)×9.81 W=2197.44 N Moment of all force at ‘B’ is zero.



MB=0 mg×S -W×1=0 mg×S=W

S=

S=.

.

S=2.49 m 2/45 In raising the pole from the position shown, the tension T in the cable must supply a moment about O of 72kNm . Determine T.


Step2: Given,moment about ‘O’ Mo=72 kN Let ‘T’ be the tension in the cable Considering triable OBC

sin60o= h=25.98 m

cos60o=

x=15 m Now considering triangle ‘ABC’

tanθ=

tanθ=.

tanθ=.

θ=tan-10.9622 θ=43.897o From ΔOAC 1800=α+θ+120o

1800=α+43.897o +120o

α=16.103o Step3:

Taking moment about ‘O’ ∑Mo=0 Mo-Tsinα×30=0 Tsin16.103o×30=72 T=8.653 kN 2/46 The force exerted by the plunger of cylinder AB on the door is 40 N directed along the line AB, and this force tends to keep the door closed. Compute the moment of this force about the hinge O. What force FC normal to the plane of the door must the door stop at C exert on the door so that the combined moment about O of the two forces is zero?




Step2: Given that Force exerted by the plunger F=40 N

tanθ= θ=14.036o

Taking moment about point ‘O’ Mo=40cos14.036o(75)+ 40sin14.036o(425) Mo=2910.43+4123.035 Mo=7033.465 Nmm (CCW) Let ‘Fc’ be the reaction force at ‘C’ Taking moment about point ‘O’ ∑Mo=0 Fc (825) ‐ Mo=0 Fc (825) ‐ 7033.465 =0 Fc (825) = 7033.465 Fc =8.525 N 2/47 The 2lb force is applied to the handle of the hydraulic control valve as shown. Calculate the moment of this force about point O.


Step2: Given,applied force F=2 lb Taking moment about ‘O’ (CCW +) ∑Mo=-2cos20o×(10sin60o+1.5)- 2sin20o×(1ocos60o) ∑Mo=-19.095-3.42 ∑Mo=-22.5 lb-in 2/48 Calculate the moment MA of the 200-N force about point A by using three scalar methods and one vector method.


Step2: Method1 Taking moment about ‘A’ MA=Fcos15o×280+ Fsin15o×400 MA=200cos15o×280+ 200sin15o×400 MA=74797.4 Nm (CW) Step3: Method2



Considering the force applied at ‘D’ CD=400tan15o CD=107.18 mm DA=AC+CD DA=280+107.18 DA=387.18 mm Taking moment about ‘A’ MA= 200cos15o×DA MA= 200cos15o×387.18 MA=74797.4 Nm (CW) Step4: Method3 Finding the perpendicular distance AE DA=387.18 mm DE=DAsin15o DE=387.18 sin15o DE=100.21 mm

AE=√

AE=√387.18 100.21 AE=374 mm Taking moment about ‘A’ MA=200×374 MA=74797.4 Nm Step5: Vector method r=200i+480j F=‐200 cos15oi+200 sin15oj MA= r × F MA=(200i+480j) × (=‐200 cos15oi+200 sin15oj) MA=74797.4k Nm 2/50 (a) Calculate the moment of the 90-N force about point O for the condition θ=15o. Also, determinethe value of θ for which the moment about O is (b) zero and (c) a maximum.


Step2: Elastic modulus of band k=60 N/m Unstretched length of band xo=0.74m OC=OA+AC OC=0.635+0.74 OC=1.375 m

θ=tan-1 .

.

θ=65.2o Step3: Change in the length of band x=BC-xo

x=1.515-0.74 Deflection of spring x=0.775m Spring force F=kx F=60×0.775 F=46.5 N Taking moment about ‘O’ (CW +) Mo=Fsinθ×OB Mo=46.5sin62.5o×0.535 Mo=26.8 Nm 2/50 (a) Calculate the moment of the 90-N force about point O for the condition θ=15o. Also, determine the value of θ for which the moment about O is (b) zero and (c) a maximum.



Soln. Given that Applied force F=90 N

a) θ=15o Step1: Free body diagram

Step2: (Anticlock wise +) Mo=Fcosθ×0.6 - Fsinθ × 0.8 Mo=90cos15o×0.6+ 90sin15o×0.8 Mo=33.52 Nm Step3: b) Moment of the force about ‘O’ is zero. Mo=0 Fcosθ×0.6 - Fsinθ × 0.8=0 cosθ×0.6=sinθ × 0.8

tanθ=.

.

θ=36.9o Step4:

c) For the moment to be maximum the applied force should be perpendicular.

tanα=.

.

α=36.9o

θ=90+α θ=90+36.9o θ=126.9o 2/51 The small crane is mounted along the side of a pickup bed and facilitates the handling of heavy loads. When the boom elevation angle is θ=40o,the force in the hydraulic cylinder BC is 4.5 kN, and this force applied at point C is in the direction from B toC (the cylinder is in compression). Determine the moment of this 4.5-kN force about the boom pivotpoint O.


Step2: Let the angle made by the force F with the horizontal be α

tanα=

tanα=.

.



α=56.2o Taking moment about ‘O’ (CW +) Mo=Fcosα×OB Mo=4.5cos56.2o ×0.36 Mo=0.902 kNm 2/52 Design criteria require that the robot exert the 90-N force on the part as shown while inserting a cylindrical part into the circular hole. Determine the moment about points A, B, and C of the force which the part exerts on the robot.


Step2: Given that Force exerted by the robot at ‘D’ P=90 N Taking moment about ‘C’ ∑MC=0 MC – P(150)=0 MC=13500 Nmm MC=13.5 Nm Taking moment about ‘B’

∑MB=0 MB – P(EF+FB)=0 MB = P(EF+FB) MB =90(150+450sin30o) MB =90(375) MB =33750 Nmm MB =33.75 Nm Step3: Taking moment about ‘A’ ∑MA=0 MA – P(EF+FB+BG)=0 MA = P(EF+FB+BG) MA =90(150+450sin30o+550sin45o) MA =90(3763.91) MA =68751.9 Nmm MA =68.752 Nm 2/53 The masthead fitting supports the two forces shown. Determine the magnitude of T which will cause no bending of the mast (zero moment) at point O.

