Question

A board of mass $m=11kg$ is placed on the floor and a man of mass $M=70kg$ is standing on the board as shown (Case-I). The coefficient of friction between the board and the floor is $\mu =0.25$. The maximum force that the man can exert on the rope so that the board does not slip on the floor is

• A. 125 N
• B. 90 N
• C. 162 N
• D. None of these

Answer 1

Answer: (B)

For man + board system:
$f=2T$ ...(i)
$T=N=(M+m)g$ ...(ii)
$N=(M+m)g-T$ ...(iii)
If board is not sliding $f\le \mu N$
$2T\le \mu [(M+m)g-T]$
$[2+\mu ]T\le \mu (M+m)g;T\le \frac{\mu (M+m)g}{(2+\mu )}$
Hence, $T_{\max }=\frac{\mu (M+m)g}{(2+\mu )}$
After substituting the values, we get
$T_{\max }=\frac{0.25(70+11)}{(2+0.25)}\times 100$
$=\frac{0.25\times 81\times 10}{2.25}=90N$