Question

Two blocks each having mass M, rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless, and M on the incline is allowed to move down, then tension in the string will be:

A.$\dfrac{2}{3}Mg\sin \theta$
B.$\dfrac{3}{2}Mg\sin \theta$
C.$\dfrac{{Mg\sin \theta }}{2}$
D.$2Mg\sin \theta$

Answer 1

Draw the free-body diagram of the blocks. Apply Newton’s second law of motion to both the blocks in the horizontal direction. These equations give the relation between the tension in the string, mass of the block, angle of inclination and acceleration due to gravity.

Formula used:
The equation for Newton’s second law of motion is
${F_{net}} = ma$ - (Eq 1)
Here, ${F_{net}}$ is the net force on the object, $m$ is the mass of the object and $a$ is the acceleration of the object.

Complete step by step answer:

Two blocks of each mass $M$ rest on a frictionless surface.

Draw the free body diagram of the blocks.

In the above free-body diagram of the blocks, $\theta$ is the angle of inclination of the inclined plane, $Mg$ is the weight of the blocks and $T$ is the tension in the string. The directions of X and Y axes for the forces on both the blocks are shown in the diagram.

Apply Newton’s second law of motion to the block on inclined plane in horizontal direction.
$Mg\sin \theta - T = Ma$ - (Eq 2)

Apply Newton’s second law of motion to the block on horizontal plane in horizontal direction.
$T = Ma$

Substitute $Ma$ for $T$ in equation (2).
$Mg\sin \theta - Ma = Ma$
$\Rightarrow Mg\sin \theta = 2Ma$
$\Rightarrow a = \dfrac{{g\sin \theta }}{2}$

Substitute $\dfrac{{g\sin \theta }}{2}$ for $a$ in equation (2).
$Mg\sin \theta - T = M\dfrac{{g\sin \theta }}{2}$

Rearrange the above equation for the tension $T$ in the string.
$T = Mg\sin \theta - \dfrac{{Mg\sin \theta }}{2}$
$\Rightarrow T = \dfrac{{Mg\sin \theta }}{2}$

Therefore, the tension in the string is $\dfrac{{Mg\sin \theta }}{2}$.

**Hence, the correct option is C.
**
Note: There is no need to apply Newton’s law in the vertical direction as the required tension in the string is along the horizontal direction.