In the arrangement shown in the figure, friction exists only between the two blocks, A and B . The coefficient of static friction μ s = 0.6 and coefficient of kinetic friction μ k = 0.4 , the masses o

Question

In the arrangement shown in the figure, friction exists only between the two blocks, $A$ and $B$ . The coefficient of static friction $μ_{s} = 0.6$ and coefficient of kinetic friction $μ_{k} = 0.4$ , the masses of the blocks $A$ and $B$ are $m_{1} = 20 kg$ and $m_{2} = 30 kg$ , respectively. Find the acceleration (in $m s^{- 2}$ ) of $m_{1}$ , if a force $F = 150 N$ is applied, as shown in the figure. [Assume that string and pulleys are massless]

TestHub · Accepted Answer

Let us assume that the two blocks move together, without slipping, relative to each other. The acceleration of the system in that case is

$a = \frac{F \cos 60 °}{m_{1} + m_{2}}$

$a = \frac{150 \times \frac{1}{2}}{20 + 30} = 1.5 m s^{- 2}$

In this case, if the frictional force acting between the two blocks is $f$ , then writing the Newton's second law of motion, for the block of mass $m_{1,}$ we get

$T - f = m_{1} a$

$150 - f = 20 \times 1.5 = 30$

$f = 120 N$

$f_{\max} = 0.6 \times 200 = 120 N$

Since $f \leq f_{\max}$ , our assumption about the two blocks moving together is correct and hence the acceleration of the blocks is $1.5 m s^{- 2}$

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