Answer:
192.6N
Step-by-step explanation:
Let's consider the forces acting on the beam:
Weight of the beam (W): It acts vertically downward and has a magnitude of W = mass * gravitational acceleration = 39.4 kg * 9.8 m/s^2.
Force exerted by the link on the beam (F_link): It acts at an angle of 90 degrees with respect to the beam and has two components: the vertical component and the horizontal component.
Tension in the cable (T): It supports the far end of the beam and acts at an angle of 90 degrees with respect to the beam. Since the angle between the beam and the cable is 90 degrees, the tension in the cable only has a vertical component.
Let's break down the forces acting on the beam:
Vertical forces:
W (weight of the beam) - T (vertical component of tension) = 0
T = W
Horizontal forces:
F_link (horizontal component of the force exerted by the link) = ?
To find the magnitude of the horizontal component of the force exerted by the link on the beam (F_link), we need to consider the equilibrium of forces in the horizontal direction.
Since the beam is inclined at an angle of θ = 33.1 degrees with respect to the horizontal, the horizontal equilibrium equation can be written as:
F_link = W * sin(θ)
Let's substitute the given values:
W = 39.4 kg * 9.8 m/s^2
θ = 33.1 degrees
F_link ≈ (39.4 kg * 9.8 m/s^2) * sin(33.1 degrees)
Using a calculator, we find that the magnitude of the horizontal component of the force exerted by the link on the beam (F_link) is approximately 192.6 N.