Question

A helicopter of mass M carrying a box of mass m at the end of a rope and is moving horizontally with constant acceleration 'a'. The acceleration due to gravity is 'g'. Neglect air resistance. The rope is stretched out from the helicopter at a constant angle $\theta$ to the vertical. What is this angle?

• A. $sin\theta =a/g$
• B. $cos\theta =a/g$
• C. $tan\theta =a/g$
• D. $sin\theta =ma/\left(Mg\right)$

Answer 1

The correct option is C $tan\theta =a/g$

FBD of the box of mass $m$ is shown in the above figure.

Along horizontal direction: $T$ $sin\theta =ma$ ........(1)

Along vertical direction: $T$ $cos\theta =mg$ .......(2)

Dividing (1) and (2) we get $tan\theta =\frac{a}{g}$