OpenStax-CNX module: m14404
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Torque about a point (check your
understanding)
∗
Sunil Kumar Singh
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 2.0
Abstract
Objective questions, contained in this module with hidden solutions, help improve understanding of
the topics covered under the module "Torque about a point".
The questions have been selected to enhance understanding of the topics covered in the module titled "
Torque
1
". All questions are multiple choice questions with one or more correct answers.
1 Understanding level (Torque about a point)
Exercise 1
(Solution on p. 6.)
A force, F, acts on a particle at a linear distance r from the origin of a coordinate system. If
force acts in coordinate plane yz, then the toque on the particle is :
∗ Version
1.3: Aug 30, 2009 2:09 am -0500
† http://creativecommons.org/licenses/by/2.0/
1 "Torque" <http://cnx.org/content/m14402/latest/>
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Torque about the origin
Figure 1:
The force acts in yz plane.
(a)
makes an angle θ with z axis
(b)
makes an angle θ with x axis
(c)
acts along x axis
(d)
acts along -x axis
Exercise 2
(Solution on p. 6.)
The torque on a particle at a position on x-axis (other than origin) is zero. If the force applied is
not zero, then force is acting :
(a)
(b)
Exercise 3
either in y or z direction
in x direction
±
(c)
in
(d)
- z direction
x direction
(Solution on p. 6.)
The bob of a pendulum of length "L" is raised to one side and released to oscillate about the mid
point. The torque about the point of suspension, at an instant when the bob makes an angle θ with respect to vertical, is :
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Pendulum
Figure 2:
The pendulum bob makes an angle θ with the vertical.
(a) mgLsinθ (b) mgLcosθ (c) mgLtanθ (d) mgLcotθ
Exercise 4
A force F = (2i + 2j 3k)
k) meters. If the particle is
(Solution on p. 7.)
Newton acts on a particle, placed at a point given by
r
= (i +
j
constrained to rotate about x-axis along a circular path, then the
magnitude of torque about the axis is :
(a) 1 (b) 2 (c) 3 (d) 4
Exercise 5
(Solution on p. 9.)
In the gure, three forces of 10 N each act on a triangular plate as shown in the gure. If C be
the center of mass of the plate, then torque, in N-m, about it is :
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Torque about center of mass
Figure 3:
Three forces are acting.
(a) 0.61 (b) − 1.33 (c) − 0.91 (d) − 1.11
Exercise 6
(Solution on p. 9.)
A force, F, acts on a particle lying in xz plane at a linear distance r from the origin of a
coordinate system. If the direction of force "F" is parallel to y-axis, then the toque on the particle
:
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Torque about the origin
Figure 4:
The particle lies in xz plane.
(a)
makes an angle θ with z axis
(b)
makes an angle θ with x - axis
(c)
acts along x axis
(d)
acts along -x axis
2 Answers
1. (c)
6. (b)
2. (c)
3. (a)
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4. (a)
5. (c)
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Solutions to Exercises in this Module
Solution to Exercise (p. 1)
In this case, vectors r and F
are in yz plane. As torque is perpendicular to this plane, it is directed
either along x or -x axis. Applying right hand rule by shifting force vector at the origin, we see that the
torque acts in x-direction.
Hence, option (c) is correct.
Solution to Exercise (p. 2)
The torque on the particle is given as :
τ = rxF
Its magnitude is given by :
τ = rF sinθ
The torque can be zero for following conditions (i) force is zero (ii) particle is at the origin and (iii) sine
of enclosed angle is zero. Since "r" and "F" are non-zero, it follows that :
θ = 0
Thus, the force on the particle is acting
±
◦
or 180
◦
x direction.
Hence, option (c) is correct.
Solution to Exercise (p. 2)
There are two forces that operate on the pendulum bob (see the left gure):
•
•
Weight of the bob (mg)
Tension in the string (T)
The line of action of tension in the string of the pendulum is along the string itself, which passes through
center of mass. It means that torque is applied by only the weight of the bob. Since the linear distance of
the weight acting on the bob from the point of suspension is given, it would be easier to nd magnitude of
torque as product of linear distance and component of force perpendicular to it.
Pendulum
(a)
Figure 5:
(b)
(a) Two forces act on the pendulum bob. (b) The component of weight perpendicular to
string
The component of weight perpendicular to the string is (see the right gure) :
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F⊥ = mgsinθ
The torque due to the weight is :
τ = LF⊥ = mgLsinθ
Hence, option (a) is correct.