Step2: Let T be the tension in the string. Taking moment about ‘O’ ∑Mo=0 5cos30o×90+5sin30o×90-

T√

×120- T√

=0

389.71+150-11.42T-22.28T=0 133.7T=539.71 T=4.04 kN 2/54 The piston, connecting rod, and crankshaft of a diesel engine are shown in the figure. The crank throw OA is half the stroke of 8in, and the length AB of the rod is 14in. For the position indicated, the rod is under a compression along AB of 3550lb. Determine the moment M of this force about the crankshaft axis O.




Step2: OA=8 in AB=14 in Compression in the rod AB, T=3550 lb. AD=OA cos30o

AD=4(0.866) AD=3.464 in OD=OA sin30o

OD=4 sin30 OD=2 in BD=√ BD=√14 3.464 BD=13.56 in OB=BD+OD OB=13.56+2 OB=15.56 in

θ=tan-1

θ=tan-1 .

.

θ=14.33o Step3: Taking moment about ‘O’

(Clockwise +) Mo=Fsinθ×OB Mo=3550 sin14.33o×15.56 Mo=13671.81 lb‐in Mo=1139 lb-ft 2/55 The 120-N force is applied as shown to one end of the curved wrench. If α=30o, calculate the moment of F about the center O of the bolt. Determine thevalue of α which would maximize the moment about O; state the value of this maximum moment.


Step2:Applied force F=120 N Taking moment about ‘O’(CW+) Mo=120cos30o(70+150+70)+ 120cos30o(25+70+70+25) Mo=41537.68Nmm Step3: For maximum moment Mo the force F should be perpendicular to the line joining AB.

tanα= α=33.2o Step4:For this condition



OA= 25 70 25 70 70 150 70

OA=346.7mm Taking moment about ‘O’ Mo=120×346.7 Mo=41603.84 Nmm 2/56 If the combined moment of the two forces about point C is zero, determine (a) the magnitude of the force P. (b) the magnitude R of the resultant of the two forces. (c) the coordinates x and y of the point A on the rim of the wheel about which the combined moment of the two forces is a maximum. (d) the combined moment MA of the two forces about A.


Step2: a) Taking moment about ‘C’. ∑MC=0 P×8+100cos60o×4-100sin60o×8=0 P=61.6 lb Step3: b) ∑Fx=Rx Rx= - 100cos60o-61.6 Rx=-111.6 lb ∑Fy=Ry

Ry=100 sin60o Ry=86.6 lb R=

R= 111.6 86.6 R=141.3 lb Step3:

c) θ=tan-1

θ=tan-1 .

.

θ=37.81o Step4:

x=8sinθ x=8sin37.81o x=4.9 in y=8cosθ x=8cos37.81o x=6.32 in Step5: Moment about ‘A’



MA=R×BA BA=OA+OB BA=8+4 cosθ BA=8+4cos37.81o BA=11.61 in Since MA=R×BA MA=143.3×11.61 MA=15771 lb-in (CW)

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.


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PROBLEMS Introductory Problems 2/57 The caster unit is subjected to the pair of 400-Nforces shown. Determine the moment associated with these forces.

2/58 A force F=60 N acts along the line AB. Determine the equivalent force–couple system at point C.

2/59 The top view of a revolving entrance door is shown. Two persons simultaneously approach the door and exert force of equal magnitudes as shown. If the resulting moment about the door pivot axis at O is 25 Nm, determine the force magnitude F.

2/60 The indicated force–couple system is applied to a small shaft at the center of the plate.Replace this system by a single force and specify the coordinate of the point on the x-axis through which the line of action of this resultalt force passes.

2/61 The bracket is spot welded to the end of the shaft at point O.To show the effect of the 900-N force on the weld, replace the force by its equivalent of a force and couple M at O. Express M in vector notation.



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2/62 As part of a test, the two aircraft engines are revved up and the propeller pitches are adjusted so as to result in the fore and aft thrusts shown. What force F must be exerted by the ground on each of the main braked wheels at A and B to counteract the turning effect of the two propeller thrusts? Neglect any effects of the nose wheel C, which is turned 90N and unbraked.

2/63 Replace the 10-kN force acting on the steel column by an equivalent force–couple system at point O.This replacement is frequently done in the design of structures.

2/64 Each propeller of the twin-screw ship develops a fullspeed thrust of 300 kN. In maneuvering the ship,one propeller is turning full speed ahead and the other full speed in reverse. What thrust P must each tug exert on the ship to counteract the effect of the ship’s propellers?

Representative Problems 2/65 A lug wrench is used to tighten a square-head bolt. If 250-N forces are applied to the wrench as shown, determine the magnitude F of the equal forces exerted on the four contact points on the 25-mm bolt head so that their external effect on the bolt is equivalent to that of the two 250-N forces. Assume that the forces are perpendicular to the flats of the bolt head.



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2/66 During a steady right turn, a person exerts the forces shown on the steering wheel. Note that each force consists of a tangential component and a radiallyinward component. Determine the moment exerted about the steering column at O.

2/67 The 180-N force is applied to the end of body OAB. If θ= 50o, determine the equivalent force–couple system at the shaft axis O.

2/68 A force F of magnitude 50 N is exerted on the automobile parking -brake lever at the position x =250mm. Replace the force by an equivalent force–couple system at the pivot point O.

2/69 The tie-rod AB exerts the 250-N force on the steering knuckle AO as shown. Replace this force by an equivalent force–couple system at O.



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2/70 The combined drive wheels of a front-wheel-drive automobile are acted on by a 7000-N normal reaction force and a friction force F, both of which are exerted by the road surface. If it is known that the resultant of these two forces makes a 15o angle with the vertical, determine the equivalent force–couple system at the car mass center G. Treat this as a two dimensional problem.

2/71 The system consisting of the bar OA,two identicalpulleys, and a section of thin tape is subjected to the two 180N tensile forces shown in the figure. Determine the equivalent force–couple system at point O.

2/72 Calculate the moment MB of the 900-N force about the bolt at B. Simplify your work by first replacing the force by its equivalent force-couple system at A.

2/73 The bracket is fastened to the girder by means of the two rivets A and B and supports the 2-kN force. Replace this force by a force acting along the centreline between the rivets and a couple. Then redistribute this force and couple by replacing it by two forces,one at A and the other at B, and ascertain the forces supported by the rivets.



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2/74 The angle plate is subjected to the two 250-N forces shown. It is desired to replace these forces by an equivalent set consisting of the 200-N force applied at A and a second force applied at B. Determine the y-coordinate of B.

