Question

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

• A. 200Kg
• B. 300Kg
• C. 400Kg
• D. 500Kg

Answer 1

The correct option is C

400Kg

From OAB,
$R^2=x^2+(R-h{)}^2$
$\Rightarrow x^2=R^2-(R-h{)}^2$
$\Rightarrow x^2=(1{)}^2-(1-0.2{)}^2$
$\Rightarrow x=0.6$
As the roller is just ready to turn on the step, the net torque on the roller about the edge of step 'A' is zero.
Torque due to normal force 'N' about A=0.
Torque due to F=Torque due to 'mg'
$\Rightarrow F(R-h)=mg\left(x\right)$
$\Rightarrow m=\frac{F(R-h)}{gx}$
$\Rightarrow m=\frac{\left(2940\right)(1-0.2)}{\left(9.8\right)\left(0.6\right)}=400kg$.