Question

In the figure shown, the minimum force $F$ applied perpendicular to the incline so that the block does not slide is equal to $50\, X$ newton. Find $X$. (The block is always in contact with the incline)

Answer 1

$f_{s}=10 \,g \sin 37^{\circ}$
$f_{s}=10 \times 10 \times \frac{3}{5}=60\, N$
$N=F-10\, g \cos 37^{\circ}$
$N=F-80$
$f_{s} \leq \mu_{s} N$
$60 \leq 0.5(F-80)$
$120 \leq F-80$
$F \geq 200$
$F_{\min }=200=50 \,X$
$\therefore X=4$