2/75 The weld at O can support a maximum of 2500 N of force along each of the n- and t-directions and a maximum of 1400 Nm of moment. Determine the allowable range for the direction θ of the 2700-N force applied at A. The angle θ is restricted to 0 ≤θ≥90o.

2/76 The device shown is a part of an automobile seatback-release mechanism. The part is subjected to the 4-N force exerted at A and a restoring moment exerted by a hidden torsional spring. Determine the y-intercept of the line of action of the single equivalent force.



.

Resultant



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PROBLEMS Introductory Problems 2/77 Calculate the magnitude of the tension T and the angle θ for which the eye bolt will be under a resultant downward force of 15 kN.

2/78 Determine the resultant R of the four forces acting on the gusset plate. Also find the magnitude of R and the angle θx which the resultant makes with the x-axis.

2/79 Determine the equivalent force–couple system at the origin O for each of the three cases of forces being applied to the edge of a circular disk. If the resultant can be so expressed, replace this force–couple system with a stand-alone force.

2/80 Determine the height h above the base B at which the resultant of the three forces acts.

2/81 Where does the resultant of the two forces act?

2/82 Under nonuniform and slippery road conditions, thetwo forces shown are exerted on the two rear-drive wheels of the pickup truck, which has a limited-slip rear differential. Determine the y-intercept of the resultant of this force system.



.

2/83 If the resultant of the two forces and couple M passes through point O, determine M.

2/84 Determine the magnitude of the force F applied to the handle which will make the resultant of the three forces pass through O.

2/85 Determine and locate the resultant R of the two forces and one couple acting on the I-beam.

2/86 A commercial airliner with four jet engines, each producing 90 kN of forward thrust, is in a steady, level cruise when engine number 3suddenly fails.Determine and locate the resultant of the three remaining engine thrust vectors. Treat this as a twodimensional problem.

2/87 Replace the three forces acting on the bent pipe by a single equivalent force R. Specify the distance x from point O to the point on the x-axis through which the line of action of R passes.



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2/88 The directions of the two thrust vectors of an experimental aircraft can be independently changed from the conventional forward direction within limits. For the thrust configuration shown, determine the equivalent force–couple system at point O. Then replace this force–couple system by a single force and specify the point on the x-axis through which the line of action of this resultant passes. These results are vital to assessing design performance.

2/89 Determine the resultant R of the three forces acting on the simple truss. Specify the points on the x- and y-axes through which R must pass.

2/90 The gear and attached V-belt pulley are turning counterclockwise and are subjected to the tooth load of 1600 N and the 800-N and 450-N tensions in the V-belt. Represent the action of these three forces by a resultant force R at O and a couple of magnitude M. Is the unit slowing down or speeding up?

2/91 The design specifications for the attachment at A for this beam depend on the magnitude and location of the applied loads. Represent the resultant of the three forces and couple by a single force R at A and a couple M. Specify the magnitude of R.

2/92 In the equilibrium position shown, the resultant of the three forces acting on the bell crank passes through the bearing O. Determine the vertical force P. Does the result depend on θ?



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2/93Two integral pulleys are subjected to the belt tensions shown.If the resultant R of these forces passes through the center O, determine T and the magnitude of R and the counterclockwise angle θ it makes with the x-axis.

2/94 While sliding a desk toward the doorway, three students exert the forces shown in the overhead view. Determine the equivalent force–couple system at point A. Then determine the equation of the line of action of the resultant force.

2/95 Under nonuniform and slippery road conditions, the four forces shown are exerted on the four drive wheels of the all-wheel-drive vehicle. Determine the resultant of this system and the x- and y-intercepts of its line of action. Note that the front and rear tracks are equal (i.e.AB=CD).

2/96 The rolling rear wheel of a front-wheel-drive automobile which is accelerating to the right is subjected to the five forces and one momentshown. The forces Ax =240 N and Ay= 2000 N are forces transmitted from the axle to the wheel, F =160 N is the friction force exerted by the road surface on the tire, N =2400 N is the normal reaction force exerted by the road surface, and W = 400 N is the weight of the wheel/tire unit. The couple M=3 Nm is the bearing friction moment. Determine and locate the resultant of the system.

2/97 A rear-wheel-drive car is stuck in the snow between other parked cars as shown. In an attempt to free the car, three students exert forces on the car at points A, B, and C while the driver’s actions result in a forward thrust of 200 N acting parallel to the plane of rotation of each rear wheel. Treating the problem as two-dimensional, determine the equivalent



.

force–couple system at the car center of mass G and locate the position x of the point on the car centreline through which the resultant passes. Neglect all forces not shown.

2/98 An exhaust system for a pickup truck is shown in the figure. The weights Wh, Wm, and Wt of the headpipe,muffler, and tailpipe are 10, 100, and 50 N, respectively, and act at the indicated points. If the exhaust-pipe hanger at point A is adjusted so that its tension FA is 50 N, determine the required forces in the hangers at points B, C, and D so that the force–couple system at point O is zero. Why is a zero force–couple system at O desirable?



.

Ch#3 Equilibrium Step1:Free body diagram Step2: Step3: Step4: Step5: 1.5'



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PROBLEMS Introductory Problems 3/1 Determine the force P required to maintain the 200-kg engine in the position for which θ=30o. The diameter of the pulley at B is negligible.


Step2: Mass m=200kg Weight W=200×9.81 =1962 N DC=2sin30o DC=1 m AD=2cos30o AD=1.732 m BD=2-AD BD=0.27 m

α=tan-1

α= tan-1 .

α=75o Step3: Applying sine’s law

=

P=1962× P=1759 N 3/2 The mass center G of the 1400-kg rear-engine car is located as shown in the figure. Determine the normal force under each tire when the car is in equilibrium.State any assumptions.


Step2: Mass of car m=1400kg Weight W=1400×9.81 W=13734 N ∑Fy=0 Since there are two front and two rear tyres of a car,therefore 2RA+2RB-13734=0 (i) Taking moment about ‘A’ ∑MA=0 2RB(1.386+0.964)-13734×1.386=0 4.7RB=19035.324 RB=4050 N (ii) Using the value of RB in (i) 2RA+2(4050)-13734=0 2RA=2817 N



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3/3 A carpenter carries a 12 lb uniform board as shown.What downward force does he feel on his shoulder at A?

