Question

Two masses A and B of 10 Kg and 5 Kg respectively are connected with a string passing over a frictionless pulley at a corner of a table a shown in the adjoining diagram. The coefficient of friction of A with the table is 0.2. The minimum mass of C that may be placed on A to prevent it from moving is equal to?

Answer 1

Given that,

Mass of A $m_A=10kg$

Mass of B $m_B=5kg$

Coefficient of friction $\mu =0.2$

Now, the tension is

$T=m_{b}g$

$T=5\times 10$

$T=50N$

Now, we consider the combination of A and C

Then,

$T=\mu R$

$T=0.2(m_A+m_C)g$

$50=0.2\times 10\times 10+0.2m_C\times 10$

$50-20=2m_c$

$m_c=15kg$

Hence, the minimum mass of $C$ is $15kg$