Question

Three forces acting on a body are shown in the figure. To have the resultant force only along the X-direction, the magnitude of minimum additional force needed is

• A. 0.5 N
• B. 1.5 N
• C. $\frac{\sqrt{3}}{4}$ N
• D. $\frac{3\sqrt{3}}{2}$ N

Answer 1

Answer: (D)

(D) $\frac{3\sqrt{3}}{2}$ N

Answer 2

To determine the magnitude of the minimum additional force needed to have the resultant force only along the X-direction, we need to ensure that the net Y-component of all forces (initial three forces plus the additional force) is zero. The minimum additional force will be one that has only a Y-component and no X-component, as any X-component would only increase its magnitude without contributing to cancelling the Y-component.

Let's resolve each of the three given forces into their X and Y components. We will use the convention that angles are measured from the nearest axis, and assign signs based on the quadrant.

1. Force of 1 N:

   • It is in the first quadrant, making an angle of 60° with the positive X-axis.
   • Y-component: $F_{1y} = 1 \sin(60^\circ) = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$ N

2. Force of 4 N:

   • It is in the second quadrant, making an angle of 30° with the positive Y-axis.
   • Y-component: $F_{4y} = 4 \cos(30^\circ) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}$ N
   • (Its X-component would be $-4 \sin(30^\circ)$)

3. Force of 2 N:

   • It is in the fourth quadrant, making an angle of 30° with the negative Y-axis.
   • Y-component: $F_{2y} = -2 \cos(30^\circ) = -2 \times \frac{\sqrt{3}}{2} = -\sqrt{3}$ N
   • (Its X-component would be $2 \sin(30^\circ)$)

Now, let's find the resultant Y-component ($R_y$) of these three forces:

$R_y = F_{1y} + F_{4y} + F_{2y}$

$R_y = \frac{\sqrt{3}}{2} + 2\sqrt{3} - \sqrt{3}$

$R_y = \frac{\sqrt{3}}{2} + \sqrt{3}$

To sum these, find a common denominator:

$R_y = \frac{\sqrt{3}}{2} + \frac{2\sqrt{3}}{2} = \frac{\sqrt{3} + 2\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$ N

To have the resultant force only along the X-direction, the total Y-component must be zero. Let the additional force be $\vec{F}_{add}$. Its Y-component must cancel out $R_y$.

So, $F_{add,y} = -R_y = -\frac{3\sqrt{3}}{2}$ N.

The magnitude of the minimum additional force needed is the magnitude of this Y-component, as its X-component must be zero for the force to be minimum.

Magnitude of minimum additional force $= |-\frac{3\sqrt{3}}{2}|$ N $= \frac{3\sqrt{3}}{2}$ N.