7 Statics

By Liew Sau Poh

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Objectives

7.1 Centre of Gravity 7.2 Equilibrium of particles 7.3 Equilibrium of rigid bodies

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Learning Outcome

(a) define centre of gravity (b) state the condition in which the

centre of mass is the centre of gravity (c) state the condition for the equilibrium

of a particle (d) solve problems involving forces in

equilibrium at a point (e) define torque as = r F

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(f) state the conditions for the equilibrium of a rigid body

(g) sketch and label the forces which act on a particle and a rigid body

(h) use the triangle of forces to represent forces in equilibrium

(i) solve problems involving forces in equilibrium.

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7. Statics

Statics equilibrium is a state where balanced forces acting on a rigid body or a particle so that it remains at rest.

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7.1 Centre of Gravity (C.G.)

The Centre of Gravity (C.G.) of a body is the point through which the whole weight of the body appears to act For a regular object, the C.G is always at the centre

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7.1 Centre of Gravity (C.G.)

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Determine the C.G of a lamina (Experimental method) Line of action of its weight must pass through the centre of gravity.

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hung freely at A

at rest, draw vertical line

hung freely at B

at rest, draw vertical line

G is centre of gravity

© Manhattan Pre ss (H.K.) Ltd.

The Centre of Gravity of a Lamina (Mathematical method)

Assumption: Lamina is made up of n particles

coordinate of each particle

coordinate of CG

The Centre of Gravity of a Lamina

Taking moments about the y-axis, Xavg = Mg( ) = (m1g1)x1 + (m2g2)x2 + (m3g3)x3 + . . . + (mngn)xn

Mg

xgmx

xgmx

n

iiii

n

iiii

1

1

)(

)(

The Centre of Gravity of a Lamina

If g = constant (in a uniform gravitational field), g1 = g2 = g3 = ... = gn = g, then Similarly, by taking moments about the x-axis,

M

xmx

n

iii

1)(

M

ymy

n

iii

1)(

The Centre of Gravity of a Lamina

Coordinates of centre of gravity: Note: These are in fact the coordinates of the centre of mass.

M

xmx

n

iii

1)(

M

ymy

n

iii

1)(

Centre of Mass (C.M.) vs. Centre of Gravity (C.G.)

Centre of mass of a body is defined as the point where the entire mass of the body acts. Centre of gravity of a body is defined as the point where the weight of the body acts.

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Example of C.G.:

If a body is uniform and regular in shape, then the c.g. is at the geometrical centre.

c.g. c.g.

c.g. c.g. Centre of cylinder

7.2 Equilibrium of particles

A body is balanced if pivoted at a point which passes through c.g..

c.g.

The uniform ruler above is balanced.

7.2 Equilibrium of particles

The condition for a particle in static equilibrium state.

(a) Resultant force is zero. If F1 = - F2

Then the resultant force, Fnet = F1 + F2

= 0 16

F1 F2

7.2 Equilibrium of particles

Thus, F1 + F2 = 0 The vector sum of the forces must be zero.

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F1 F2

Example 1

Let three forces F1, F2, and F3 acting on a particle O, which is in equilibrium, as shown:

Then, (components x) F1(x) + F2(x) + F3(x) = 0, and (components y) F 1(y) + F2(y) + F3(y) = 0

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F1 F2

F3

O F1(x)

F3(x)

F2(x)

F1(y )

F3(y)

F2(y)

Alternative method

Vector sum of the forces must be zero. F1 + F2 + F3 = 0

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F1 F2

F3

O

F1

F2 F3 A

B

C

Closed polygon

Consider three forces acting on a particles is in equilibrium. The forces can be joined to form a polygon. The resultant force = 0 20

F2

F1 F3

Closed polygon

If the forces acting on a particle is not in equilibrium, a resultant force is existing.

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The length of the sides representing the magnitude of the forces. The direction of force is represented by its arrow.

F1 F3

F2

F4 Resultant, F 22

F1

F2 F3

F1

F2

F3

Resultant F

Resultant Force, F = F1 + F2 + F3

7.3 Equilibrium of a Rigid Body

The following shows a rod in balance, which is so called in static equilibrium. The weight of the rod is supported by the reaction, R

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Support

weight

R

7.3 Equilibrium of a Rigid Body

Two conditions are necessary: No resultant force (R = weight) No resultant torque (Moment = 0)

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weight

R

Toque

If a force, F is applied on the one end of the rod, it will start to rotate about G. A torque is produced by the force, where torque, = F r

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weight

r

F

G

Toque Torque can be produced by two parallel forces of the magnitude, but opposite direction of a distance from the rotating axis. The pair of forces = a couple

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F1

F2

d = 1 m d = 1 m

Torque The a couple (torque), if exists, will cause a moment (torque) about the rotating axis, which is given by = Fr = F(2x) 0 Hence the body is not in equilibrium.