Step1:Free body diagram

Step2:Weight of the board W=12 lb Let NA and NB be the reactions at points A and B respectively.Taking moment about B, ∑MB=0 W×(2+1)-NA×2=0 12×3-2NA=0 2NA=36 NA=18 lb 3/4 In the side view of a 70kg television resting on a24kg cabinet, the mass centers are labeled G1 and G2. Determine the force reactions at A and B. (Note that the mass center of most televisions is located well forward because of the heavy nature of the front portion of picture tubes.)


Step2: Weight of TV, W1=70×9.81 W1=686.7 N Weight of the cabinet W2=24×9.81 W2=235.44 N Taking moment about A, ∑MA=0 NB ×0.7-W1×0.6-W2×0.35=0 0.7NB -686.7×0.6-235.44×0.35=0 0.7NB-412.02-82.404=0 0.7NB=706 N Now ∑Fy=0 NA+NB-W1-W2=0 NA+706-686.7-235.44=0 NA=216 N



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3/5 The roller stand is used to support portions of long boards as they are being cut on a table saw. If the board exerts a 25-N downward force on the roller C, determine the vertical reactions at A and D. Note that the connection at B is rigid, and that the feet A and D are fairly lengthy horizontal tubes with a nonslip coating.


Step2: Downward force at ‘C’ is 25N

Let RA and RB be the reactions at A and B. Taking moment about A, ∑MA=0 (CCW+) RD×0.355-25×0.235=0 RD=16.55 N Now ∑Fy=0 RA+RD-25=0 RA+16.55-25=0 RA=8.45 N 3/6 The 450-kg uniform I-beam supports the load shown.Determine the reactions at the supports.


Step2: Weight of the I-beam W1=450×9.81 W1=4414.5 N Weight of the drum W2=220×9.81 W2=2158.2 N RA+RB=4414.5+2158.2 RA+RB=6572.7 (i) Taking moment about A, ∑MA=0 (CW+) 4414.5×4+2158.2×5.6-RB×8=0 RB=3718 N Putting this value in (i) RA+3718=6572.7 RA=2854.7 N



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3/7 Calculate the force and moment reactions at the bolted base O of the overhead traffic-signal assembly. Each traffic signal has a mass of 36 kg,while the masses of members OC and AC are 50 kg and 55 kg, respectively.


Step2: Weight of each traffic signal is 80 lb. Weight of members OC and AC are 110 lb and 120 lb respectively. Let Ox,Oy and Mo=0 be the reactions and reactive moment at ‘O’ respectively. Considering forces along x-axis ∑Fx=0 Ox =0 Considering forces along y-axis ∑Fy=0 Oy-80-80-110-120=0 Oy=390 lb Taking moment about ‘O’ ∑Mo=0 (CCW+) 80(15+3+12)+80(3+12)+120×12-Mo=0 Mo=5040 lb-ft (CW)

3/8 The 20-kg homogeneous smooth sphere rests on the two inclines as shown.Determine the contact forces at A and B.


Step2: Mass of the sphere, m=20kg Weight of sphere, W=20×9.81 W=196.2 N Considering forces along x-axis ∑Fx=0 RA cos150-RB sin30o=0 0.966NA-0.5NB=0 (i) Considering forces along y-axis ∑Fy=0 RA sin150+RB cos30o-196.2=0 0.966NA+0.5NB=196.2 0.259NA=196.2-0.866NB

NA=758.06-3.344NB (ii) Putting above value in (i) 0.966(758.06-3.344NB)-0.5NB=0 732.29-3.23NB-0.5NB=0 3.73NB=732.29 NB=196.3 Using this value in (i) 0.966NA-0.5(196.3)=0 0.966NA=98.15 NA=101.6 N



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3/9 A 120 lb crate resets on the 60 lb pickup tailgate. Calculate the tension T in each of the two restraining cables, one of which is shown. The centers of gravity are at G1 and G2. The crate is located midway between the two cables.


Step2: Weight of the crate W1=120 lb Weight of the tailgate W2=60 lb

tanθ=. .

θ=tan-10.9796 θ=44.4o Taking moment about ‘O’ ∑Mo=0 (CCW+) 120(14)+60(9.5)-2Tsin44.4o(12.25)=0 1680+570-17.14T=0 T=131.27 lb 3/10 A portable electric generator has a mass of 160 kg with mass center at G. Determine the upward force F

necessary to reduce the normal force at A to onehalf its nominal (F = 0) value.


Step2: Weight of the generator W=160×9.81 W=1569.6 N Let F be the force required. Let RA and RO be the reactions at A and O respectively. For F=0 Taking moment about ‘O’ ∑Mo=0 (CCW+) -RA×0.46+1596.6×0.135=0 RA=460.64 N

For RA=230.32 N

Taking moment about ‘O’ ∑Mo=0 (CCW+) 1569.6(0.135)-230.32(0.46)-F(0.7)=0 -0.7F=-105.95 F=151.36 N



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3/11 With what force magnitude T must the person pull on the cable in order to cause the scale A to read 2000 N? The weights of the pulleys and cables are negligible. State any assumptions.


Step2: Weight of the block is 1000 lb and reading is 500 lb. Let T be the tension in the cable. Considering forces along y-axis ∑Fy=0 5T+500-1000=0 T=100 lb 3/12 The device shown is designed to aid in the removal of pull-tab tops from cans. If the user exerts a 40-N force at A, determine the tension T in the portionBC of the pull tab.


Step2: Force applied F=40 N Let the tension in portion ‘BC’ be ‘T’ Let Ro be reaction at ‘O’ Taking moment about ‘O’ ∑Mo=0 (CCW+) Tcos45o(32)+Tsin45o(36) - 40cos10o(78+32)-40sin10o(27)=0 T=94.06 N 3/13 A woodcutter wishes to cause the tree trunk to fall uphill, even though the trunk is leaning downhill.With the aid of the winch W, what tension T in the cable will be required? The 1200lb trunk has a center of gravity at G. The felling notch at O is sufficiently large so that the resisting moment there is negligible.



.


Step2: Weight of the trunk W=1200 lb Let ‘T’ be the tension in the cable Taking moment about ‘O’ ∑Mo=0 (CCW+) 1200(12sin5o)-T(10+4)cos15o=0 T=81.21 lb 3/14 To facilitate shifting the position of a lifting hook when it is not under load, the sliding hanger shown is used. The projections at A and B engage the flanges of a box beam when a load is supported, and the hook projects through a horizontal slot in the beam. Compute the forces at A and B when the hook supports a 300-kg mass.


