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PHYSICS CHAPTER 5
The branch of mechanics concerned with bodies that are acted upon by balanced forces and couples so that
they remain at restat rest or in in unaccelerated motionunaccelerated motion.
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CHAPTER 5: CHAPTER 5: StaticStatic
(4 Hours)(4 Hours)
PHYSICS CHAPTER 5
2
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine the equilibrium of a particle and the equilibrium of a particle and statestate the the
condition for equilibrium.condition for equilibrium. State two types of equilibrium, i.e. static (State two types of equilibrium, i.e. static (vv=0) and =0) and
dynamic (dynamic (aa=0).=0). SketchSketch polygon of forces to represent forces in polygon of forces to represent forces in
equilibrium.equilibrium.
Learning Outcome:5.1 Equilibrium of a particle (1 hour)5.2 Polygon of forces (1 hour)
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PHYSICS CHAPTER 5
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5.1 Equilibrium of a particle5.1.1 Concurrent forces is defined as the forces whose lines of action pass through a forces whose lines of action pass through a
single common pointsingle common point. (whether inside or outside of the body) The forces cause the translational motion translational motion on the body. Figure 5.1 and Figure 5.2 show the examples of concurrent
forces.
2F
3F
1F
3F
2F
1F
Figure 5.1Figure 5.1 Figure 5.2Figure 5.2
PHYSICS CHAPTER 5
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5.1.2 Equilibrium of a particle is defined as the vector sum of all forces acting on a particle vector sum of all forces acting on a particle
(point) must be zero(point) must be zero. The equilibrium of a particle ensures the body in translational translational
equilibriumequilibrium and its conditioncondition is given by
This is equivalent to the three independent scalar equations along the direction of the coordinate axes,
There are two types of equilibrium of a particle. It is StaticStatic equilibrium (v=0) body remains at restrest (stationarystationary). DynamicDynamic equilibrium (a=0) body moving at a uniform uniform
(constant) (constant) velocityvelocity.
0nettFF
Newton’s first law of motion
0 , 0 , 0 zyx FFF
PHYSICS CHAPTER 5
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5.1.3 Problem solving strategies for equilibrium of a particle
The following procedure is recommended when dealing with problems involving the equilibrium of a particle: Sketch a simple diagramSketch a simple diagram of the system to help
conceptualize the problem. Sketch a separate free body diagramSketch a separate free body diagram for each body. Choose a convenient coordinate axesChoose a convenient coordinate axes for each body and
construct a tableconstruct a table to resolve the forces into their components.
Apply the condition for equilibrium of a particleApply the condition for equilibrium of a particle in component form :
SolveSolve the component equationsequations for the unknowns.
0xF 0yFand
PHYSICS CHAPTER 5
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5.2 Polygon of forcesCase 1:Case 1: A particle in equilibrium as a result of two forces acting on it as
shown in Figure 5.3.
They are equal in magnitude but opposite in the direction, thus
Case 2:Case 2: A particle in equilibrium as a result of
three forces acting on it as shown in
Figure 5.4.
Figure 5.3Figure 5.3
Figure 5.4Figure 5.4
1F
2F
i.e. 0 21 FFF 0 xF OR 0 yF
2F
1F
3F
PHYSICS CHAPTER 5
7
They are form a closed triangle of forces, thus
Case 3:Case 3: A particle in equilibrium as a result of
four forces acting on it as shown in
Figure 5.5.
They will form a closed polygon of forces, thus
2F
1F
3F
0 321 FFFF
i.e. 0 xF and 0 yF
04321 FFFFF
1F
2F
3F
4F
1F
2F
3F
4F
Figure 5.5Figure 5.5
i.e. 0 xF and 0 yF
PHYSICS CHAPTER 5
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A load of 250 kg is hung by a crane’s cable. The load is pulled by a horizontal force such that the cable makes a 30 angle to the vertical plane. If the load is in the equilibrium, calculate
a. the magnitude of the tension in the cable,
b. the magnitude of the horizontal force. (Given g =9.81 m s2)
Solution :Solution :
Example 1 :
30
F F
Free body diagram of the load :
gm
T
yT3060
xT
kg 250m
PHYSICS CHAPTER 5
9
Solution :Solution :
11stst method : method :
a.
