Question

A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration $a$, directed horizontally as shown in figure. The other end of the string is being pulled with constant acceleration $a$ (relative to car) vertically downward. The tension in the string is equal to (assume $\theta$ remains constant)

• A. $m\sqrt{g^2+a^2}$
• B. $m\sqrt{g^2+a^2}-ma$
• C. $m\sqrt{g^2+a^2}+ma$
• D. $m(g+a)$

Answer 1

The correct option is C $m\sqrt{g^2+a^2}+ma$

Let us draw a free body diagram of the body, in the frame of the car.

Writing force equation in a direction along the string, we have
$T-ma\sin \theta -mg\cos \theta =ma....\left(1\right)$
and, along the direction, perpendicular to the string
$ma\cos \theta =mg\sin \theta$
$\Rightarrow \tan \theta =\frac{a}{g}$
or, $\sin \theta =\frac{a}{\sqrt{a^2+g^2}},\text{and}\cos \theta =\frac{g}{\sqrt{a^2+g^2}}$

Putting the obtained value of $\sin \theta$ and $\cos \theta$ in eq $\left(1\right)$ we get
$T-m\sqrt{a^2+g^2}=ma$
$\Rightarrow T=ma+m\sqrt{a^2+g^2}$

Hence, $\left(C\right)$ is the correct answer.