Question

In the arrangement shown in the figure, friction exists only between the two blocks, $A$ and $B$ . The coefficient of static friction ${\mu }_s=0.6$ and coefficient of kinetic friction ${\mu }_k=0.4$ , the masses of the blocks $A$ and $B$ are $m_1=20kg$ and $m_2=30kg$ , respectively. Find the acceleration (in $m{s}^{-2}$ ) of $m_1$ , if a force $F=150N$ is applied, as shown in the figure. [Assume that string and pulleys are massless]

Answer 1

Answer: 1.5

Let us assume that the two blocks move together, without slipping, relative to each other. The acceleration of the system in that case is
$a=\frac{Fcos60^{\circ }}{m_1+m_2}$
$a=\frac{150\times 1/2}{20+30}=1.5m{s}^{-2}$
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
$T-f=m_{1}a$
$150-f=20\times 1.5=30$
$f=120N$
$f_{max}=0.6\times 200=120N$
Since $f\le f_{max}$ , our assumption about the two blocks moving together is correct and hence the acceleration of the blocks is $1.5m{s}^{-2}$