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

To solve this problem, we will determine the net force acting on the body when the force \(F_1\) is removed, and then use that net force to find the acceleration.

1. According to the problem, the body is initially in equilibrium under the action of three forces \(F_1\), \(F_2\), and \(F_3\). Therefore, these forces balance each other:

\(\mathbf{F_1 + F_2 + F_3 = 0}\)

2. It is given that \(F_2 = 6 \, \text{N}\) and \(F_3 = 8 \, \text{N}\), and they act perpendicularly to each other. Let's assume \(F_2\) is acting along the x-axis and \(F_3\) is acting along the y-axis.
3. Since the body is in equilibrium initially, the resultant of \(F_2\) and \(F_3\) is balanced by \(F_1\), that is:

\(\mathbf{F_1 = -(F_2 + F_3)}\)

4. Now, the magnitude of the resultant of \(F_2\) and \(F_3\) can be found using the Pythagorean theorem because they are perpendicular:

\(R = \sqrt{F_2^2 + F_3^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{N}\)

5. Once \(F_1\) is removed, the net force on the body is \(R = 10 \, \text{N}\).
6. Using Newton's second law, we can find the acceleration of the body:

\(\mathbf{F = ma}\)

\(10 = 5 \times a\)

\(a = \frac{10}{5} = 2 \, \text{m/s}^2\)

7. Thus, the acceleration of the body when \(F_1\) is removed is \(2 \, \text{m/s}^2\).

Therefore, the correct answer is 2 m/s².