Step2: Weight of the block W=300×9.81 W=2943 N Let RA and RB be reactions at A and B. Taking moment about ‘A’ ∑MA=0 (CCW+) 2943×0.4 - RB×0.6=0 RB=1962 N Considering forces along y-axis ∑Fy=0 RA-RB-W=0 RA=1962+2923 RA=4905 N 3/15 Three cables are joined at the junction ring C. Determine the tensions in cables AC and BC caused by the weight of the 30-kg cylinder.



.


Step2: Let TAC,TAB and TDC be the tension in the cables AB,BC and CD respectively. Weight of the cylinder W=30×9.81 W=294.3 N Here TDC=294.3 N ∑Fx=0 -TACcos45o+294.3cos15o-TBCcos60o=0 0.707TAC+0.5TBC=284.27 (i) ∑Fy=0 TACsin45o+294.3sin15o-TBCsin60o=0 0.707TAC=0.866TBC-76.17 TAC=1.225 TBC-107.74 (ii) Using (ii) in (i) 0.707(1.225TBC-107.74)+0.5TBC=284.27 0.866TBC-76.17+0.5TBC=284.27 1.366TBC=360.44 TBC=263.87 N Using above value in (ii) TAC=1.225(263.87)-107.74 TAC=215.49 N 3/16 A 700-N axial force is required to remove the pulley from its shaft. What force F must be exerted on the handle of each of the two prybars? Friction at the contact points B and E is sufficient to prevent slipping;friction at the pulley contact points C and F is negligible.


Step2: Total axial force ‘F’ requires is 700N Since two prybars are acting at both sides of the pully,the axial force by each prybar would be 350N. Taking moment about ‘E’ ∑ME=0 (CCW+)

- ×0.038+Fcos5(0.25)+Fsin5(0.031)=0

-13.3+0.249F+0.0027F=0 0.252F=13.3 F=52.84N 3/17 The uniform beam has a mass of 50 kg per meter of length. Compute the reactions at the support O.The force loads shown lie in a vertical plane.




.

Step2: Let Ox and Oy be the reactions at ‘O’ along x and y directions respectively. Let Mo,be the reaction moment at ‘O’ Self load of the beam portions.i.e F1=50×9.81×(1.8+0.6) F1=1177.2 N F1=1.117 kN F2=50×9.81×(0.6+0.6) F2=588.6 N F2=0.589 kN Step3: Taking moment about ‘O’ ∑Mo=0 Mo-F1×1.2-3×2.8-F2(2.4+0.6cos30o)+4 – 1.4(2.4cos30o+1.2)=0 Mo-1.177×1.2-5.4-0.589×2.916+4-1.4×3.278=0 Mo-1.41-5.4-1.72+4-4.59=0 Mo=9.12 kNm ∑Fx=0 Ox+1.4sin30o=0 Ox+0.7=0 Ox=-0.7 kN ∑Fy=0 Oy-F1-F2-3-1.4cos30o=0 Oy-1.177-0.589-3-1.21=0 Oy=5.98 kN

Representative Problems 3/18 A pipe P is being bent by the pipe bender as shown. If the hydraulic cylinder applies a force of magnitude F = 24 kN to the pipe at C, determine the magnitude of the roller reactions at A and B.

Soln. Step1:Free body diagram Soln. Step1:Free body diagram

Step2: Applied force 24 kN Let RA and RB be the reactions at A and B. Here RA=RB ∑Fy=0 24-RAcos15o-RBcos15o=0 2RAcos15o=24 RA=12.42 kN RA= RB=12.42 kN 3/19 The uniform 15m pole has a mass of 150 kg and is supported by its smooth ends against the vertical walls and by the tension T in the vertical cable.Compute the reactions at A and B.


Step2: Length of the pole, L=15m Weight of the pole, W=150×9.8 W=1471.5 N



.

Let ‘T’ be the tension in the cable. Let RA and RB be the horizental reactions at A and B respectively. OB2+122=152

OB=√15 12 OB=9 m ∑Fx=0 RA=RB

∑Fy=0 T-1471.5=0 T=1471.5 N Step3:

From similar triangles ADC and ABO

=

= AC=4 m Now consider trianglesAFE and ABO

=

.=

AE=6 m Taking moment about ‘A’ (CCW+) ∑MA=0 RB×9-1471.5×6+1471.5×4=0 RB=327 N RA=RB=327 N

3/20 Determine the reactions at A and E if P = 500 N. What is the maximum value which P may have for static equilibrium? Neglect the weight of the structure compared with the applied loads.


Step2: Let Ax,Ay and Ex be the reactions at A and E respectively. ∑Fx=0 Ax+Ex-4000sin30o=0 Ax+Ex=2000 (i) ∑Fy=0 Ay-4000cos30o+500=0 Ay=3464.1-500=0 Ay=2964.1 N Taking moment about ‘A’ ∑MA=0 500×8+Ex×3-4000cos30o(4)=0 3Ex=16000cos30o-4000 Ex=3285.46 N Using above value in (i) Ax+3285.46=2000 Ax=-1285.46 N For maximum P, Ex=0 Taking moment about ‘A’ ∑MA=0 P×8-4000cos30o(4)=0 8P=13856.4 P=1732 N 3/21 While digging a small hole prior to planting a tree, a homeowner encounters rocks. If he exerts a horizontal 225-N force on the prybar as shown, what is the horizontal force exerted on rock C? Note that a small ledge on rock C supports a vertical force reaction there. Neglect friction at B. Complete solutions (a) including and(b)excluding the weight of the 18kg prybar.



.


Step2: Let (Fc)x and (Fc)y be the horizontal and vertical reactions at ‘C’. Let FB be the force exerted by rock at ‘B’ (a) Considering the weight of the prybar. Taking moment about ‘C’ ∑MC=0

50×68+40×34tan20o-FB =0

3400+495-8.51FB=0 8.51FB=3895 FB=457.7 lb Taking force along x-axis (Fc)x+50- FBcos20o=0 (Fc)x= 457.7cos20o-50 (Fc)x=380 lb Step3: (b) Excluding the weight of the prybar. Taking moment about ‘C’ ∑MC=0

50×68-FB =0

FB=399 lb Taking forces along x-axis. ∑Fx=0 FC- FBcos20o+50=0 FC- 399cos20o+50=0 FC=325 lb

3/22 Determine the force P required to begin rolling the uniform cylinder of mass m over the obstruction of height h.