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F (external force) x = 1 m x = 1 m

F 28

More about Torque

Torque is the tendency of a force to rotate an object about an axis. Torque, , is a vector quantity. Consider an object pivoting about the point P by the force F being exerted at a distance r.

F

d

P r

Moment arm

Line of Action

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More about Torque

The line that extends out of the tail of the force vector is called the line of action. The perpendicular distance from the pivoting point P to the line of action is called Moment arm.

F

d

P r

Moment arm

Line of Action

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More about Torque Magnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm.

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2211 dFdF

sinrF Fd F

d

P r

Moment arm

Line of Action

F Sin

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More about Torque When there are more than one force being exerted on certain points of the object, one can sum up the torque generated by each force vectorially. The convention for sign of the torque is positive if rotation is in counter-clockwise and negative if clockwise.

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2211 dFdF

sinrF Fd

F1

d1

P

d2

F2

7.3 Equilibrium of a Rigid Body

If a mass of 1 kg is put on one end, the rod will rotate in clockwise direction. While the force, F1 = 10N The torque, = Fr = (10)(1) = 10 Nm

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Support F1 =10 N

r = 1 m

7.3 Equilibrium of a Rigid Body If another mass of 1 kg is applied on the other end, the system will be in static equilibrium again, where R = F1 + F2, or R + F1 + F2 = 0.

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Support F2 = 10N F1 =10N

1 m 1 m R = 20N

7.3 Equilibrium of a Rigid Body

In this case, the sum of moment/torque = F1r1 F2r2 = (10)(1) (10)(1) = 0.

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Support F2 = 10N F1 =10N

1 m 1 m

Direction of Rotation

Example 1

The figure shows a rod of length 2 m and weight 1 kg resting on supports at end A and B. Find the reaction at A and B. (g = 10 ms-2)

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A B

0.5 m

Solution

Let R1 = reaction at point A, R2 = reaction at point B, F = force due to weight of the rod.

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F = 10N A B

0.5 m R1 R2 0.5 m

Solution

Since the system is in equilibrium, Resultant Force = 0

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F = 10N A B

0.5 m R1 R2 0.5 m

Solution

Sum of moment about A, M = 0 R2 (0.5 + 1) 10(1) = 0 R2 = 10/1.5 =6.67 N From (1), R1 = 10 6.67 =3.33 N

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F = 10N A B

0.5 m R1 R2 0.5 m 1 m

Solution

OR, Sum of moment about B, M = 0 R1 (0.5 + 1) 10(0.5) = 0 R1 = 5/1.5 = 3.33 N R2 = 10 3.33 = 6.67N

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F = 10N A B

0.5 m R1 R2 0.5 m 1 m

Example 2 A uniform horizontal beam with a length of l = 8.00 m and a weight of Wb = 200 N is attached to a wall by a pin connection and supported by a cable as shown. A person of weight Wp = 600 N stands a distance d = 2.00 m from the wall.

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Example 2

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Find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam.

Solution

Analyze Draw a free body diagram Use the pivot in the problem (at the wall) as the pivot

This will generally be easiest

Note there are three unknowns (T, R, q)

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Solution The forces can be resolved into components in the free body diagram.

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Solution

Apply the two conditions of equilibrium to obtain three equations Solve for the unknowns

Taking moment about R,

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Nm

mNmNl

lWdW

T

lWdWlT

bp

bpz

31353sin)8(

)4)(200()2)(600(sin

)2

(

0)2

())(sin(

Solution

Considering sum of Fx and sum of Fy,

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0sinsin0coscos

bpy

x

WWTRFTRF

NNT

R

TTWW

TTWW

RR

bp

bp

5817.71cos

53cos)313(coscos

7.71sin

sintan

sinsin

tancossin

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Example 3

A safe whose mass is M = 430 kg is hanging by a rope from a boom with dimensions a = 1.9 m and b = 2.5 m as shown in the figure. The uniform beam has a mass m of 85 kg; the mass of the cable and rope are negligible.

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a

b

Example 3

a) What is the tension Tc in the cable; i. e., what is the magnitude of the force Tc on the beam from the horizontal cable?

b) What is the force at the hinge?

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a

b

Solution

Consider the sum of Fx and sum of Fy,

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; 0,, chchxextxnet TFTFFFMgTmgMgFFF rvyextynet ; 0,,

Tc

mg

Mg

Fh

Tr

Fv

Solution

Taking moment about the hinge

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Tc

mg

Mg

Fh

Tr

Fv

NabgmM

abmg

abMgT

MgbaTbmg

bTaTbmg

c

c

rc

TrTcmgvhzextznet

6093)2

(2

02

02

00

,,

Solution

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Tc

mg

Mg

Fh

Tr

Fv

NFFF

NmgMgFNTF

vh

v

ch

7912

50476093

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Summary

Centre of Gravity Centre of Mass

Statics

Resultant Force = 0 Triangle of forces

Equilibrium of Particles

Torque, = r F Resultant force = 0 Resultant torque = 0

Equilibrium of Rigid Bodies

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