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CHAPTER 3
RESULTANTS OF COPLANAR FORCE SYSTEMS
Fy
Fx
R
Fy
Fx
Fy
Fx
R
Finding a Resultant ForceParallelogram law is carried out to find the resultant force
Resultant, R = ( P1 + P2 )
VECTOR ADDITION OF NONORTHOGONAL FORCES
The screw eye is subjected to two forces, F1 and F2. Determine the magnitude and direction of the resultant force.
By Parallelogram Law
select 1 force triangleto analyze
SOLUTION
Step 1 - Law of Cosines to find FR
NN
NNNNFR
2136.2124226.0300002250010000
115cos1501002150100 22
Step 2 - Law of Sines to find angle θ
8.39
9063.06.212
150sin
115sin
6.212
sin
150
N
N
NN
8.54
158.39
Angle of φ relative to X-axis
Scalar Notation x and y axes are designated positive and
negative Components of forces expressed as
algebraic scalars
sin and cos FFFF
FFF
yx
yx
RESULTANT BY COMPONENTS
F2 = F2x + F2y
F1 = F1x + F1y
FRx = F1x + F2x
FRy = F1y + F2y
FRx = ΣFx
FRy = ΣFy
–Magnitude of FR can be found by Pythagorean Theorem
Rx
RyRyRxR F
FFFF 1-22 tan and
Determine x and y components of F1 and F2 acting on the boom.
NNNF
NNNF
y
x
17317330cos200
10010030sin200
1
1
By similar triangles we have
N10013
5260
N24013
12260
2
2
y
x
F
F
NNF
NF
y
x
100100
240
2
2
12
5tan 1
FRX = Σ Fx = - 100N + 240N = 140 N
FRY = Σ Fy = 173N -100N = 73 N
NFR
15873140 22 01 5.27
14073
tan
140 N
73 NFR
The link is subjected to two forces F1 and F2. Determine the magnitude and orientation of the resultant force.
N
NNFR629
8.5828.236 22
RESULTANT FORCE
9.67
8.236
8.582tan 1
N
N ORIGINAL FORCE SYSTEM
RESULTANT OF THREE OR MORE CONCURRENT FORCES
F1
XFF
1
45
F1x
F1yF1
NNF
FX
6805
48504
51
YFF
1
35
NF
FY
510335
1
OR
01 13.5334
tan
θ=90-α=36.870
F1x=850Cos 36.87 = 680N
F1y=850Sin 36.87 = 510N
F2
F2y
F2x
30
F2x= -625sin30 = -312.5N625N
F2y= -625cos30 = -541N 750
45
F3
F3x
F3y
F3x=-750cos45 = -530N
F3y=750sin45= 530NFRX = F1x + F2x +F3x = 680N – 312.5N -530N = -162.5N
FRY = F1y + F2y +F3y = -510N - 541N + 530N = -521N
22 )521()5.162( RF
= 546N
FR
θ =01 67.72
5.162521
tan
TRANSLATION VS ROTATION
TRANSLATION
ROTATION
ROTATING TRANSLATION
M = F x d
Characteristics of MomentsCharacteristics of Moments
F
Fp
Fa
If these wheel nuts must be tightened to 85Nm, what is the force F?Angle = 35 degrees, d = 420mm.
From M = F * d, then F = M / d
Perpendicular distance = d * cos (35)
F = 85 / (0.42 * cos(35)) = 247.06 N
Example - MomentExample - Moment
A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at 0. Determine (a) the moment of 100-lb force about 0.(b) the magnitude of the horizontal force applied at A which will create the same moment about 0.(c ) the smallest force applied at A which will create same moment (d) how far from the shaft a 240 lb vertical force must act to create the same moment about 0. (e) whether any one of the forces obtain in parts (a), (b), (c ) and (d) is equivalent to the original force.
Example - MomentExample - Moment
(a) The perpendicular distance from ) to the line of action of 100-lb force is
0
24 in. cos 60 12 in.
M 100 lb 12 in. 1200 lb-in
d
Fd
The magnitude of the moment about 0
The force rotates the lever in clockwise about 0 and M0 is perpendicular to the plane.
Example - MomentExample - Moment
(b)Horizontal force
00
24 in. sin 60 20.8 in.
