Question

For a free body diagram shown in the figure, the four forces are applied in the ' $x$ ' and ' $y$ ' directions What additional force must be applied and at what angle with positive $x$-axis so that the net acceleration of body is zero?

• A. $\sqrt{2} N , 45^{\circ}$
• B. $\sqrt{2} N , 135^{\circ}$
• C. $\frac{2}{\sqrt{3}} N , 30^{\circ}$
• D. $2 N , 45^{\circ}$

Answer 1

The Correct Option is A

To ensure the net acceleration of the body is zero, the net force acting on the body must also be zero. This means the vector sum of all forces must be zero.

Given forces:

• 5 N along the positive x-axis
• 6 N along the negative x-axis
• 7 N along the positive y-axis
• 8 N along the negative y-axis

Firstly, calculate the resultant force in the x-direction \(F_x\):

\(F_x = 5\, \text{N} - 6\, \text{N} = -1\, \text{N}\)

Secondly, calculate the resultant force in the y-direction \(F_y\):

\(F_y = 7\, \text{N} - 8\, \text{N} = -1\, \text{N}\)

For equilibrium, an additional force \(F_a\) must be applied such that it cancels out the resultant force. Thus:

\(F_{\text{net}} = \sqrt{F_x^2 + F_y^2} = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}\, \text{N}\)

The direction of this additional force, \(\theta\), with the positive x-axis is given by:

\(\theta = \tan^{-1} \left(\frac{F_y}{F_x}\right) = \tan^{-1} \left(\frac{-1}{-1}\right) = 45^\circ\)

Therefore, the additional force required for zero net acceleration is \(\sqrt{2}\, \text{N}\) at an angle of \(45^\circ\) with the positive x-axis.

Correct Answer: \(\sqrt{2}\, \text{N},\, 45^\circ\)