Question

If a body of mass 5 kg is in equilibrium due to forces \(F_1\), \(F_2\) and \(F_3\). \(F_2\) and\(F_3\) are perpendicular to each other. If \(F_1\) is removed then find the acceleration of the body. Given \(F_2\)=6N and \(F_3\)=8N

• A. 2 m/s\(^{2}\)
• B. 3 m/s\(^{2}\)
• C. 4 m/s\(^{2}\)
• D. 5 m/s\(^{2}\)

Answer 1

The Correct Option is A

Since the body is in equilibrium, the net force acting on it is zero. Hence, we have:
F₁ + F₂ + F₃ = 0
If we remove F₁, the body will no longer be in equilibrium and will experience acceleration due to the remaining forces F₂ and F₃.
The magnitude of the net force acting on the body is:
\(|Fnet| = \sqrt(F2^2 + F3^2) = \sqrt{6^2 + 8^2} = 10N\)
The direction of the net force is given by the angle between F2 and F3:
\(tanθ = \frac{F_3}{F_2} = \frac{8}{6}\)
\(θ = tan^(-1)(\frac{8}{6}) = 53.13 ^{\circ}\)
So, the net force is acting at an angle of 53.13 degrees with F2.
Now, we can use Newton's second law to find the acceleration of the body:
Fnet = ma
\(a = \frac{Fnet}{m} = \frac{10}{5} = 2 \frac{m}{s^2}\)
Therefore, the acceleration of the body is \(2 \frac{m}{s^2}\). 
Answer. A