Explanation

(c) [image:5_image264.emf]
Here, mass of the block, $m = 200\ kg$
Coefficient of static friction, $\mu_{s} = 0.5 = \frac{1}{2}$
Angle of incline plane, $\theta = 45{^\circ}$
Maximum force that each man can apply, $F = 500\ N$
Let $N$ number of man are required for the block to just start moving up the plane $NF = mg\sin\theta + f$
$= mg\sin\theta + \mu_{s}R$
$= mg\sin\theta + \mu_{s}\ mg\cos\theta = mg\ \lbrack\sin\theta + \mu_{s}\cos\theta\rbrack$
$= 200 \times 10\left\lbrack \sin{45{^\circ} + \frac{1}{2}\cos{45{^\circ}}} \right\rbrack$
$= 200 \times 10\left\lbrack \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} \right\rbrack = \frac{200 \times 10 \times 3}{2\sqrt{2}}$
$\therefore N = \frac{200 \times 10 \times 3}{2\sqrt{2} \times 500} = 5$