Question

A vertical disc of mass 5 $kg$ and radius $50cm$ rests against a step of height $25cm$ as shown in figure. What minimum horizontal force applied perpendicular to the axle will make the disc to climb the step? Take $g=$ $10m/s^2$

• A. 50 N
• B. $50\sqrt{3}N$
• C. 25 N
• D. none of these

Answer 1

Answer: (B)

$y=50-25=25cm$
$x=\sqrt{50^2-25^2}=25\sqrt{3}$
$Fy=5gx$
$\Rightarrow (25)=5\times 10(25\sqrt{3})$
$\Rightarrow F=50\sqrt{3}N$