Since the load is in the equilibrium, then
Thus
b. By substituting eq. (2) into eq. (1), therefore
0xF 060cos TF
kg 250m
Force x-component (N) y-component (N)
gm
0 9.81250 mg
F
F 0
T 60cosT 60sinT
N 2833T
2453
0F
(1)
(2) 0yF 0245360sin T
060cos2833 F N 1417F
PHYSICS CHAPTER 5
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30
Solution :Solution :
22ndnd method : method :
a. Since the load is in the equilibrium, then a closed triangle of
forces can be sketched as shown below.
b. 30sinT
F
30cos9.81250
T
kg 250m
N 2833T
30cosT
mg
30sin2833
F
N 1417F
F
gm T
From the closed triangle of forces, hence
PHYSICS CHAPTER 5
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Calculate the magnitude and direction of a force that balance the three forces acted at point A as shown in Figure 5.6.
Example 2 :
N 121FN 202F
N 303F
30.055.0
45.0 A
Figure 5.6Figure 5.6
PHYSICS CHAPTER 5
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Solution :Solution :
To find a force to balance the three forces means the system must be in equilibrium hence
N 30 N; 20 N; 12 321 FFF
Force x-component (N) y-component (N)
1F 55.0cos12
F
xF yF
6.8855.0sin12
9.83
2F 30.0cos20
17.330.0sin20
10.0
3F 45.0cos30
21.245.0sin30
21.2
0 xF
021.217.36.88 xFN 31.6xF
PHYSICS CHAPTER 5
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Solution :Solution :
The magnitude of the force,
and its direction,
0 yF021.210.09.83 yF
N 1.37yF
222y
2x FFF 1.3731.6
N 31.6F
x
y1
F
Fθ tan
31.6
1.37tan 1θ
2.48θ from the +x-axis anticlockwise
PHYSICS CHAPTER 5
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A window washer pushes his scrub brush up a vertical window at
constant speed by applying a force F as shown in Figure 5.7. The brush weighs 10.0 N and the coefficient of kinetic friction is
k= 0.125. Calculate
a. the magnitude of the force F ,
b. the normal force exerted by the window on the brush.
Example 3 :
F
50.0
Figure 5.7Figure 5.7
PHYSICS CHAPTER 5
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Solution :Solution :
a. The free body diagram of the brush :
The brush moves up at constant speed (a=0) so that
Thus
0.125 ;N 10.0 kμW
W
F
N
kf
constant constant speedspeed
Force x-component (N) y-component (N)
F
50.0cosF
kf
0Nμk
50.0sinF
W
0 10.0
N
N 0
N0.125
0amF
50.0cosFN 0 xF (1)
(2)10.00.12550.0sin NF 0 yF
50.0
PHYSICS CHAPTER 5
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Solution :Solution :
a. By substituting eq. (1) into eq. (2), thus
b. Therefore the normal force exerted by the window on the brush
is given by
N 14.6F
10.050.0cos0.12550.0sin FF
50.0cosFN 50.0cos14.6N
N 9.39N
PHYSICS CHAPTER 5
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Exercise 5.1 :
Use gravitational acceleration, g = 9.81 m s2
1.
The system in Figure 5.8 is in equilibrium, with the string at the centre exactly horizontal. Calculate
a. the tensions T1, T2 and T3.
b. the angle .
ANS. : 49 N, 28 N, 57 N; 29ANS. : 49 N, 28 N, 57 N; 29
Figure 5.8Figure 5.8
PHYSICS CHAPTER 5
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Exercise 5.1 :2.
A 20 kg ball is supported from the ceiling by a rope A. Rope B pulls downward and to the side on the ball. If the angle of A to the vertical is 20 and if B makes an angle of 50 to the vertical as shown in Figure 5.9, Determine the tension in ropes A and B.
