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 :
Torque about the origin 

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 xdirection.
Hence, option (c) is correct.
Exercise 2
The torque on a particle at a position on xaxis (other than origin) is zero. If the force applied is not zero, then force is acting :
Solution
The torque on the particle is given as :
Its magnitude is given by :
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 nonzero, it follows that :
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 :
Pendulum 

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.
Pendulum  


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

Exercise 4
A force F = (2i + 2j – 3k) Newton acts on a particle, placed at a point given by r = (i + j – k) meters. If the particle is constrained to rotate about xaxis along a circular path, then the magnitude of torque about the axis is :
Solution
The torque about the origin of coordinate system is given by :
τ =  i j k 
 1 1 1 
 2 2 3 
The particle, here, moves about the axis of rotation in xdirection. In this case, the particle is restrained to rotate or move about any other axis. Thus, torque in rotation about xaxis is equal to the x component of torque about the origin. Now, the vector xcomponent of torque is :
The magnitude of torque about xaxis for rotation is :
Hence, option (a) is correct.
Note : We can check the result, considering individual force components. We first identify A (1,1) in xyplane, then we find the position of the particle, B(1,1,1) by moving “1” in negative zdirection as in the figure below.
Torque on the particle 

The component in xdirection does not constitute a torque about xaxis as the included angle is zero.
The component of force in ydirection 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 xaxis due to component of force in y  direction is anticlockwise and hence is positive :
The component of force in zdirection is Fz = 3 N at a perpendicular distance y = 1 m. The torque about xaxis due to the component of force in z  direction is clockwise and hence is negative :
Since both torques are acting along same axis i.e. xaxis, we can obtain net torque about xaxis by arithmetic sum :
The magnitude of net torque about xaxis is :
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 Nm, about it is :
Torque about center of mass 

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 :
Similarly, the perpendicular force component at “D” is :
The torque due to force at “B” about center of mass, “C”, is clockwise and hence is negative :
Torque due to force at “D” about center of mass, “C”, is anticlockwise and hence is positive :
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 :
Hence, option (c) is correct.