Question

A small metallic sphere of mass $m$ is suspended from the ceiling of a car accelerating on a horizontal road with constant acceleration $a$. The tension in the string attached with metallic sphere is:
(A) $mg$
(B) $m\left( {g + a} \right)$
(C) $m\left( {g - a} \right)$
(D) $m\sqrt {{g^2} + {a^2}} $

Answer 1

Hint: Tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable. On squaring the vertical and horizontal components of tension, we can find the answer.
Formula Used: The formulae used in the solution are given here.
For any angle $\theta $, it can be said,
${\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete Step by Step Solution: Tension can be defined as an action-reaction pair of forces acting at each end of the said elements. While considering a rope, the tension force is felt by every section of the rope in both the directions, apart from the endpoints. The endpoints experience tension on one side and the force from the weight attached. Throughout the string, the tension varies in some circumstances.
It has been given that a small metallic sphere of mass $m$ is suspended from the ceiling of a car accelerating on a horizontal road with constant acceleration $a$.
Since there is movement of the car, the string will accelerate thereby making some angles. Let $\theta $ be the angle made by the string while moving from its original position to displaced position. On resolving the tension $T$, when it makes sine and cosine components.
The vertical component of the tension $T$ is given by, $T\cos \theta = mg$ where $g$ is the acceleration due to gravity.
The horizontal component of the tension $T$ is given by, $T\sin \theta = ma$.
We know that, for any angle $\theta $, it can be said,
${\sin ^2}\theta + {\cos ^2}\theta = 1$
Thus, on squaring the vertical and horizontal components of tension, we get,
${\left( {T\cos \theta } \right)^2} + {\left( {T\sin \theta } \right)^2} = {\left( {mg} \right)^2} + {\left( {ma} \right)^2}$.
Simplifying the equation further, ${T^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = {m^2}\left( {{g^2} + {a^2}} \right)$
Now, since ${\sin ^2}\theta + {\cos ^2}\theta = 1$, we have,
${T^2} = {m^2}\left( {{g^2} + {a^2}} \right)$
$ \Rightarrow T = m\sqrt {{g^2} + {a^2}} $

Hence, the correct answer is Option D.

Note: Tension is a pulling force and not a pushing force as ropes can’t push effectively. Attempting to push the rope will cause the rope to go slack losing the tension it possesses. This might sound obvious but while illustrating the forces acting on an object, people often draw the force of tension going in the wrong direction. Hence it is important to remember that tension can only pull on an object but not push against it.