First slide

Friction

Question

In the figure, the coefficient of friction between the floor and the block B is 0.1. The coefficient of friction between the blocks B and A is 0.2. The mass of A is m/2 and of B is m. What is the maximum horizontal force F which can be applied to the block B so that two blocks move together?

Moderate

A

$0 .15 mg$
B

$0 .05 mg$
C

$0.1 mg$
D

$0 .45 mg$

Solution

Here, $m_{A} = \frac{m}{2}, m_{B} = m$
$μ_{A} = 0 .2, μ_{B} = 0 .1$
Let both the blocks are moving with common acceleration a. Then,
$a = \frac{μ_{A} m_{A} g}{m_{A}} = μ_{A} g = 0 .2 g$
and $F - μ_{B} (m_{B} + m_{A}) g = (m_{B} + m_{A}) a$
$F = (m_{B} + m_{A}) a + μ_{B} (m_{B} + m_{A}) g$
$= (m + \frac{m}{2}) (0 .2 g) + (0 .1) (m + \frac{m}{2}) g$
$= (\frac{3}{2} m) (0 .2 g) + (\frac{3}{2} m) (0 .1 g) = \frac{0 .9}{2} mg = 0 .45 mg$

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