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.

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.

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

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.

Figure 3

Pendulum
(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 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" :

Figure 4: The pendulum bob makes an angle “θ” with the vertical.

Pendulum

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

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.

Figure 5: The position of particle along with components of force

Torque on the particle

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

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.