Chapter 4 : statics4-1 Torque• Torque, t , is the tendency of a force to rotate

an object about some axis• is an action that causes objects to rotate.• Torque is not the same thing as force.• For rotational motion, the torque is what is

most directly related to the motion, not the force.

What property of the applied force

causes the door to open?

Definition of torque:

r

F

where r is the vector from the reference point (generally either the pivot point or the center of mass) to the point of application of the force F.

||r F sin

where q is the angle between the vectors r and F.

Cross Product Direction

• Curl right-hand fingers in direction of q

• Right-hand thumb points in direction of cross-product

• Not commutative

A

B

A

B

q

AB = –(BA)

Direction of Torque• Torque is a vector quantity– Direction determined by axis of twist– Perpendicular to both r and F– In 2-d, torque points into or out of paper– Clockwise torques point into paper.

Defined as negative– Counter-clockwise torques point out of paper.

Defined as positive

Torque

t = r x F

Lever arm length (m)

Force (N)

Torque (N.m)

Fulcrum• the point of support, or axis, about which a lever

may be made to rotate

Lever Arm

Shortest distance from line of action to point of interest

What is the direction of the torque about point O from force F2?

.A B. C. D. E. F.

The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.

In which of the cases is the torque on the nut the biggest? 1. Case 1 2. Case 2 3. Case 3

Calculate a torque

• A force of 50 newtons is applied to a wrench that is 30 centimeters long.

Calculate the torque if the force is applied perpendicular to the wrench so the lever arm is 30 cm.

You are installing a new spark plug in your car, and the manual specifies that it be tightened to a torque of 37 N ·

m. Using the data in the drawing, determine the magnitude F of the force that you must exert on the

wrench.

F sin(50) 0.28 = 37∙ ∙F = 172.5 N

example

Couples

One way of producing rotation without translation of an object is to apply a pair of equal

and opposite forces with offset lines of action . Consider the situation shown. Forces F1 and F2

are equal in magnitude, opposite in direction, and their lines of action are offset by distance l. The moment arm of F1,2 is d1,2, and d1+d2=l. Then the net torque about the pivot point is:

net 1 2 1 2( )d F d F d d F lF The net torque does not depend on the location of the pivot point. Therefore, the

couple exerts the same torque t=lF about any point on the object .

The net torque is the sum of all the torques produced by all the forcesRemember to account for the direction of the tendency for rotation

Counterclockwise torques are positiveClockwise torques are negative

Net Torque

Example 1:

Given:

weights: w1= 500 Nw2 = 800 N

lever arms: r1=4 m r2=2 m

Find:

St = ?

1. Draw all applicable forces

(500 )(4 ) ( )(800 )(2 )

2000 1600

400

N m N m

N m N m

N m

2. Consider CCW rotation to be positive

500 N 800 N

4 m 2 m

Rotation would be CCW

N

Determine the net torque:

Example 2:

Given:

weights: w1= 500 Nw2 = 800 N

lever arms: d1=4 mSt = 0

Find:

d2 = ?

1. Draw all applicable forces and moment arms

2

2 2

(800 )(2 )

(500 )( )

800 2 [ ] 500 [ ] 0 3.2

RHS

LHS

N m

N d m

N m d N m d m

500 N 800 N

2 md2 m

According to our understanding of torque there would be no rotation and no motion!

N’ y

What does it say about acceleration and force?

Thus, according to 2nd Newton’s law SF=0 and a=0! NN

NNNFi

1300'

0)800(')500(

4-2 equilibrium of rigid bodies

• First Condition of Equilibrium• The net external force must be zero

– This is a necessary, but not sufficient, condition to ensure that an object is in complete mechanical equilibrium

– This is a statement of translational equilibrium• Second Condition of Equilibrium

• The net external torque must be zero

• This is a statement of rotational equilibrium

0

0 and 0x y

F

F F

%%%%%%%%%%%%%%

0

If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque

A- 6 KgB -2 KgC – 30 KgD – 1 Kg

example

(Θ = 90 ⁰)

• When an object is in rotational equilibrium, the net torque applied to it is zero.

• Rotational equilibrium is often used to determine unknown forces.

• What are the forces (FA, FB) holding the bridge up at either end?

ExampleGiven M = 120 kg.Neglect the mass of the beam.

a) Find the tension in the cable

b) What is the force between the beam and the wall

Solutiona) Given: M = 120 kg, L = 10 m, x = 7 m

Find: T

0formula Basic

0 MgxTL

Solve for T =824 N

Axis

Solutionb) Given: M = 120 kg, L = 10 m, x = 7 m, T = 824 N Find f

Solve for f = 353 N

f

0formula Basic

yF

0 fMgT

Alternative Solutionb) Given: M = 120 kg, L = 10 m, x = 7 m Find f

Solve for f = 353 N

f

Axis

0)( fLxLMg

0formula Basic

Another ExampleGiven: W=50 N, L=0.35 m,

x=0.03 mFind the tension in the muscle

0formula Basic

0 WLFx

F = 583 N

x

L

W

ExampleConsider the 400-kg beam shown below. Find TR

SolutionFind TR

to solve for TR, choose axis here

X = 2 m

L = 7 mTR = 1 121 N

0formula Basic

0 LTMgX R