Test Index

BITSAT 2016 Paper

Section: Physics

Question : 2 of 150

Marks: +1, -0

Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in figure.

The wedge is inclined at 45° to the horizontal on both the sides. If the coefficient of friction between the block A and he wedge is 2/3 and that between the block B and the wedge is 1/3 and both the blocks A and B are released from rest, the acceleration of A will be

$-1\,\text{ms}^{-2}$
$1.2\,\text{ms}^{-2}$
$0.2\,\text{ms}^{-2}$
zero

Validate

Solution:

\mu_{AW}= \;\; \frac{2}{3} ; \mu_{BW}= \;\; \frac{1}{3}

Diagram shows the various forces acting on the masses and their resolution in the direction of motion. Let us consider the two masses to be a system. The forces trying to move the system such that A moves upwards and B moves downwards

= \;\; \frac{2mg}{\sqrt{2}} - \;\; \frac{mg}{\sqrt{2}} = \;\; \frac{mg}{\sqrt{2}}

The forces trying to stop this motion (i.e., maximum frictional force)

= f_A + f_B = \mu_A N_A + \mu_B N_B

= \;\; \frac{2}{3} \times \;\; \frac{mg}{\sqrt{2}} + \;\; \frac{1}{3} \times \;\; \frac{2mg}{\sqrt{2}} = \;\; \frac{4}{3} \;\; \frac{mg}{\sqrt{2}}

Since the stopping force is more therefore the mass system will not move

⇒ acceleration of the system is zero.

Go to Question:

Prev Question

Next Question