Name: Conservation of Linear Momentum
Uploaded: 2019-07-19
Duration: 3 min 5 s
Description: In their study of natural phenomena over many centuries, scientists have identified certain fundamental laws. One such law is the law of conservation of mome...

Question

Consider the situation shown in figure given below. Both the pulleys and the string are light and all the surfaces are frictionless.

(a) Find the acceleration of the mass $M$ .

(b) Find the tension in the string.

(c) Calculate the force exerted by the clamp on the pulley $A$ in the figure.

Accepted Answer

The pulleys are assumed to be massless and since $F = m a$ , the net force on the pulley $B$ needs to be zero, because its acceleration cannot be zero, since it accelerates and moves as per the motion of the system of the blocks.

Let the tension in the string connected to the block of mass $2 M$ be $T .$ Since the net force on the pulley should be zero and the strings are ideal and massless, the tension in each part of the string connecting the pulley $B$ on its right will be $T' = \frac{T}{2}$ . Hence, the tension in the string connected to the block of mass $M$ will also be $\frac{T}{2} .$

(a) Let the block of mass $M$ move a distance $x'$ downwards when the block of mass $2 M$ moves a distance $x$ rightwards. Clearly $x' = 2 x \Rightarrow a' = 2 a$ , since acceleration is the second-order derivative of displacement.

Hence, the block of mass $M$ moves with an acceleration twice as that of the block of mass $2 M .$

Writing the force equation for the block of mass $2 M,$ we get,

$T = 2 M a \dots (1)$

Also, for the block of mass $M$ ,

$M g - \frac{T}{2} = M (a') \dots (2)$

Using $a' = 2 a$ and solving these two equations simultaneously, we get,

$a = \frac{g}{3}, a' = \frac{2}{3} g$ ,

which are the respective rightward and downward accelerations of the blocks of masses $2 M$ and $M .$

(b) Putting the value $a = \frac{g}{3}$ in the equation $(1)$ , we get, $T = \frac{2}{3} M g$ .

(c) Let the force on the pulley by the clamp be $F .$ The other forces on the pulley are the two tensions $\frac{T}{2}$ each in the horizontal and the vertical part of the same rope that goes around it. The pulley being in rest, the two tensions must balance the force on the pulley by the clamp.

Hence, $F = \sqrt{({(\frac{M g}{3})}^{2} + {(\frac{M g}{3})}^{2}) + (2 \times (\frac{M g}{3}) (\frac{M g}{3}) \cos 90 °)} = \frac{\sqrt{2}}{3} M g$ .

Consider the situation shown in figure given below. Both the pulleys and the string are light and all the surfaces are frictionless.

(a) Find the acceleration of the mass $M$ .

(b) Find the tension in the string.

(c) Calculate the force exerted by the clamp on the pulley $A$ in the figure.

Important Questions on Newton's Laws of Motion

Learn and Prepare for any exam you want !

Consider the situation shown in figure given below. Both the pulleys and the string are light and all the surfaces are frictionless. (a) Find the acceleration of the mass M. (b) Find the tension in the string. (c) Calculate the force exerted by the clamp on the pulley A in the figure.

Important Questions on Newton's Laws of Motion

Learn and Prepare for any exam you want !

Consider the situation shown in figure given below. Both the pulleys and the string are light and all the surfaces are frictionless.

(a) Find the acceleration of the mass $M$ .

(b) Find the tension in the string.

(c) Calculate the force exerted by the clamp on the pulley $A$ in the figure.