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Course by: Sunil Kumar Singh. E-mail the author

Module by: Sunil Kumar Singh. E-mail the author

Summary: 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 ". All questions are multiple choice questions with one or more correct answers.

## Understanding level (Torque about a point)

### Exercise 1

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 :

(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 (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

#### Solution

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.

### Exercise 2

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) either in “y” or “z” direction (b) in x direction (c) in ± x direction (d) - z direction (a) either in “y” or “z” direction (b) in x direction (c) in ± x direction (d) - z direction

#### Solution

The torque on the particle is given as :

τ = r x F τ = r x F

Its magnitude is given by :

τ = r F sin θ τ = r F sin θ

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 ° or 180 ° θ = 0 ° or 180 °

Thus, the force on the particle is acting ± x direction.

Hence, option (c) is correct.

### Exercise 3

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 :

(a) m g L sin θ (b) m g L cos θ (c) m g L tan θ (d) m g L cot θ (a) m g L sin θ (b) m g L cos θ (c) m g L tan θ (d) m g L cot θ

#### Solution

There are two forces that operate on the pendulum bob (see the left figure):

• 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 find magnitude of torque as product of linear distance and component of force perpendicular to it.

The component of weight perpendicular to the string is (see the right figure) :

F = m g sin θ F = m g sin θ

The torque due to the weight is :

τ = L F = m g L sin θ τ = L F = m g L sin θ

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" :

τ = L F = AB x F = L sin θ x m g = m g L sin θ τ = L F = AB x F = L sin θ x m g = m g L sin θ

### Exercise 4

A force F = (2i + 2j – 3k) Newton acts on a particle, placed at a point given by r = (i + jk) meters. If the particle is 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 (a) 1 (b) 2 (c) 3 (d) 4

#### Solution

The torque about the origin of coordinate system is given by :

τ = r x F τ = r x F


τ =   | i	j	 k |
| 1	1	-1 |
| 2	2	-3 |


τ = [ ( 1 x - 3 ) - ( 2 x - 1 ) ] i + [ ( - 1 x 2 ) - ( 1 x - 3 ) ] j + [ ( 1 x 2 } - ( 1 x 2 ) ] k τ = ( - 3 + 2 ) i + ( - 2 + 3 ) j τ = - i + j τ = [ ( 1 x - 3 ) - ( 2 x - 1 ) ] i + [ ( - 1 x 2 ) - ( 1 x - 3 ) ] j + [ ( 1 x 2 } - ( 1 x 2 ) ] k τ = ( - 3 + 2 ) i + ( - 2 + 3 ) j τ = - i + j

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 :

τ x = - i τ x = - i

The magnitude of torque about x-axis for rotation is :

τ x = 1 N m τ x = 1 N m

Hence, option (a) is correct.

Note : We can check the result, considering individual force components. We first identify A (1,1) in xy-plane, then we find the position of the particle, B(1,1,-1) by moving “-1” in negative z-direction as in the figure below.

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 :

τ xy = z F y = 1 x 2 = 2 N m τ xy = z F y = 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 :

τ xz = - y F z = - 1 x 3 = - 3 N m τ xz = - y F z = - 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 :

τ x = τ xy + τ xz = 2 - 3 = - 1 N m τ x = τ xy + τ xz = 2 - 3 = - 1 N m

The magnitude of net torque about x-axis is :

τ x = = 1 N m τ x = = 1 N m

### Exercise 5

In the figure, three forces of 10 N each act on a triangular plate as shown in the figure. If “C” be the center of mass of the plate, then torque, in N-m, about it is :

(a) 0.61 (b) - 1.33 (c) - 0.91 (d) - 1.11 (a) 0.61 (b) - 1.33 (c) - 0.91 (d) - 1.11

#### Solution

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 :

F B⊥ = 10 sin 45 ° = 10 2 N F B⊥ = 10 sin 45 ° = 10 2 N

Similarly, the perpendicular force component at “D” is :

F D⊥ = 10 sin 30 ° = 5 N F D⊥ = 10 sin 30 ° = 5 N

The torque due to force at “B” about center of mass, “C”, is clockwise and hence is negative :

τ CB = - 0.2 x 10 2 = - 2 N m τ CB = - 0.2 x 10 2 = - 2 N m

Torque due to force at “D” about center of mass, “C”, is anticlockwise and hence is positive :

τ CD = 0.1 x 5 = 0.5 N m τ CD = 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 find the net torque as algebraic sum :

τ C = - 2 + 0.5 = - 0.91 N m τ C = - 2 + 0.5 = - 0.91 N m

Hence, option (c) is correct.

## Exercise 6

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 :

(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 (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

### Solution

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 = r x F τ = r x F

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 fingers 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.

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 figure above.

Hence, option (b) is correct.


1. (c)   2. (c)    3. (a)    4. (a)    5.  (c)
6. (b)


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