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 in figure. The other end of the string is pulled with constant acceleration $a$ vertically. The tension in the string is equal is

• 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 $m\sqrt{g^2+a^2}+ma$

Free body diagram of the bob can be shown as


As this frame is non-inertial, so to apply Newton's laws of motion we will consider a pseudo force in opposite direction of the acceleration of the frame.
By using force balance on the bob, the net force on the bob comes out to be
$ma=T-m\sqrt{g^2+a^2}$
∴ $T=ma+m\sqrt{g^2+a^2}$