ANS. : 134 N; 300 NANS. : 134 N; 300 N
Figure 5.9Figure 5.9
PHYSICS CHAPTER 5
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Exercise 5.1 :3.
A block of mass 3.00 kg is pushed up against a wall by a force
P that makes a 50.0 angle with the horizontal as show in Figure 5.10. The coefficient of static friction between the block and the wall is 0.250. Determine the possible values for the
magnitude of P that allow the block to remain stationary.
ANS. : 31.8 N; 48.6 NANS. : 31.8 N; 48.6 N
Figure 5.10Figure 5.10
PHYSICS CHAPTER 5
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define and useDefine and use torque, torque, .. State and useState and use conditions for equilibrium of rigid body: conditions for equilibrium of rigid body:
Examples of problems :Examples of problems :
Fireman ladder leaning on a wall, see-saw, pivoted / Fireman ladder leaning on a wall, see-saw, pivoted / suspended horizontal bar.suspended horizontal bar.
Sign convention for moment or torque :Sign convention for moment or torque :
+ve+ve : anticlockwiseanticlockwise
veve : clockwiseclockwise
Learning Outcome:
5.3 Equilibrium of a rigid body (2 hours)
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0 , 0 , 0 τFF yx
PHYSICS CHAPTER 5
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5.3 Equilibrium of a rigid body5.3.1 Non-concurrent forces is defined as the forces whose lines of action do not pass the forces whose lines of action do not pass
through a single common point. through a single common point. The forces cause the rotational motionrotational motion on the body. The combination of concurrent and non-concurrent forces cause
rolling motionrolling motion on the body. (translational and rotationaltranslational and rotational motion)
Figure 5.11 shows an example of non-concurrent forces.
2F
3F
1F
Figure 5.11Figure 5.11
4F
PHYSICS CHAPTER 5
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5.3.2 Torque (moment of a force), The magnitude of the torquemagnitude of the torque is defined as the product of a the product of a
force and its perpendicular distance from the line of action force and its perpendicular distance from the line of action of the force to the point (rotation axis)of the force to the point (rotation axis).
OR
Because of
where r : distance between the pivot point (rotation axis) and the point of application of force.
Thus
Fdτ
force theof magnitude :Farm)(moment distancelar perpendicu : d
torque theof magnitude : τwhere
sinrd
sin FrrF
and between angle : where
OR Fr
PHYSICS CHAPTER 5
23
It is a vector quantityvector quantity. The dimension of torque is
The unit of torqueunit of torque is N mN m (newton metre), a vector productvector product unlike the joule (unit of work)joule (unit of work), also equal to a newton metre, which is scalar productscalar product.
Torque is occurred because of turning (twisting) effects of the turning (twisting) effects of the forces forces on a body.
Sign convention of torque: PositivePositive - turning tendency of the force is anticlockwiseanticlockwise. NegativeNegative - turning tendency of the force is clockwiseclockwise.
The value of torque dependstorque depends on the rotation axisrotation axis and the magnitude of applied forcemagnitude of applied force.
22TMLdF
PHYSICS CHAPTER 5
24
Case 1 :Case 1 : Consider a force is applied to a metre rule which is pivoted at
one end as shown in Figures 5.12a and 5.12b.
Figure 5.12aFigure 5.12a
F
F
θ
Figure 5.12bFigure 5.12b
Pivot point (rotation axis)
Fdτ
θrd sin
θFrFdτ sin
(anticlockwise)
(anticlockwise)r
Point of action of a force
Line of action of a force
d
PHYSICS CHAPTER 5
25
O
Figure 5.13Figure 5.132θ
Case 2 :Case 2 : Consider three forces are applied to the metre rule which is
pivoted at one end (point O) as shown in Figures 5.13.
