First slide

Two block problem

Question

ln the arrangement shown in figure, $m_{A} = m_{B} = 2 kg .$ String is massless and pulley is frictionless. Block B is resting on a smooth horizontal surface, while friction coefficient between blocks A and B is $μ = 0.5$ The maximum horizontal force F that can be applied so that block.A does not slip over the block B is

Moderate

A

25 N
B

40 N
C

30 N
D

20 N

Solution

Maximum value of friction on A, $f_{\max} = μmg$ = $0.5 \times 2 \times 10 = 10 N$
Hence, the block B is subjected to this frictional force in the forward direction. Since the block A does not slip on B, both of them move with a common acceleration. The acceleration of B is the same as the acceleration of A.
To determine acceleration of 'a', we find that $m_{B} a_{\max}$ = $f_{\max}$
$a_{\max} = \frac{10}{2} = 5 m / s^{2}$
$F_{\max} = (m_{A} + m_{B}) a_{\max} = 4 \times 5 = 20 N$

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