Question

A roller of mass $300kg$ and of radius $50cm$ lying on the horizontal floor is resting against a step of height $20cm$. The minimum horizontal force to be applied on the roller passing through its centre to turn the roller on to the step is:
(A) $950N$
(B) $1960N$
(C) $2940N$
(D) $3920N$

Answer 1

Hint: To solve this problem, we need to equate the values of the torque. Then from the values given in the question and substituting them in the equation we will get the values of the force.
Formulas used: We will be using the formulas to balance the existing forces on the roller, $\tau = F \times {r_ \bot }$ , where $\tau $ is the torque experienced by a point, $F$ is the force experienced by the object to produce the torque, and ${r_ \bot }$ is the distance perpendicular to the force applied.

Complete Step by Step answer:
We know that a body that is spherical in shape when applied a force will travel in a linear and rolling motion simultaneously. When the body is performing rolling motion it experiences a torque. Also, we know that the body experiences motion only when the forces acting on it are not completely balanced.
Considering the above situation,

We can see that the edge of the step is around $30cm$ from the center of the roller. Also a force $F$ is being applied on the roller to help it role through the step. Also the highest point on the step that stands at $20cm$ from the circumference of the sphere experiences a torque. The other force acting on the roller in this rotational frame is the gravitational force. The body experiences motion in the desired direction only when the torque is either balanced or exceeded.
The torque due to force $F$ will be $\tau = F \times 30$ . Similarly the torque due to the gravitational force of the body will be, $\tau = mg \times \sqrt {{{\left( {50} \right)}^2} - {{\left( {30} \right)}^2}} $ Since the forces are required to either be balanced or exceeding to experience motion,
$F \times 30 = mg \times \sqrt {{{\left( {50} \right)}^2} - {{\left( {30} \right)}^2}} $
$F \times 30 = mg\left( {40} \right)$
Substituting the values of $m = 300kg$ and $g = 9.8m/{s^2}$ , $F \times 30 = 300 \times 9.8 \times \left( {40} \right)$
Solving for $F$ we get,
$F = \dfrac{{300 \times 9.8 \times \left( {40} \right)}}{{30}}$
$ \Rightarrow F = 3920N$
Thus, the minimum force required to move the roller through the step will be $F = 3920N$ .

Hence the correct answer is option D.

Note: We have used $\tau = mg \times \sqrt {{{\left( {50} \right)}^2} - {{\left( {30} \right)}^2}} $ above because we know the torque is product of distance and the force acting perpendicular to it. The force $mg$ acts perpendicular to, the right-angled leg of the triangular formed by joining the two vectors.