Question

A roller of radius $1m$ is on horizontal floor in contact with the edge of a step of height $20cm$. A horizontal force $2940N$ is applied on the roller passing through its centre by which the roller is just ready to turn on to the step. The mass of the roller is:

• A. $300kg$
• B. $400kg$
• C. $600kg$
• D. $800kg$

Answer 1

The correct option is B $400kg$

The applied force exerts a clockwise torque about point of contact P with the obstacle.

The weight exerts an anti-clockwise torque about the point of contact P with the obstacle.

Given,

Radius $R=OB=1m$

Step height $h=AB=0.2m$
Torque due to applied force $F\left(2940N\right)$ is
${\tau }_F=2940\times OA=2940\times (1-0.2)=2940\times 0.8$
Torque due to weight: ${\tau }_{weight}=mg\times AP$
But, $AP=\sqrt{O{P}^2-O{A}^2}=\sqrt{1^2-{0.8}^2}=0.6$
${\tau }_F$ has to just exceed the ${\tau }_{weight}$.
Equating both torques,
$2940\times 0.8=mg\times 0.6$
$\Rightarrow m=\frac{2940\times 0.8}{9.8\times 0.6}=400kg$