MM
1200 lb-in57.7 lb
20.8 in.
d
Fd Fd
F
Since the moment about 0 is 1200 lb-in the resulting F
Example - MomentExample - Moment
(c) Smallest Force, since M=Fd, the smallest value of F occurs when d is a maximum. It will be perpendicular to 0A
00
24 in.
MM
1200 lb-in50 lb
24 in.
d
Fd Fd
F
Example - MomentExample - Moment
(d)A 240-lb vertical force In this case the force is given determine the distance
00
MM
1200 lb-in5.0 in.
240 lb
0B cos 60 5.0 in. 0B 10.0 in.
Fd dF
d
Example - MomentExample - Moment
(e) None of the force in parts b, c, and d is equivalent of original 100-lb force. Although they have the same moment about 0, they have different x and y components .
ExampleExample
A vertical force P of magnitude 60 lb is applied to the crank at A. Knowing that = 75o, determine the moment P alone each of the coordinate axes.
75
60 lb
d
d = 8 cos 15 =7.727
8
T = 60lb x 7.727 in = 463.6 in-lb
Fp =60Sin75 = 57.96 lb
Fa = 60cos75 = 15.53 lb
Tx = 57.96 lb x 8 in = 463.6 in-lb
T2 = 15.53 lb x 5 in = 77.65 ft-lb
75
60 lbFp
Fa
75
The mechanism shown is used to raise a crate of supplies from a ship's hold. The crate
has total mass 56.0 . A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius 0.310 . The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius 0.120 , the cylinder turns, and the crate is raised.
What magnitude of the force applied tangentially to the rotating crank is required to
raise the crate with an acceleration of 0.750 ? (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)
Varigon’s TheoremVarigon’s Theorem
As with the summation of force combining to get resultant force
Similar resultant comes from the addition of moments
1 2 nR F F F ��������������������������������������������������������
0 R 1 1 2 2 n nM R d F d F d F d ����������������������������������������������������������������������
Moment of F1
Line of action of F1 passes through Point O.No moment is generated
Moment of F2 75 lbs
Moment of F3 60 lbs
Sum of All Moments about O
RESULTANTS OF PARALLEL FORCES
FOR MAGNITUDE
FOR LOCATION
100 lb
2
200 lb
7
A B
FR = -100lb + -200lb = -300lb
x FR
ΣMA= -100lb x 2ft + -200lb x 7 ft = -200 lb-ft - 1400 lb-ft = -1600 lb-ft
MR=ΣMA AND MR = FR x X
MA
-300lb x (X) = -1600 lb-ft
X = -1600/-300 = 5.3 ft
ΣFy = Resultant = -800lb + 600lb – 1200lb -400lb = -1800lb
Sum MA = 1800lb( x) = -600lb(3ft) + 1200lb(5ft) + 400lb(9ft) = -1800lb-ft + 6000lb-ft + 3600lb-ft MA = 7800lb-ft
x = 7800lb-ft / 1800lb = 4.33ft
DISTRIBUTED LOADS
UNIFORMLY DISTRIBUTED LINE LOADS
Brimham Rocks, North Yorkshire England
NONUNIFORMLY DISTRIBUTED LINE LOADS
b
centroid is two thirds away fromthe vertex and 1/3 away from the right angle.
(third from the right)
magnitude of distributed load comes fromarea of triangle = ½ b x h
Location of Concentrated load will be 1/3 the base length away from right angle
Value of Resultant force fromΣF
=1/3(4.5m= 1.5 m
Sum the moments about A
Solve for xbar
TRAPEZOID
Magnitude of Resultants
Lines of Action
Resultant of Dist. Loads
Resultant Point
HYDROSTATIC FORCES
calculate pressure at depth
By Pascal’s Law
P = F/A or F = PA
The distributed load looks like this
Calculated Load Intensity
Resultant
Location of Resultant
FORCE COUPLES
T=F X 2d
d d
F
F
MR
Since the 2 forces form a couple
Magnitude is the same at any point in plane
About O
About C
About B
or, I could just find theresultant of the 2 couples
Compute X resultant
Compute Y resultant
Sketch components and resultant
Compute resultant
Compute Angle