28. A 20 kg beam is supported by a cable as shown in the diagram below.
a. Find the tension in the cable supporting the beam.
b. Find the magnitude of the force exerted by the wall on the beam.

Question

Accepted Answer

∻Solution∻ ‖ a ⋮ To find the tension in the cable, we first analyze the forces acting on the beam. The weight of the beam is given by $$W = mg = 20 \, ext{kg} imes 9.81 \, ext{m/s}^2 = 196.2 \, ext{N}$$ acting downwards. The beam is in static equilibrium, so the sum of the vertical forces must equal zero. The vertical component of the tension can be expressed as $$F_{t_y} = F_t \sin(45^\circ)$$. Therefore, we have the equation: $$F_t \sin(45^\circ) = W$$. Substituting the values, we get: $$F_t \cdot \frac{\sqrt{2}}{2} = 196.2 \, ext{N}$$. Solving for $$F_t$$ gives us $$F_t = \frac{196.2 \, ext{N}}{\frac{\sqrt{2}}{2}} = 196.2 \cdot \frac{2}{\sqrt{2}} = 196.2 \sqrt{2} \approx 277.1 \, ext{N}$$. ‖ ‖ b ⋮ To find the force exerted by the wall on the beam, we need to consider the horizontal forces. The horizontal component of the tension is given by $$F_{t_x} = F_t \cos(45^\circ)$$. Since the beam is in equilibrium, the force exerted by the wall, $$F_w$$, must balance this horizontal component: $$F_w = F_{t_x} = F_t \cos(45^\circ) = 196.2 \sqrt{2} \cdot \frac{\sqrt{2}}{2} = 196.2 \cdot \frac{2}{2} = 196.2 \, ext{N}$$. ‖ ∻Answer∻ ⚹ Tension in the cable: approximately 277.1 N; Force exerted by the wall: 196.2 N ⚹ ∻Key Concept∻ ⚹ Static Equilibrium: In static equilibrium, the sum of all forces and the sum of all torques acting on an object must be zero. This leads to equations that can be solved to find unknown forces. ⚹ ∻Explanation∻ ⚹ By applying the conditions of static equilibrium, we can determine the tension in the cable and the force exerted by the wall on the beam. The calculations show how the forces balance out in both vertical and horizontal directions.⚹

Sia