Question

A balloon and a mass are attached to a rod that is pivoted at P.

Answer 1

Answer:

The correct option is;

B Move both the balloon and mass 10 cm to the right

Explanation:

Given that the system is in equilibrium, we have;

Force of balloon = $F_b$↑

Force of mass = $F_m$ ↓

The direction of the balloon is having an upward motion which gives a clockwise moment or motion to the rod while the direction of the force of the mass weight is downwards, giving the rod an anticlockwise moment

for the rod to rotate clockwise, the moment of the balloon should be larger than that of the rod

At the present equilibrium we have;

$F_b$ × 30 =  $F_m$ × 20

Therefore;

$F_m$ = 1.5×$F_b$

Moving both balloon and mass 10 cm to the right gives;

The moment of the balloon = $F_b$ × (30 - 10)  = $F_b$ × 20 = 20×$F_b$,

The moment of the mass =  $F_m$ × (20 - 10) =  $F_m$ × 10

When we substitute  $F_m$ = 1.5×$F_b$ in the moment equation for the mass, we have;

The moment of the mass = $F_m$ × 10 = 1.5×$F_b$ ×10 = 15×$F_b$

Therefore, the balloon now has a larger momentum than that of the mass and the rod will rotate clockwise.