Step2:considering the triangle ABC

BC=

BC=√ 2 BC=√2

sinθ=

sinθ=2 2

Taking moment about ‘C’ ∑MC=0 P(r-h)-mgrsinθ=0

P(r-h)-mgr2 2

=0

P=mg2 2



.

3/23 A 35-N axial force at B is required to open the springloaded plunger of the water nozzle. Determine the required force F applied to the handle at A and the magnitude of the pin reaction at O. Note that the plunger passes through a vertically-elongated hole in the handle at B, so that negligible vertical force is transmitted there.


Step2: Let ‘F’ be the force applied at handle. Let Ox and Oy be the horizontal and vertical reactions at point ‘O’. Given that 35N of force is the reactive force at ‘B’ Taking moment about ‘O’ ∑Mo=0 35×18-Fcos10o(38)- Fsin10o(44)=0 45.06F=630 F=13.98 N Ste3: ∑Fx=0 Ox-Fcos10o-35=0 Ox-13.98cos10o-35=0 Ox=48.8 N ∑Fy=0 Oy-Fsin10o=0 Oy=2.43 N Magnitude of reaction at ‘O’

O= =√48.8 2.43

O=48.86 N

3/24 A person holds a 30-kg suitcase by its handle as indicated in the figure. Determine the tension in each of the four identical links AB.


Step2: Weight of the suitcase W=60 lb Number of links is 4. Let ‘T’ be the tension in each link,so there are 2T force in the left and right. ∑Fy=0 60-4Tsin35o=0 Tsin35o=60 T=26.15 lb 3/25 A block placed under the head of the claw hammer as shown greatly facilitates the extraction of the nail. If a 200-N pull on the handle is required to pull the nail, calculate the tension T in the nail and the magnitude A of the force exerted by the hammer head on the block. The contacting surfaces at A are sufficiently rough to prevent slipping.



.


Step2: Force applied on the handle F=50 lb Let ‘T’ be the tension in the nail. Let Ax and Ay be the horizontal and vertical reactions at ‘A’ respectively. Taking moment about ‘A’ ∑MA=0 50×8-2T=0 T=200 lb ∑Fx=0 Fcos20o -Ax=0 Ax = 50cos20o Ax = 46.98 lb ∑Fy=0 Fsin20o +Ay-T=0 Ay =T- 50sin20o Ay =200- 50sin20o Ay =182.9 lb The magnitude of the force ‘A’ exerted by the hammer head of the block.

A=

A=√46.98 182.9 A=188.8 lb 3/26 The indicated location of the center of mass of the 3600 lb pickup truck is for the unladen condition. If a load whose center of mass is x = 400 mm behind the rear axle is added to the truck, determine the mass mL for which the normal forces under the front and rear wheels are equal.


Step2: Let the load weight be WL Weight of the truck W=3600 lb Let the normal forces at A and B be RA and RB respectively. Given that the normal forces under the front and rear wheels are equal. RA=RB

Taking moment about ‘A’ ∑MA=0 3600(45)+ WL(45+67+16)- RB(45+67)=0 162000+ 128WL-112RB =0 (i) Considering forces along y-axis. ∑Fy=0 RA+RB-3600-WL=0 2RB =3600+ WL (ii) Using (ii) in (i) 162000+ 128WL-56(3600+WL) =0 162000+128WL-210600-56WL=0 72WL=39600 WL=550 lb Putting above value in (ii) 2RB =3600+550 2RB =4150 RA=RB=2075 lb

3/27 The wall-mounted 2.5-kg light fixture has its mass center at G. Determine the reactions at A and B and also calculate the moment supported by the adjustment thumbscrew at C. (Note that the lightweight frame ABC has about 250 mm of horizontal tubing, directed into and out of the paper, at both A and B.)



.


Step2: Weight of the light W=2.5×9.81 W=24.53 N Let Ax,Ay and Bx be the reactions at A and B respectively. Taking moment about ‘A’ ∑MA=0 (CW+) Bx×230-24.53×300=0 Bx=32 N Considering fores along x-axis ∑Fx=0 Ax-Bx=0 Ax-32=0 Ax=32 N Step3:Free body diagram of fixture only

Taking moment about ‘C’due to weight ‘W’ ∑MC=0

MC=24.53×100 MC=2453 Nmm MC=24.53 Nm 3/28To test the validity ofaerodynamic assumptions made in the design of the aircraft, its model is being tested in a wind tunnel. The support bracket is connected to a force and moment balance, which is zeroed when there is no airflow. Under test conditions,the lift L, drag D, and pitching moment MG act as shown. The force balance records the lift, drag, and a moment MP. Determine MG in terms ofL, D, and MP.


Step2: Taking moment about ‘P’ ∑MP=0 MP-MG-Ld-Dh=0 MG=MP-Ld-Dh=0 3/29 The chain binder is used to secure loads of logs,lumber, pipe, and the like. If the tension T1 is 2 kN when θ=30o, determine the force P required on the lever and the corresponding tension T2 for this position.Assume that the surface under A is perfectly smooth.



.


Step2: Tension T1=2 kN θ=30o Let Ay be the reaction at support ‘A’ Taking moment about ‘A’ ∑MA=0 (CW+) P×600-2×100sin30o=0 P=0.1667 kN Considering fores along x-axis ∑Fx=0 T2-T1+P sin30o=0 T2-25+0.1667×0.5=0 T2=1.92 kN 3/30 The device shown is designed to apply pressure when bonding laminate to each side of a countertop near an edge. If a 120-N force is applied to the handle, determine the force which each roller exerts on its corresponding surface.


Step2: Force applied at handle F=30 lb Let the reaction forces at B and C are RB

and RC respectively. Considering triangle BCO

OB=3.5tan45o OB=3.5 BC=√ BC=√3.5 3.5 BC=4.95 Taking moment about ‘C’ ∑MC=0 B×3.5-30×5.5cos45o=0 3.5B=116.67 B=33.43 lb Considering fores along y-axis ∑Fy=0 C-B-30=0 C-33.34-30=0 C=63.34 lb 3/31 The two light pulleys are fastened together and form an integral unit.They are prevented from turning about their bearing at O by a cable wound securely around the smaller pulley and fastened to point A.Calculate the magnitude R of the force supported by the bearing O for the applied 2kN load.



