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

Answer: (A)

2 m/s$^{2}$

Answer 2

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