Caution : If the line of action of a force is through the rotation axisline of action of a force is through the rotation axis
then
1F
1θ
111 θrd sin
321 ττττ O
Therefore the resultant (nett) torque is
3F
2F 1r
0sin 333333 θrFdFτ
222 θrd sin
111111 θrFdFτ sin222222 θrFdFτ sin
2r
2211 dFdFτ O
θFrτ sin0τ
and 0θ
Simulation 5.1
PHYSICS CHAPTER 5
26
Determine a resultant torque of all the forces about rotation axis, O in the following problems.
a.
Example 4 :
m 5
N 102F
m 5 N 301F
m 3
m 3
N 203F
m 10
m 6O
PHYSICS CHAPTER 5
27
b.
Example 4 :
m 5
N 102F
m 5
N 301F
m 3
m 3
N 254F
N 203F
m 10
α
m 6O β
PHYSICS CHAPTER 5
28
m 5m 5
m 10
m 6O
Solution :Solution :
a.
Force Torque (N m), o=Fd=Frsin
1F
90330
2F
50510
N 102FN 301F
N 203F
m 31d
m 52d
3F
0The resultant torque:
m N 405090 Oτ(clockwise)(clockwise)
PHYSICS CHAPTER 5
29
m 5
m 10
m 3
m 6
m 5
Solution :Solution :
b.
Force Torque (N m), o=Fd=Frsin
1F
90330
2F
51.50.515520sin βrF33F
0 The resultant torque:51.590 Oτ
(clockwise)(clockwise)
N 102F
N 301F
0.51553
3sin
22
β
O
N 203F
N 254Fα
β
m 31dβ
m 5r
4F
0
3d
m N 38.5 Oτ
PHYSICS CHAPTER 5
30
5.3.3 Equilibrium of a rigid body Rigid bodyRigid body is defined as a body with definite shape that a body with definite shape that
doesn’t change, so that the particles that compose it stay in doesn’t change, so that the particles that compose it stay in fixed position relative to one another even though a force is fixed position relative to one another even though a force is exerted on itexerted on it.
If the rigid body is in equilibriumrigid body is in equilibrium, means the body is translational and rotational equilibriumtranslational and rotational equilibrium.
There are two conditionstwo conditions for the equilibrium of forces acting on a rigid body. The vector sum of all forces acting on a rigid body must The vector sum of all forces acting on a rigid body must
be zero.be zero.
0nettFF
OR
0 , 0 , 0 zyx FFF
PHYSICS CHAPTER 5
31
The vector sum of all external torques acting on a rigid The vector sum of all external torques acting on a rigid body must be zero about any rotation axisbody must be zero about any rotation axis.
This ensures rotational equilibriumrotational equilibrium. This is equivalent to the three independent scalar
equations along the direction of the coordinate axes,
Centre of gravity, CG Centre of gravity, CG is defined as the point at which the whole weight of a body the point at which the whole weight of a body
may be considered to actmay be considered to act. A force that exerts on the centre of gravityexerts on the centre of gravity of an object will
cause a translational motiontranslational motion.
0nettτ
0 , 0 , 0 zyx τττ
PHYSICS CHAPTER 5
32
Figures 5.14 and 5.15 show the centre of gravity for uniformcentre of gravity for uniform (symmetric) objectobject i.e. rod and sphere
rodrod – refer to the midway point between its endmidway point between its end.
spheresphere – refer to geometric centregeometric centre.
2
l
2
l
CG
CGl
Figure 5.14Figure 5.14
Figure 5.15Figure 5.15
PHYSICS CHAPTER 5
33
5.3.4 Problem solving strategies for equilibrium of a rigid body
The following procedure is recommended when dealing with problems involving the equilibrium of a rigid body: Sketch a simple diagramSketch a simple diagram of the system to help
conceptualize the problem. Sketch a separate free body diagramSketch a separate free body diagram for each body. Choose a convenient coordinate axesChoose a convenient coordinate axes for each body and
construct a tableconstruct a table to resolve the forces into their components and to determine the torque by each force.
