Concept:
A twisting or turning force is called torque. Torque
is affected by the force applied, and the length of
the lever arm.
Concept Diagrams:
EXAMPLE 1: The longer the lever arm, then for the same force, the greater the
torque.
Applied Force
Lever Arm Length
Applied Force
Lever Arm Length
Applied Force
Lever Arm Length
EXAMPLE 2: The greater the applied force, then for the same lever arm length,
the greater the torque.
Applied Force
Applied Force
Applied Force
Lever Arm Length
Lever Arm Length
Lever Arm Length
EXAMPLE 3: The larger the gear, then for the same force, the greater the torque.
Lever Arm Length
Applied Force
Lever Arm Length
Applied Force
Lever Arm Length
Applied Force
Example Concept Questions:
1. A mechanic plans to tighten a nut with a socket wrench. If the specifications call for a
very large torque, such as 100 N*m, should he choose a longer wrench or a shorter
one? Why?
Solution:
The mechanic should choose the longer wrench. Since the longer wrench would allow a greater lever
arm length, the torque could be achieved with less applied force. Very small torques might be better
achieved with a smaller wrench, to allow for better control of the torque, and prevent breaking or
stripping nuts, bolts, etc.
2. Mechanic A cannot produce enough torque to loosen a bolt by using a particular
wrench using a maximum effort. Mechanic B suggests placing a length of metal pipe
on the handle of the wrench, extending its length. Will this strategy work? Why or
why not?
Solution:
The solution suggested by mechanic B will be effective. Placing a pipe on the handle of the wrench will
yield a greater lever arm length, thus producing more torque from the same amount of force. The
danger of greatly extending the length of the lever arm is that the greater torque can more easily break
or strip nuts, bolts, etc.
3. For the same magnitude of torque supplied to the axle, will a larger or smaller
diameter wheel exert more force on the street? Why?
Solution:
The smaller tire will exert more force on the road. This means that the car with smaller wheels will be
able to accelerate from rest at a greater rate than a car with larger wheels (car pushes on road, road
pushes on car, causing it to move forward). However, the car with the smaller tires will reach a smaller
maximum speed than the car with the larger wheels.
Equations and Constants for Math Exercises:
 = F x l
g = 9.8 m/s2
Where = torque (N*m), F = force(N), and l = lever arm length(m)
Acceleration due to gravity on Earth
Example Mathematical Questions:
1. A wrench is turned with a force of 125 N applied at a point that is 0.33m from the axis
of rotation (where the bolt, etc., is). What is the torque delivered by the wrench?
Solution:
This exercise asks us to find the torque . To use the equation  = F x l , we will need
the applied force F in Newtons, and the lever arm length l in meters. The force is given
here as 125 N, and the lever arm length is given as 0.33 m. Therefore, the torque is:
 = (125 N)(0.33) = 41.25 N*m
2. A mechanic is tightening a bolt with a wrench is producing 115 N*m of torque. The
distance from his grip to the bolt is 0.2 m long. With what force is the mechanic
turning the wrench?
Solution:
This exercise asks us to find the force F. To use the equation  = F x l , we will first need
to rearrange the equation to solve for F. The rearranged equation is: F = t / l
The applied force F in Newtons is equal to the torque (given as 115 N*m), and the lever
arm length (given as 0.2 m). Therefore, the force is:
F = 115 N*m = 575 N
0.2 m
3. A person with a mass of 75 kg is hanging from one end of a 2 m horizontal flag pole.
How much torque is being delivered to the other end of the pole?
Solution:
This exercise asks us to find the torque . To use the equation  = F x l ,
we will need the applied force F in Newtons, and the lever arm length l in
meters. The force in this case is the person’s weight, which is calculated as
weight x acceleration due to gravity. The lever arm length is given as 2 m.
Therefore, the torque is:
 = (75 kg x 9.8 m/s2)(2 m) = 1470 N*m
Practice Concept Questions:
(Answers are in back of text)
1. A mechanic has a choice between using a 0.15 m long wrench and a 0.25 m long
wrench. Which would produce more torque using the same force? Why?
2. If mechanic A has a 0.20 m long wrench, and mechanic B also has a 0.20 m long wrench,
how might mechanic B apply more torque to a stubborn bolt?
a. Push or pull harder
b. Begin turning the wrench from a vertical position
c. Mechanic B cannot apply more torque
3. A mechanic has a 1 m long pry bar. Assuming that the mechanic is able to push or pull
the bar with the same amount of force, then to apply the most torque using the bar, he
should:
a. Push upward instead of downward
b. “Choke up” on the bar, gripping it closer to the end that is doing the prying
c. Grip the bar at the very end, farthest away from end that is doing the prying
4. A mechanic is attempting to remove a stubborn screw. Should the mechanic choose a
screwdriver with a small-diameter handle, or one with a large-diameter handle? Why?
5. Which delivers the most torque; a single wrench with lever length L pushed or pulled
with force F, or a spider wrench pushed or pulled with force 2F at ½ L ? Why?
Practice Mathematical Questions:
Do the following problems. Include units. (Answers are in back of text)
1. A wrench is turned with a force of 35 N applied at a point that is 0.25m from the axis of
rotation (where the bolt, etc., is). What is the torque produced?
2. A wrench that produces 0.33 N*m of torque is 0.2 m long. With what force is the
wrench turned?
3. A wrench that produces 9 N*m of torque is turned with a force of 6 N. How long is the
wrench?
Concept: If the force applied to a lever arm is not
perpendicular to the lever arm, the torque
generated will be less than if the force was applied
at a right angle to the lever arm.
Equations and Constants:
 = F l
Where = torque (N*m), and F l = force perpendicular to the lever arm
length(m)
Previously, the force applied to a lever arm was always perpendicular to
the lever arm. Now, that will not necessarily be the case.
= 400
l =2m
In the above example, the
force F being applied to a lever arm is not
=
perpendicular to the lever arm, but is being applied at an angle of 40 .
0
The force labeled F’ is the force to be used in the torque equation.
Using trig, F’ can be calculated as Fsinwhich in this case is:
20 N x sin400 = 12.86.
The torque applied to the axis is now:
12.86 N x 2 m = 25.7 N*m