Torque
Physics 6A
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Torque
Torque is what causes angular acceleration
(just like a force causes linear acceleration)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Torque
Torque is what causes angular acceleration
(just like a force causes linear acceleration)
For a torque to be applied to an object, there needs to be
a force that acts at some distance away from a pivot point.
For example, consider tightening a bolt with a wrench.
Which of the 3 forces shown will tighten the bolt?
Pivot Point
FA
FC
FB
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Torque
Torque is what causes angular acceleration
(just like a force causes linear acceleration)
For a torque to be applied to an object, there needs to be
a force that acts at some distance away from a pivot point.
For example, consider tightening a bolt with a wrench.
Which of the 3 forces shown will tighten the bolt?
Pivot Point
FA
Force B will tend to rotate the bolt
clockwise, which will tighten it.
FC
FB
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Torque
Torque is what causes angular acceleration
(just like a force causes linear acceleration)
For a torque to be applied to an object, there needs to be
a force that acts at some distance away from a pivot point.
For example, consider tightening a bolt with a wrench.
Which of the 3 forces shown will tighten the bolt?
Pivot Point
FA
Force B will tend to rotate the bolt
clockwise, which will tighten it.
Notice that force A will tend to rotate the
bolt counter-clockwise, loosening it.
What does force C do?
FC
FB
Prepared by Vince Zaccone
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Torque
Torque is what causes angular acceleration
(just like a force causes linear acceleration)
For a torque to be applied to an object, there needs to be
a force that acts at some distance away from a pivot point.
For example, consider tightening a bolt with a wrench.
Which of the 3 forces shown will tighten the bolt?
Pivot Point
FA
Force B will tend to rotate the bolt
clockwise, which will tighten it.
Notice that force A will tend to rotate the
bolt counter-clockwise, loosening it.
What does force C do?
FC
FB
Force C doesn’t cause any rotation at all –
there is no torque generated by force C.
Why not?
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Torque
Torque is what causes angular acceleration
(just like a force causes linear acceleration)
For a torque to be applied to an object, there needs to be
a force that acts at some distance away from a pivot point.
For example, consider tightening a bolt with a wrench.
Which of the 3 forces shown will tighten the bolt?
Pivot Point
FA
Force B will tend to rotate the bolt
clockwise, which will tighten it.
Notice that force A will tend to rotate the
bolt counter-clockwise, loosening it.
What does force C do?
FC
FB
Force C doesn’t cause any rotation at all –
there is no torque generated by force C.
Why not?
Force C points directly at the pivot point –
no torque is created in this case.
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Torque
Formula for torque:
  F  r  sin()
This is really just the magnitude of the torque. The
angle in the formula is between the force and the radius
(from the pivot point to where the force is applied).
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Torque
Formula for torque:
  F  r  sin()
FA
Pivot Point
This is really just the magnitude of the torque. The
angle in the formula is between the force and the radius
(from the pivot point to where the force is applied).
θ
r
Take a look at the diagram – r and θ are shown for force A.
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Torque
Formula for torque:
  F  r  sin()
FA
Pivot Point
This is really just the magnitude of the torque. The
angle in the formula is between the force and the radius
(from the pivot point to where the force is applied).
θ
r
Take a look at the diagram – r and θ are shown for force A.
There are 2 ways to interpret the formula.
•If you group the Fsin(θ) together, that represents the
component of the force that is perpendicular to the radius. To
get the most torque, the force should be applied perpendicular
(can you see why from the formula?)
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Torque
Lever Arm
Formula for torque:
  F  r  sin()
FA
Pivot Point
This is really just the magnitude of the torque. The
angle in the formula is between the force and the radius
(from the pivot point to where the force is applied).
θ
r
Take a look at the diagram – r and θ are shown for force A.
There are 2 ways to interpret the formula.
•If you group the F sin(θ) together, that represents the
component of the force that is perpendicular to the radius. To
get the most torque, the force should be applied perpendicular
(can you see why from the formula?)
•The other option is to group the r sin(θ) together and call it the
“lever arm” for the force. Think of this as the shortest distance
from the pivot point to where the force is applied. This is the
effective radius of the force. Again, to get maximum torque the
angle should be 90°.
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Example: Find the torque of each force shown with respect
to the pivot point at the left end of the 2m long rod.
F2=30N
F1 is applied at the right end, and F2 is at the center.
120°
50°
F1=20N
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Example: Find the torque of each force shown with respect
to the pivot point at the left end of the 2m long rod.
F2=30N
F1 is applied at the right end, and F2 is at the center.
120°
50°
We can simply use our definition of torque here.
F1=20N
  F  r  sin()
1  (20N)  (2m)  sin(50 )  30.6N  m
2  (30N)  (1m)  sin(120 )  26.0N  m
Notice the sign convention:
Counter-clockwise torque is positive.
Clockwise torque is negative.
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Torque
We mentioned earlier that torques produce angular
accelerations. We have a formula for this relationship:
  I  
This is really just Newton’s 2nd law applied to rotational
motion. The moment of inertia, I, takes the place of the
mass, and we use angular acceleration instead of linear.
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Example: Find the angular acceleration of the 2m long, uniform
rod (mass=5kg) when it is subject to the 2 forces shown.
F2=30N
F1 is applied at the right end, and F2 is at the center.
120°
50°
This is just like the last problem, so we can use the results here.
We need to add up all the torques on the rod, then solve for .
F1=20N
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Example: Find the angular acceleration of the 2m long, uniform
rod (mass=5kg) when it is subject to the 2 forces shown.
F2=30N
F1 is applied at the right end, and F2 is at the center.
120°
50°
This is just like the last problem, so we can use the results here.
We need to add up all the torques on the rod, then solve for .
  I  
1  2  ( 13 ML2 )()
F1=20N
Look up this formula for the moment of
inertia of a rod, with the axis at the end.
 30.6N  m  26.0N  m  ( 13 )  (5kg)  (2m)2  
  .69 rad
s2
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Static Equilibrium
Sometimes an object is subject to several forces, but it does not accelerate.
This is when the object is in equilibrium. We have done problems like this
before, but we neglected the rotational motion. To incorporate this, we
simply need to add a torque formula to our typical force formulas.
Here’s an example:
A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m
from the far end, as shown. It is supported by a hinge at the wall, and a metal
wire running from the wall to the far end.
Find the tension in the wire, and find the horizontal and vertical components of
the force that the hinge exerts on the beam.
60°
1.5m
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A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m
from the far end, as shown. It is supported by a hinge at the wall, and a metal
wire running from the wall to the far end.
Find the tension in the wire, and find the horizontal and vertical components of
the force that the hinge exerts on the beam.
We need to draw a diagram of all the forces, then
write down force and torque equations:
Fx  0
Fy  0
  0
60°
T
Hy
Hx
1.5m
2500N
3500N
T=Tension in wire
Hx and Hy are the components of the
force that the hinge exerts on the beam.
Prepared by Vince Zaccone
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A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m
from the far end, as shown. It is supported by a hinge at the wall, and a metal
wire running from the wall to the far end.
Find the tension in the wire, and find the horizontal and vertical components of
the force that the hinge exerts on the beam.
We need to draw a diagram of all the forces, then
write down force and torque equations:
60°
T
Fx  0
Hy
Hx  Tx  0
Hx  T  cos(30 )  0
30°

