Question

The mass of the wedges shown in Fig. is M = 4 kg and that of the block is m = 1 kg. The horizontal surface beneath the wedge is smooth while the coefficient of friction between vertical surface of the wedge and block is equal to m= 0.1. Taking $g=9.8m{s}^{-2}$ and assuming pulleys to be massless and frictionless, calculate maximum possible value of force F. upto which the block will remain stationary relative to the wedge.

• A. 20 N
  
• B. 10 N
• C. 50 N
• D. 40 N

Answer 1

Answer: (B)

10 N

$N=Mg+F$ - (i)

$F-N^{\prime }=Ma$ - (ii)

$N^{\prime }=ma$ - (iii)

$F=\mu ma+mg$ - (iv)

From (ii) \& (iii) $F=ma+Ma$

$\therefore a=\frac{F}{m+M}$

$\therefore F=\frac{\mu mF}{m+M}+mg=10N$