.


Step2: Weight acting on the pulley W=2 kN Let ‘T’ be the tension in the cable at ‘A’ Let Rx and Ry be the reactions at ‘O’ Considering the triangle

sinθ= θ=22.62o Taking moment about ‘O’ ∑Mo=0 (CCW+) 2×200-T×125=0 T=3.2 kN Considering fores along x-axis ∑Fx=0 3.2cos22.62o+Rx=0 Rx=-2.95 kN Considering fores along y-axis ∑Fy=0 Ry-2-3.2sin22.62o=0

Ry=3.23 kN Magnitude of the reaction force ‘R’

R=

R= 2.95 3.23 R=4.37 kN 3/32 In a procedure to evaluate the strength of the triceps muscle, a person pushes down on a load cell with the palm of his hand as indicated in the figure. If the load-cell reading is 160 N, determine the vertical tensile force F generated by the triceps muscle. The mass of the lower arm is 1.5 kg with mass center at G. State any assumptions.


Step2: Weight of the lower arm is 3.2 lb Let ‘F’ be the force in the triceps muscle load at the palm is 35 lb Taking moment about ‘O’ ∑Mo=0 (CCW+) 35×(6+6)-3.2×6-F×1=0 -F= - 401 F=401 lb



.

3/33 A person is performing slow arm curls with a 10-kg weight as indicated in the figure. The brachialis muscle group (consisting of the biceps and brachialis muscles) is the major factor in this exercise. Determine the magnitude F of the brachialis-musclegroup force and the magnitude E of the elbow joint reaction at point E for the forearm position shown in the figure. Take the dimensions shown to locate the effective points of application of the two muscle groups; these points are 200 mm directly above E and 50 mm directly to the right of E. Include the effect of the 1.5-kg forearm mass with mass center at point G. State any assumptions.


Step2: Weight of the lower arm is 3.2 lb Let ‘F’ be the force is the Brichiates muscle load in the palm is 20 lb. tanθ=2 8⁄ θ=14.04o Taking moment about ‘E’ ∑ME=0 (CCW+) Fcos14.04o(2)-G×6-20×14=0

1.94F-3.2×6-280=0 F=154.23 lb Considering fores along x-axis ∑Fx=0 Ex-Fsin14.04o=0 Ex-154.23sin14.04o=0 Ex=37.4 lb Considering fores along y-axis ∑Fy=0 Ey-3.2-20+Fcos14.04o=0 Ey=-126.4 lb Resultant force at ‘E’

E=

E= 37.4 126.4 E=131.82 lb 3/34 A woman is holding a 3.6-kg sphere in her hand with the entire arm held horizontally as shown in the figure. A tensile force in the deltoid muscle prevents the arm from rotating about the shoulder joint O; this force acts at the 21o angle shown.Determine the force exerted by the deltoid muscle on the upper arm at A and the x- and y-components of the force reaction at the shoulder joint O. The mass of the upper arm is mU =1.9 kg, the mass of the lower arm is mL = 1.1 kg, and the mass of the hand is mH = 0.4 kg; all the corresponding weights act at the locations shown in the figure.




.

Step2: Weight of the upper arm Wu=1.9×9.81 N Weight of the lower arm Wl=1.1×9.81 N Weight of the hand Wh=0.4×9.81 N Weight of the sphere W=3.6×9.81 N Let ‘T’ be the tensile force acting in the deltoid muscle. Let Ox and Oy be the reactions at joint ‘O’ Taking moment about ‘O’ ∑Mo=0 (CCW+) Fsin21o(125)-1.9×9.81×130-1.1×9.81 ×412- (3.6+0.4) ×9.81×6.35=0

F=.

F=710 N Considering fores along x-axis ∑Fx=0 Ox-Fcos21o=0 Ox =662.8 N Considering fores along y-axis ∑Fy=0 Oy+710sin21o-1.9×9.81-1.1×9.81-0.4×9.81-3.6×9.81=0 Oy=-185.7 N 3/35 With his weight W equally distributed on both feet, a man begins to slowly rise from a squatting position as indicated in the figure. Determine the tensile force F in the patellar tendon and the magnitude of the force reaction at point O, which is the contact area between the tibia and the femur. Note that the line of action of the patellar tendon force is along its midline. Neglect the weight of the lower leg.


Step2: Let ‘F’ be the force in the patellar tendon. Let Ox and Oy be the reactions at ‘O’. Let 2⁄ be the weighton one leg. Taking moment about ‘O’ ∑Mo=0 (CCW+)

F×50 -

×225=0

F=2.25W Considering fores along x-axis ∑Fx=0 Fcos55o-Ox=0 2.25cos55o-Ox=0 Ox=1.29W Considering fores along y-axis ∑Fy=0

Fsin55o+

+Oy=0

2.25sin55o+ 0.5W+Oy=0 Oy=-2.34W Magnitude of the reaction at ‘O’

O=

O= 1.29 2.34 O=2.67W 3/36 The elements of an on-off mechanism for a table lamp are shown in the figure. The electrical switch S requires a 4N force in order to depress it. Whatcorresponding force F must be exerted on the handleat A?



.


Step2: The force at ‘S’ is 0.9 lb Let Ox and Oy be the reactions at ‘O’. The force has been replaced by a force-couple system at B.

Where M=F(2.4cos15o) Taking moment about ‘O’ ∑Mo=0 (CW+) 0.9×1.2- F×(2.4cos15o)-F×(3.6cos15o)=0

1.8-2.318-3.477F=0 F=0.186 lb 3/37 The uniform 18-kg bar OA is held in the position shown by the smooth pin at O and the cable AB. Determine the tension T in the cable and the magnitude and direction of the external pin reaction at O.


Sstep2: Weight of the bar ‘OA’ W=18×9.81 Let ‘T’ be the tension in the cable AB Let Ox and Oy be the reactions at ‘O’.

tanθ=.

. .

tanθ=.

.

θ=33.7o

Taking moment about ‘O’ ∑Mo=0 (CW+) Tsin33.7o×1.5cos60o-Tcos33.7o×1.5sin60o

+ 18×9.81×.

cos60o=0

0.42T-1.08T+66.22=0 0.66T=66.22 T=100 N Step3: Considering fores along x-axis ∑Fx=0 Ox-100c0s33.7o=0 Ox=83.45 N



.