Apply the condition for equilibrium of a rigid bodyApply the condition for equilibrium of a rigid body :
SolveSolve the equationsequations for the unknowns.
0xF 0yF; and 0τ
PHYSICS CHAPTER 5
34
A hanging flower basket having weight, W2 =23 N is hung out over
the edge of a balcony railing on a uniform horizontal beam AB of length 110 cm that rests on the balcony railing. The basket is
counterbalanced by a body of weight, W1 as shown in Figure 5.16. If
the mass of the beam is 3.0 kg, calculate
a. the weight, W1 needed,
b. the force exerted on the beam at point O.
(Given g =9.81 m s2)
Example 5 :
1W2W
A BO35 cm 75 cm
Figure 5.16Figure 5.16
PHYSICS CHAPTER 5
35
Solution :Solution :
The free body diagram of the beam :
Let point O as the rotation axis.
N 23 ;kg 3 2Wm
0.75 mA B
OCG
1W
2W
N
gm
0.35 m
0.55 m 0.55 m
0.20 m0.20 m
Force y-comp. (N) Torque (N m), o=Fd=Frsin
1W
1W
gm 9.813 5.880.2029.4
11 WW 0.750.75
2W
23 8.050.3523
N
N 029.4
PHYSICS CHAPTER 5
36
Solution :Solution :
Since the beam remains at rest thus the system in equilibrium.
a. Hence
b.
N 2.891W
0 yFand
0Oτ
05.888.050.75 1W
N 55.3N
029.423 NW1
029.4232.89 N
PHYSICS CHAPTER 5
37
A uniform ladder AB of length 10 m and mass 5.0 kg leans against a smooth wall as shown in Figure 5.17. The height of the end A of the ladder is 8.0 m from the rough floor.
a. Determine the horizontal and vertical
forces the floor exerts on the end B of
the ladder when a firefighter of mass
60 kg is 3.0 m from B.
b. If the ladder is just on the verge of
slipping when the firefighter is 7.0 m
up the ladder , Calculate the coefficient
of static friction between ladder and
floor.
(Given g =9.81 m s2)
Example 6 :
A
B
smooth wall
rough floor
Figure 5.17Figure 5.17
PHYSICS CHAPTER 5
38
Solution :Solution :
a. The free body diagram of the ladder :
Let point B as the rotation axis.
kg 60 ;kg 5.0 fl mm
A
B
CG
gm f
1N
gml
2N
α
m 8.0m 10
m 3.0
m 5.0
Forcex-comp.
(N)y-comp.
(N)
Torque (N m),
B=Fd=Frsin
gml
1N1N
0.810
8sin α
sf
gm f
49.1 0.6
10
6sin β
2N
sf
0
5890
2N
0
0
m 6.0
αβ βsin5.049.1
β
β
147
0
βsin3.05891060
αN1 sin101N8
0
0 sf
PHYSICS CHAPTER 5
39
Solution :Solution :
Since the ladder in equilibrium thus
0 Bτ
081060147 1N
N 1511N
0 xF
0 s1 fNN 151sfHorizontal force:Horizontal force:
0 yF
058949.1 2NN 6382NVertical force:Vertical force:
PHYSICS CHAPTER 5
40
m 10
A
B
m 8.0
m 6.0
m 5.0 α
β
Solution :Solution :
b. The free body diagram of the ladder :
Let point B as the rotation axis.
0.6sin 0.8;sin βα
gm f
gml
sf
m 7.0
Forcex-comp.
(N)y-comp.
(N)
Torque (N m),
B=Fd=Frsin
gml
1N1N
2s Nμ
gm f
49.1
2N
sf
0
5890
2N
0
0
α
βsin5.049.1
β
β147
0
βsin7.05892474
αN1 sin101N8
0
0
2N
1N
PHYSICS CHAPTER 5
41
Solution :Solution :
Consider the ladder stills in equilibrium thus
0 Bτ
082474147 1N
N 3281N
0 xF
0 2s1 NμN
0 yF
058949.1 2NN 6382N
0638328 sμ0.514sμ
PHYSICS CHAPTER 5
42
Figure 5.18Figure 5.18
A floodlight of mass 20.0 kg in a park is supported at the end of a 10.0 kg uniform horizontal beam that is hinged to a pole as shown in Figure 5.18. A cable at an angle 30 with the beam helps to support the light.
a. Sketch a free body diagram of the beam.
b. Determine
i. the tension in the cable,
ii. the force exerted on the beam by the
pole.