Hx
1.5m
This could also be sin(60)
We will save this equation and come back to it later.
2500N
3500N
T=Tension in wire
Hx and Hy are the components of the
force the hinge exerts on the beam.
Prepared by Vince Zaccone
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A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m
from the far end, as shown. It is supported by a hinge at the wall, and a metal
wire running from the wall to the far end.
Find the tension in the wire, and find the horizontal and vertical components of
the force that the hinge exerts on the beam.
We need to draw a diagram of all the forces, then
write down force and torque equations:
60°
T
Fy  0
Hy  Ty  2500N  3500N  0
Hy  T  sin(30 )  2500N  3500N  0
Hy
30°

Hx
1.5m
This could also be cos(60)
We will save this equation and come back to it later.
2500N
3500N
T=Tension in wire
Hx and Hy are the components of the
force the hinge exerts on the beam.
Prepared by Vince Zaccone
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A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m
from the far end, as shown. It is supported by a hinge at the wall, and a metal
wire running from the wall to the far end.
Find the tension in the wire, and find the horizontal and vertical components of
the force that the hinge exerts on the beam.
We need to draw a diagram of all the forces, then
write down force and torque equations:
60°
T
  0
Hy
Before we can fill in the torque equation we need to
choose a pivot point. A convenient choice is where
the hinge attaches to the beam. This simplifies the
torque equation because the 2 unknown hinge forces
will not create any torque about that point.
Also, remember the sign convention – clockwise
torques are negative and counterclockwise positive.
30°
Hx
Pivot point here
  0
 (2500N)  (2m)  (3500N)  (2.5m)  (T)(4m)  sin(30 )  0
T  6875N

1.5m
2500N
3500N
T=Tension in wire
Hx and Hy are the components of the
force the hinge exerts on the beam.
Now we can go back and substitute this value into the
other equations to find the hinge forces.
Hx  T  cos(30 )  0
Hy  T  sin(30 )  2500N  3500N  0
Hx  5954N
Hy  2564N
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