Considering fores along y-axis ∑Fy=0 Oy-18×9.81-100sin33.7o=0 Oy=232 N Magnitude of the reaction at ‘O’

O=

O= 83.45 232 O=246 N 3/38 A person attempts to move a 20-kg shop vacuum by pulling on the hose as indicated. What force F will cause the unit to tip clockwise if wheel A is against an obstruction?


Step2: Weight of the cart W=40 lb Let Ax and Ay be the horizontal and vertical reactions at ‘A’ respectively. Let ‘F’ be the extended force. The reaction ‘R’ at B is zero at the point trepping. Taking moment about ‘A’ ∑MA=0 40×7-Fcos15o×(15+8)+Fsin15o×3=0 280-21.44F=0 F=13.06 lb

3/39 The exercise machine is designed with a lightweight cart which is mounted on small rollers so that it is free to move along the inclined ramp. Two cables are attached to the cart-one for each hand. If the hands are together so that the cables are parallel and if each cable lies essentially in a vertical plane, determine the force P which each hand must exert on its cable in order to maintain equilibrium position.The mass of the person is 70 kg, the ramp angle θ is 15o, and the angle β is 18o. In addition, calculate the force R which the ramp exerts on the cart.


Step2: Weight of the person W=70×9.81 N Inclination of the ramp θ=15o Let ‘R’ be the force exerted by the ramp on the cart. Let ‘P’ be the force exerted by each hand. Considering the forces along the inclined plane. ∑Fx=0 70×9.81×sin15o-2P-2Pcos18o=0 177.73-3.90P=0 P=45.6 N Considering the forces perpendicular to the plane. R-70×9.81×sin15o-2Psin18o=0 R=663.3+2×45.6 sin18o R=691N 3/40 The device shown is used to test automobile-engine valve springs. The



.

torque wrench is directly connected to arm OB. The specification for the automotive intake-valve spring is that 370 N of force should reduce its length from 50 mm (unstressed length) to 42 mm. What is the corresponding reading M on the torque wrench, and what force F exerted on the torque-wrench handle is required to produce this reading? Neglect the small effects of changes in the angular position of arm OB.


Step2: Force exerted by the spring on ‘OB’=83 lb Let Ox and Oy be the reactions at ‘O’. Taking moment about ‘O’ ∑Mo=0 M-83×6=0 M=498 l.in Considering free body diagram of ‘OA’

Taking moment at ‘O’ M=Fcos20o×15 498=14.095F F=35.3 lb



.

Truss:A framework composed of members joined at their ends to form arigid structure is called a truss. e.g; Trusses of roofing system, bridge trusses, transmission towers etc. Types of trusses: There are two types of trusses. i) Plane truss ii) Space truss Plane Truss: When the members of the truss lie essentially in a single plane, the truss is called a plane truss. Space Truss: When the members of the truss do not lie in the same plane, the truss is called space truss. Type of plane trusses depending upon the arrangement of members.

Commonly used bridge trusses.

SimpleTruss:A truss which is constructed from a basic triangular structure to such a manner that to increase new elements, two members and one joint is added, is known as simple truss.

Compound Truss: A compound truss is formed by connecting two or more simple trusses.

Complex Truss: It is a truss which cannot be classified as simple or compound.

Assumptions in the analysis and design of trusses.

1-Members are joined together by smooth pins, although the members are riveted, bolted and welded. 2-The centre line of all the members are concurrent at the joint. 3-All the loads are only applied at the joints and the weight of the members is assumed negligible so there will be only axial forces (tension or compression).

Stability and Determinacy of Trusses. Stability: External Stability: A truss will be external unstable if all the reactions are concurrent or parallel.



.

Internal Stability: A truss will be internally stable if it is not liable to collapse. A simple truss is always internally stable. Indeterminacy:A truss will be indetermina-te in which all support reactions and internal forces cannot be calculated only by available equilibrium equations for a given system of forces. Degree of indeterminacy: The access of total number of reactive components or access of members over the available equilibrium equations is known as degree of determinacy. It is convenient to consider stability and determinacy as follows: 1-With respect to reactions. e.g; external stability and determinacy. 2-With respect to members. e.g; internal stability and determinacy. 3-The combination of internal and external conditions. External Determinacy: A determinate structure should have at least three reactions. Ne=r-N Where r=number of reactions N=number of equilibrium equations available. Internal Determinacy: It can be checked if minimum number of reacting components necessary for the external determinacy and stability are known. m+r=2j where m=number of members r= j= The structure will be stable and determinate if m+r > 2j Sample Problem 4/1 Compute the force in each member of the loaded cantilever truss by the method of joints.


Step2: Taking moment about E, ∑ME=0 5T-20(5)-30(10)=0 5T=400 T=80 kN ∑FX=0 80cos30o-EX=0 EX=69.3 kN ∑FY=0 80sin30o+EY-30-20=0 EY=10 kN Step3: Considering the joint A



.

∑FY=0 AB Sin60o-30=0 0.866AB=30 AB=34.6 kN (T) ∑FX=0 -AC+AB cos60o=0 -AC+34.6(0.5)=0 AC=17.32 kN (C) Step4: Considering joint B

∑FY=0 BC Sin60o-AB Sin60o=0 0.866BC-34.6(0.866)=0 BC=34.6 kN(C) ∑FX=0 BD-AB cos600-BC cos60o=0 BD-34.6(0.5)-34.6(0.5)=0 BD=34.6 kN (T) Step5: Considering joint C

∑FY=0 CD Sin60o-BC Sin60o-20=0 CD(0.866)-34.6(0.866)-20=0 0.866CD-30-20=0 0.866CD=50 CD=57.7 kN (T) ∑FX=0 BC cos60o-CD cos60o-AC-CE=0 34.6(0.5)-57.7(0.5)-17.32-CE=0 CE=63.5 kN (C) Step6: Finally consider joint E

∑FY=0 -DE Sin60o+10=0 0.866DE=10 DE=11.55 kN ∑FX=0 DE cos60o+63.5-69.3=0 11.55(0.5)-5.8=0 0=0 checks

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EngineeringMechanics by J L Meriam 6th Solutions by Khalid Yousaf