(Given g =9.81 m s2)
Example 7 :
PHYSICS CHAPTER 5
43
Solution :Solution :
a. The free body diagram of the beam :
b. Let point O as the rotation axis.
kg 10.0 ;kg 20.0 bf mm
Force x-comp. (N) y-comp. (N) Torque (N m), o=Fd=Frsingm f
l1960 196
O CG
gm f
gmb
T
S
30
l
l0.5
gmb
ll 49.10.598.1 0 98.1
T
TlTl 0.530sin 30cosT 30sinT
S
xS yS 0
PHYSICS CHAPTER 5
44
Solution :Solution :
b. The floodlight and beam remain at rest thus
i.
ii.
0 Oτ
00.549.1196 TlllN 490T
0 xF
0cos xS30T
N 424xS
0 yF
030sin98.1196 yST
N 49.1yS
PHYSICS CHAPTER 5
45
Solution :Solution :
b. ii. Therefore the magnitude of the force is
and its direction is given by
N 427S
2y
2x SSS
22S 49.1424
x
y
S
Sθ 1tan
6.61θ
424
49.1tan 1θ
from the +x-axis anticlockwise
PHYSICS CHAPTER 5
46
Exercise 5.2 :
Use gravitational acceleration, g = 9.81 m s2
1.
Figure 5.19 shows the forces, F1 =10 N, F2= 50 N and F3=
60 N are applied to a rectangle with side lengths, a = 4.0 cm
and b = 5.0 cm. The angle is 30. Calculate the resultant torque about point D.
ANS. : -3.7 N mANS. : -3.7 N m
D
AB
C γ
1F
3F
2F
Figure 5.19Figure 5.19
a
b
PHYSICS CHAPTER 5
47
Figure 5.20Figure 5.20
Exercise 5.2 :2.
A see-saw consists of a uniform board of mass 10 kg and length 3.50 m supports a father and daughter with masses 60 kg and 45 kg, respectively as shown in Figure 5.20. The fulcrum is under the centre of gravity of the board. Determine
a. the magnitude of the force exerted by the fulcrum on the
board,
b. where the father should sit from the fulcrum to balance the
system.
ANS. : 1128 N; 1.31 mANS. : 1128 N; 1.31 m
PHYSICS CHAPTER 5
48
3.
A traffic light hangs from a structure as show in Figure 5.21. The uniform aluminum pole AB is 7.5 m long has a mass of 8.0 kg. The mass of the traffic light is 12.0 kg. Determine
a. the tension in the horizontal massless cable CD,
b. the vertical and horizontal components of the force exerted
by the pivot A on the aluminum pole.
ANS. : 248 N; 197 N, 248 NANS. : 248 N; 197 N, 248 N
Figure 5.21Figure 5.21
Exercise 5.2 :
PHYSICS CHAPTER 5
49
4.
A uniform 10.0 N picture frame is supported by two light string
as shown in Figure 5.22. The horizontal force, F is applied for holding the frame in the position shown.
a. Sketch the free body diagram of the picture frame.
b. Calculate
i. the tension in the ropes,
ii. the magnitude of the horizontal force, F .
ANS. : 1.42 N, 11.2 N; 7.20 NANS. : 1.42 N, 11.2 N; 7.20 N
Exercise 5.2 :
Figure 5.22Figure 5.22
F
50.0
cm 15.0
cm 30.0
50
PHYSICS CHAPTER 5
THE END…Next Chapter…
CHAPTER 6 :Circular motion