# A man of mass 60 kg is pulling a mass M by an inextensible string passing over a frictionless (massless) pulley. The coefficient of friction between man and the ground is $\mu=\large\frac{1}{2}$. The maximum value of M that can be pulled by the man without slipping on ground is

a) 51 kg

b) 26 kg

c) 46 kg

d) 32 kg

Tension in the rope T
$T=Mg$
N normal reaction between man and ground
$N=60 g-T \sin 60$
$f=\mu N=T \cos 60 ^{\circ}$
$N=600-10 M \large\frac{\sqrt 3}{2}$
$f=\large\frac{1}{2}$$\bigg[600-10 M \large\frac{\sqrt 3}{2} \bigg]$$=M \times 10 \times \large\frac{1}{2}$
Solving we get $M=32.15 kg$
Hence d is the correct answer
edited May 27, 2014