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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

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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: 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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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