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?

We need to find the additional force (both magnitude and direction) that must be applied to the body so that its net acceleration becomes zero, meaning the body remains in translational equilibrium.

1. Find the Net Existing Force

From the free body diagram ,let's break down the given forces along the horizontal ($$x$$) and vertical ($$y$$) axes:

• Along the $$x$$-axis:

  $$\Sigma F_x = 5\text{ N} - 6\text{ N} = -1\text{ N}$$

  (This means there is a net force of $$1\text{ N}$$ pointing to the left).
• Along the $$y$$-axis:

  $$\Sigma F_y = 7\text{ N} - 8\text{ N} = -1\text{ N}$$

  (This means there is a net force of $$1\text{ N}$$ pointing downward).

Expressing the net existing force vector ($$\vec{F}_{\text{net}}$$):

$$\vec{F}_{\text{net}} = -1\hat{i} - 1\hat{j}$$

2. Determine the Balancing Force Vector ($$\vec{F}_{\text{add}}$$)

For the net acceleration of the body to be zero, the total vector sum of all forces must be zero:

$$\vec{F}_{\text{net}} + \vec{F}_{\text{add}} = 0 \implies \vec{F}_{\text{add}} = -\vec{F}_{\text{net}}$$

$$\vec{F}_{\text{add}} = -(-1\hat{i} - 1\hat{j}) = 1\hat{i} + 1\hat{j}$$

3. Calculate Magnitude and Direction

• Magnitude:

  $$|\vec{F}_{\text{add}}| = \sqrt{(1)^2 + (1)^2} = \sqrt{2}\text{ N}$$

• Direction ($\theta$ with the positive $$x$$-axis):

  $$\tan \theta = \frac{F_y}{F_x} = \frac{1}{1} = 1$$

  Since both the $$x$$ and $$y$$ components of our additional force are positive, the vector lies in the first quadrant:

  $$\theta = \tan^{-1}(1) = 45^\circ$$

Conclusion

The additional force that must be applied is $$\sqrt{2}\text{ N}$$ at an angle of $$45^\circ$$ with the positive $$x$$-axis (Option A).