We have a light rod that is held horizontally by two vertical strings at points $A$ and $B$, with forces $P$ and $Q$ acting at these points, respectively. There are two additional vertical forces acting on the rod: 21 N at a point $C$ (4 m from $A$) and 25 N at a point $D$ (10 m from $C$ or 6 m from $B$).

To find the magnitudes of forces $P$ and $Q$, we use the following principles:

1. Equilibrium Condition:
   • The rod is in static equilibrium, so the sum of the vertical forces must be zero.
   • The sum of the moments (torques) about any point must also be zero.

Step 1: Vertical Force Equilibrium

The sum of the upward forces ($P$ and $Q$) must equal the sum of the downward forces (21 N and 25 N):

$P + Q = 21 \, \text{N} + 25 \, \text{N} = 46 \, \text{N}$

Step 2: Moment Equilibrium (taking moments about point $A$)

The moment (torque) due to a force is given by $\text{Force} \times \text{Distance from the point of rotation}$.

To eliminate $P$, let's take moments about point $A$:

$\text{Moment of } P = 0 \quad (\text{since it's applied at point } A)$ $\text{Moment of } Q = Q \times 20 \, \text{m} \quad (\text{distance from } A \text{ to } B \text{ is } 20 \, \text{m})$ $\text{Moment of 21 N} = 21 \, \text{N} \times 4 \, \text{m}$ $\text{Moment of 25 N} = 25 \, \text{N} \times 16 \, \text{m} \quad (\text{distance from } A \text{ to } D \text{ is } 16 \, \text{m})$

Since the system is in equilibrium, the sum of the moments must be zero:

$Q \times 20 = 21 \times 4 + 25 \times 16$ $Q \times 20 = 84 + 400 = 484$ $Q = \frac{484}{20} = 24.2 \, \text{N}$

Step 3: Solving for $P$

Now that we know $Q = 24.2 \, \text{N}$, we can substitute this value into the vertical force equilibrium equation:

$P + 24.2 = 46$ $P = 46 - 24.2 = 21.8 \, \text{N}$

Final Answer:

• ( P = 21.8 ,