Question

Three coplanar equilibrium forces \( P = 40\text{ N} \), \( Q = x\text{ N} \) and \( R = y\text{ N} \) are acting on a body and the body is said to be in equilibrium when the angle included between each other is \( 120^\circ \), then \( x = \)

• A. \( 20\text{ N} \)
• B. \( 40\text{ N} \)
• C. \( 60\text{ N} \)
• D. \( 80\text{ N} \)

Hint:

When three concurrent coplanar forces maintain static equilibrium and have completely identical angles of \( 120^\circ \) separating them, the force system is perfectly symmetrical. This symmetry means all three force magnitudes must be exactly equal to each other! Thus, \( P = Q = R = 40\text{ N} \) instantly.

Answer 1

The Correct Option is B

Step 1: Set up the three forces at their actual directions.
Since the three forces are \( 120^\circ \) apart all around, place \( P = 40\text{ N} \) along \( 0^\circ \), \( Q = x\text{ N} \) along \( 120^\circ \), and \( R = y\text{ N} \) along \( 240^\circ \), measured from the positive x-axis.

Step 2: Write the vertical equilibrium equation.
For the body to stay in equilibrium, the vertical components must cancel: \[ \sum F_y = 0: \quad P\sin(0^\circ) + Q\sin(120^\circ) + R\sin(240^\circ) = 0 \] \[ 0 + Q\left(\frac{\sqrt{3}}{2}\right) + R\left(-\frac{\sqrt{3}}{2}\right) = 0 \quad \Rightarrow \quad Q = R \]
Step 3: Write the horizontal equilibrium equation.
\[ \sum F_x = 0: \quad P\cos(0^\circ) + Q\cos(120^\circ) + R\cos(240^\circ) = 0 \] \[ P + Q\left(-\frac{1}{2}\right) + R\left(-\frac{1}{2}\right) = 0 \quad \Rightarrow \quad P = \frac{Q+R}{2} \] Since \( Q = R \) from Step 2, this becomes \( P = Q \), so \[ x = Q = P = \boxed{40\text{ N}} \]
which matches option (B). The symmetry also makes physical sense: when three forces are spread exactly \( 120^\circ \) apart and still balance, they must all be equal in size.