Question

Three forces \(F_1, F_2\) and \(F_3\) acting on a body of mass \(m\) keep the body stationary. If the forces \(F_1\) and \(F_2\) are mutually perpendicular, the acceleration of the body when the force \(F_3\) is removed is

• A. \(\frac{F_3}{m}\)
• B. \(\frac{F_1F_2}{m}\)
• C. \(\frac{(F_1 - F_2)}{m}\)
• D. \(\frac{F_1}{m}\)
• E. \(\frac{F_2}{m}\)

Hint:

If forces balance initially, removing one gives net force equal to that force.

Answer 1

The Correct Option is A

Concept: For equilibrium: \[ \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0 \]

Step 1: Rearrange equation.
\[ \vec{F}_3 = -(\vec{F}_1 + \vec{F}_2) \]

Step 2: After removing \(F_3\).
Net force: \[ \vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = -\vec{F}_3 \]

Step 3: Acceleration.
\[ a = \frac{F_{net}}{m} = \frac{F_3}{m} \]