Question

\( F_A, F_B, \) and \( F_C \) are three forces acting at point \( P \) as shown in the figure. The whole system is in equilibrium state. The magnitude of \( F_A \) is:

• A. 100 N
• B. 83.3 N
• C. 73.3 N
• D. 89.6 N

Hint:

In problems involving equilibrium, always resolve the forces into components and apply the equilibrium conditions (both horizontal and vertical) to solve for unknown forces.

Answer 1

The Correct Option is C

In this problem, three forces \( F_A \), \( F_B \), and \( F_C \) are acting on point \( P \) in such a way that the system is in equilibrium. In equilibrium, the vector sum of all forces must be zero, meaning both the horizontal and vertical components of the forces must balance.
Given the directions and angles, we can resolve the forces into horizontal and vertical components.
Step 1: Resolve the forces \( F_A \) and \( F_B \) into components.
- \( F_A \) has an angle of 30° with the horizontal, so its components are: - Horizontal component: \( F_A \cos(30^\circ) \) - Vertical component: \( F_A \sin(30^\circ) \) - \( F_B \) has an angle of 45° with the horizontal, so its components are: - Horizontal component: \( F_B \cos(45^\circ) \) - Vertical component: \( F_B \sin(45^\circ) \) - \( F_C \) is acting vertically downward with a magnitude of 100 N, so its components are: - Horizontal component: 0 N (since it is vertically downward), - Vertical component: \( F_C = 100 \, \text{N} \) Step 2: Apply the equilibrium conditions.
For horizontal equilibrium, the sum of the horizontal components of the forces must be zero: \[ F_A \cos(30^\circ) + F_B \cos(45^\circ) = 0 \] For vertical equilibrium, the sum of the vertical components of the forces must be zero: \[ F_A \sin(30^\circ) + F_B \sin(45^\circ) - F_C = 0 \] Step 3: Solve the system of equations.
From the horizontal equilibrium equation: \[ F_A \cos(30^\circ) = - F_B \cos(45^\circ) \] \[ F_A \times \frac{\sqrt{3}}{2} = - F_B \times \frac{\sqrt{2}}{2} \] Simplifying: \[ F_A = F_B \times \frac{\sqrt{2}}{\sqrt{3}} = F_B \times 0.8165 \] Now, using the vertical equilibrium equation: \[ F_A \sin(30^\circ) + F_B \sin(45^\circ) = F_C \] Substituting the known values: \[ F_A \times \frac{1}{2} + F_B \times \frac{\sqrt{2}}{2} = 100 \] Substitute \( F_A = 0.8165 F_B \) into the equation: \[ 0.8165 F_B \times \frac{1}{2} + F_B \times \frac{\sqrt{2}}{2} = 100 \] Simplifying: \[ 0.40825 F_B + 0.7071 F_B = 100 \] \[ 1.11535 F_B = 100 \] Solving for \( F_B \): \[ F_B = \frac{100}{1.11535} = 89.7 \, \text{N} \] Now, substituting \( F_B = 89.7 \, \text{N} \) into the equation for \( F_A \): \[ F_A = 0.8165 \times 89.7 = 73.3 \, \text{N} \] Thus, the magnitude of \( F_A \) is \( 73.3 \, \text{N} \).