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Torque
Introduction
Lever Arm or
Moment Arm |
 The
perpendicular distance from the axis of rotation to the line along which
the force acts.
It is  in the diagram
to the left.
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Tau,  |
Product of the Force times the Lever
arm.
= F
The torque can also be calculated using the component of the force
perpendicular to the radius,  .
= r
If we look at both drawings, we find the equation can be simplified
as
= rF Sin
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Rotational equivalent of Newton's First Law states that a
freely rotating body will continue to rotate with constant angular velocity as
long as no net torque acts to change that motion.
The unit for torque is m . N. Although
it has the same units as Work and Energy, it is not equivalent. This is
why we write Work and Energy as N . m. 1 N . m is
equivalent to 1 Joule, J. However, torque is never expressed as
Joules.
Table of Contents
Problems
| 1. |
 Calculate
the net torque about the axle of the wheel shown to the left. Assume
that a friction torque of 0.40 m . N opposes the motion.
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| 2. |
The coefficient of static friction between
the tires and pavement is 0.75. Calculate the minimum torque that
must be applied to the 70.0 cm diameter tire of a 1500. kg car in order to
"lay rubber". Assume each wheel supports an equal share of
the weight.
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| 3. |
A person exerts a force of 33.3 N on the end
of a door 85.0 cm wide. What is the magnitude of the torque if the
force is exerted (a) perpendicular to the door, and (b) at a 60o
angle to the face of the door?
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| 4. |
 Calculate
the torque around the axis of rotation and label whether it is clockwise
or counter-clockwise. |
Table of Contents
Answers
| 1. |
Calculate
the net torque about the axle of the wheel shown to the left. Assume
that a friction torque of 0.40 m . N opposes the motion.
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All of the forces are perpendicular to the
axle. The angle between where the force is applied to the axis of
rotation is not a factor. It is important to label the torque as
clockwise and counter-closewise. The total torque is the sum of the
torques.
 1
= r  = 0.300 m x 20.0
N = 6.00 m . N
Clockwise r
=
2
= r = 0.300 m x 25.0
N = 7.50 m . N
Counter-clockwise
1
= r = 0.100 m x 35.0
N = 3.50 m . N
Clockwise
= 6.00 m
. N + 3.50 m . N - 7.50 m . N
+ 0.40 m . N = 2.40 m . N
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| 2. |
The coefficient of static friction between
the tires and pavement is 0.75. Calculate the minimum torque that
must be applied to the 70.0 cm diameter tire of a 1500. kg car in order to
"lay rubber". Assume each wheel supports an equal share of
the weight.
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Fw = 1500. kg x 9.80 m s
-2 = 14,700 N
FN per tire = 14,700 N
/ 4 = 3680 N
Ffr =  FN
= 0.75 x 3680 N = 2760 N
Ffr is parallel to the road which
is perpendicular to the axis of rotation.
= r = 0.350 m x 2760
N = 966 m . N
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| 3. |
A person exerts a force of 33.3 N on the end
of a door 85.0 cm wide. What is the magnitude of the torque if the
force is exerted (a) perpendicular to the door, and (b) at a 60o
angle to the face of the door?
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(a)
= r = 0.850 m x 33.3
N = 28.3 m . N
 (b)
= r F Sin =
0.850 m x 33.3 N x Sin 60o = 24.5 m . N
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| 4. |
Calculate
the torque around the axis of rotation and label whether it is clockwise
or counter-clockwise.
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1
= r = 0.550 m x 50.0
N = 27.5 m . N
Clockwise
= r F Sin =
0.420 m x 35.0 N x Sin 20.0o = 5.03 m . N
Clockwise
= r F Sin =
0.320 m x 30.0 N x Sin 45.0o = 6.79 m . N
Clockwise
= r F Sin =
0.750 m x 40.0 N x Sin 35.0o = 17.2 m . N
Clockwise
= 27.5 m
. N + 5.03 m . N - 6.79 m . N
+ 17.2 m . N = 56.5 m . N
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Table of Contents
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