Question

A board of mass $m=11\, kg$ is placed on the floor and a man of mass $M=70\, kg$ 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:
$f=2 T$ ...(i)
$T=N=(M+m) g$ ...(ii)
$N=(M+m) g-T$ ...(iii)
If board is not sliding $f \leq \mu N$
$2 T \leq \mu[(M+m) g-T]$
$[2+\mu] T \leq \mu(M+m) g ; T \leq \frac{\mu(M+m) g}{(2+\mu)}$
Hence, $T_{\max }=\frac{\mu(M+m) g}{(2+\mu)}$
After substituting the values, we get
$T_{\max }=\frac{0.25(70+11)}{(2+0.25)} \times 100$
$=\frac{0.25 \times 81 \times 10}{2.25}=90\, N$