Question

A body of mass \( M \) is at equilibrium under the action of four forces \( F_1, F_2, F_3 \), and \( F_4 \). If \( F_1 \) is removed from the body, then the body moves with an acceleration of:

• A. \( \frac{F_1}{M} \)
• B. \( \frac{F_1 + F_2}{2M} \)
• C. \( \frac{F_1 + F_3}{2M} \)
• D. \( \frac{F_1 + F_4}{M} \)
• E. \( \frac{F_4}{M} \)

Hint:

For an object in equilibrium, the sum of all the forces acting on it is zero. When one of the forces is removed, the remaining net force determines the acceleration of the body.

Answer 1

The Correct Option is A

Since the body is initially at equilibrium under the action of the four forces, the net force on the body is zero. 
This gives us the equation for equilibrium: \[ F_1 + F_2 + F_3 + F_4 = 0 \] When \( F_1 \) is removed, the remaining forces acting on the body are \( F_2, F_3, \) and \( F_4 \). 
The net force on the body is now: \[ F_{{net}} = F_2 + F_3 + F_4 \] From the equilibrium condition, we know that \( F_1 = -(F_2 + F_3 + F_4) \).
Therefore, when \( F_1 \) is removed, the remaining net force is: \[ F_{{net}} = F_1 \] Thus, the acceleration \( a \) of the body is given by Newton's second law: \[ a = \frac{F_{{net}}}{M} = \frac{F_1}{M} \] 
Therefore, the body moves with an acceleration of \( \frac{F_1}{M} \) after \( F_1 \) is removed.
Thus, the correct answer is: \[ \boxed{\frac{F_1}{M}} \]