Chapter 4 : statics
4-1 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:
t  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.
| t | r F sin 
where  is the angle between the vectors r and F.
Cross Product Direction
A

A
B
• Curl right-hand fingers in
direction of 
• Right-hand thumb points in
direction of cross-product
• Not commutative
B
AB = –(BA)
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
Lever arm length (m)
Torque (N.m)
t=rxF
Force (N)
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:
t net  d1F  d2 F  (d1  d2 )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.
Net Torque
The net torque is the sum of all the torques produced by all the forces
Remember to account for the direction of the tendency for rotation
Counterclockwise torques are positive
Clockwise torques are negative
Example 1:
N
Determine the net torque:
4m
2m
Given:
weights: w1= 500 N
w2 = 800 N
lever arms: r1=4 m
r2=2 m
500 N
800 N
1. Draw all applicable forces
2. Consider CCW rotation to be positive
Find:
St = ?
t  (500 N )(4 m)  ()(800 N )(2 m)
 2000 N  m  1600 N  m
 400 N  m
Rotation would be CCW
Example 2:
N’
d2 m
y
2m
Given:
weights: w1= 500 N
w2 = 800 N
lever arms: d1=4 m
St = 0
Find:
d2 = ?
500 N
800 N
1. Draw all applicable forces and moment arms
t
t
RHS
  (800 N )(2 m)
LHS
 (500 N )(d 2 m)
800  2 [ N  m]  500  d 2 [ N  m]  0
 d 2  3.2 m

According to our understanding of torque there
would be no rotation and no motion!
What does it say about acceleration and force?
 F (500 N )  N '(800 N )  0
i
Thus, according to 2nd Newton’s law SF=0 and a=0!
N '  1300 N
4-2 equilibrium of rigid bodies
• First Condition of Equilibrium
• The net external force must be zero
SF  0
SFx  0 and SFy  0
– 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
St  0
• This is a statement of rotational equilibrium
If the object is in equilibrium, it does not matter where
you put the axis of rotation for calculating the net
torque
example
A- 6 Kg
B -2 Kg
C – 30 Kg
D – 1 Kg
(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?
Example
Given 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
Solution
a) Given: M = 120 kg, L = 10 m, x = 7 m
Find: T
Basic formula
t  0
Axis
TL  Mgx  0
Solve for T
= 824 N
Solution
b) Given: M = 120 kg, L = 10 m, x = 7 m, T = 824 N
Find f
Basic formula
F
y
0
T  Mg  f  0
Solve for f
= 353 N
f
Alternative Solution
b) Given: M = 120 kg, L = 10 m, x = 7 m
Find f
Basic formula
t  0
Axis
Mg ( L  x)  fL  0
Solve for f
= 353 N
f
Another Example
Given: W=50 N, L=0.35 m,
x=0.03 m
Find the tension in the muscle
W
x
Basic formula
t  0
L
Fx WL  0
F = 583 N
Example
Consider the 400-kg beam
shown below. Find TR
Solution
Find TR
Basic formula
t  0
to solve for TR, choose axis here
 MgX  TR L  0
TR = 1 121 N
X=2m
L=7m