Question

A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hands of a person standing in the car. The car is moving with constant acceleration $a$ directed horizontally as shown. The other end of the string is pulled with constant acceleration $a$ vertically down. Find the tension in the string.

• 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 $m\sqrt{g^2+a^2}+ma$

Using the concept of pseudo force in the frame of the car, we have

Perpendicular to string, we get
$ma\cos \theta =mg\sin \theta \Rightarrow \tan \theta =\frac{a}{g}$
Along the string, we get
$T-mg\cos \theta -ma\sin \theta =ma$
$\Rightarrow T=m\sqrt{g^2+a^2}+ma$