Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

A cubical block of mass M and edge a slides down a rough inclined plane of
inclination \[\theta\] with uniform velocity. The torque of the normal force on the block about its centre has magnitude
 \[
A)Zero \\
B)Mga \\
C)Mga\sin \theta \\
D)\dfrac{{Mga\sin \theta }}{2} \;
\]

seo-qna
Last updated date: 13th Jun 2024
Total views: 371.1k
Views today: 6.71k
Answer
VerifiedVerified
371.1k+ views
Hint: First, draw a diagram, and show all the forces acting on the block along with their directions. The forces which are acting on the block are the gravitational force and its components, the normal force and the force opposing the motion of the body which is the frictional force. Using these forces and their equations, we can obtain the torque which is the cross product of force and distance.

Complete step by step solution:

seo images


Here, the block has a mass M and it is sliding down the inclined plane having an angle . The block has a side length value a, thus the value of the side length at the centre of the block would be a/2.
Now, the gravitational force on the block is acting downwards towards the ground and it is equal to mg. The two components of the gravitational force are the \[\sin \theta \] and the \[\cos \theta \] components, which are perpendicular to each other. There is also the friction component of the force acting on the block in the direction opposite to the motion of the block.

The normal force acts above the block and according to the Newton’s third law of motion, it is opposed by the cosine function of the gravitational force as
shown in the figure. Now, torque for a body is given by the formula as below:
\[\tau =r\times F\]
Here, F is the force; r is the distance of the line joining the force and \[\tau\] is the torque.
Now, from the diagram, for the motion of the body,
\[f=Mg\sin \theta \]
And hence the torque would be
 \[
\tau =r\times Mg\sin \theta \\
\tau =\dfrac{a}{2}Mg\sin \theta \;
\]
Hence, option (D) is the correct answer.

Notes:Here, even though there is torque acting on the block, the motion of the block is linear towards the bottom end of the inclined plane and not rotational. This is not only because of the sine component of the gravitational force, but also because of the cosine component and the block being in contact with the inclined plane.