Note :
The torque about the suspension point can also be determined, using moment arm. In this case
moment arm i.e. perpendicular distance between suspension point and line of action of force is "AB" :
Pendulum
Figure 6:
The pendulum bob makes an angle θ with the vertical.
τ = L⊥ F = AB x F = Lsinθ x mg = mgLsinθ
Solution to Exercise (p. 3)
The torque about the origin of coordinate system is given by :
τ = rxF
τ =
| i j k |
| 1 1 -1 |
| 2 2 -3 |
⇒
τ
[ ( −1x2) −
=
[ (1x − 3) − (2x − 1) ]
i
+
(1x − 3) ] j+ [ (1x2} − (1x2) ] k ⇒ τ = ( −3 + 2) i + ( −2 +
The particle, here, moves about the axis of rotation in x-direction. In this case, the particle is restrained
to rotate or move about any other axis. Thus, torque in rotation about x-axis is equal to the x- component
of torque about the origin. Now, the vector x-component of torque is :
τx = − i
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The magnitude of torque about x-axis for rotation is :
τx = 1 N m
Hence, option (a) is correct.
Note :
We can check the result, considering individual force components. We rst identify A (1,1) in
xy-plane, then we nd the position of the particle, B(1,1,-1) by moving -1 in negative z-direction as in the
gure below.
Torque on the particle
Figure 7:
The position of particle along with components of force
The component in x-direction does not constitute a torque about x-axis as the included angle is zero.
The component of force in y-direction is Fy = 2 N at a perpendicular distance z = 1 m. For determining
torque about the axis, we multiply perpendicular distance with force. We apply appropriate sign, depending
on the sense of rotation about the axis. The torque about x-axis due to component of force in y - direction
is anticlockwise and hence is positive :
τxy = zFy = 1 x 2 = 2 N m
The component of force in z-direction is Fz = 3 N at a perpendicular distance y = 1 m.
The torque
about x-axis due to the component of force in z - direction is clockwise and hence is negative :
τxz = − yFz = − 1 x 3 = − 3 N m
Since both torques are acting along same axis i.e.
x-axis, we can obtain net torque about x-axis by
arithmetic sum :
⇒ τx = τxy + τxz = 2 − 3 = − 1 N m
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The magnitude of net torque about x-axis is :
⇒ τx = = 1 N m
Solution to Exercise (p. 3)
We note here that line of action of the force acting on the horizontal face passes through center of mass.
Thus, it does not constitute a torque about the center of mass.
Further, we see that the linear distances of the point of action of the forces are given.
Therefore, it
would be easy to apply the formula for the magnitude of torque, which makes use of perpendicular force
component.
Now, the perpendicular force component at B is :
FB⊥ = 10sin45
◦
=
10
√
2
N
Similarly, the perpendicular force component at D is :
FD⊥ = 10sin30
◦
= 5N
The torque due to force at B about center of mass, C, is clockwise and hence is negative :
τCB = − 0.2 x
10
√
2
= −
√
2 Nm
Torque due to force at D about center of mass, C, is anticlockwise and hence is positive :
⇒ τCD = 0.1 x 5 = 0.5 N m
The torques are along the same direction i.e. perpendicular to the surface of the plate. Therefore, we
can nd the net torque as algebraic sum :
⇒ τC = −
√
2 + 0.5 = − 0.91 N m
Hence, option (c) is correct.
Solution to Exercise (p. 4)
First thing that we note here is that the plane formed by operands is not same as any of the coordinate
planes. Now, the torque on the particle is given as :
τ = rxF
From the relation, we come to know that the vector product τ " is perpendicular to the plane formed by
the vectors r and F.
Let us imagine force vector is moved to origin keeping the direction same. Applying Right hand cross
product rule, we move from vector r from its position to that of F such that the curl of ngers is along
the direction of movement. The direction of thumb downward tells us that torque is acting perpendicular to
plane and below the surface i.e. making an acute angle with x-axis.
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Torque about the origin
Figure 8:
Direction of torque
We observe here that torque τ and position vector r are perpendicular to each other. On the other
hand, z and x axes are perpendicular to each other. Since the angle between perpendiculars on two lines
is same, we conclude that torque acts in xz plane, making an angle θ with the x-axis as shown in the gure
above.
Hence, option (b) is correct.
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