A uniform horizontal beam 5.00 m long and weighting 2.88 102 N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal (Figure (a)). If a person weighing 5.99 102 N stands 1.40 m from the wall, find the magnitude of the tension vector T in the cable and the force vector R exerted by the wall on the beam.

Question

A uniform horizontal beam 5.00 m long and weighting 2.88 102 N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal (Figure (a)). If a person weighing 5.99 102 N stands 1.40 m from the wall, find the magnitude of the tension vector T in the cable and the force vector R exerted by the wall on the beam.

Answer 1

To find the magnitude of the tension vector T in the cable and the force vector R exerted by the wall on the beam, we can use the principles of equilibrium.

Let's start by resolving the forces acting on the beam in the horizontal and vertical directions.

In the horizontal direction:
- There is no horizontal force acting on the beam.

In the vertical direction:
- The weight of the beam (2.88 * 10^2 N) acts downward.
- The tension in the cable (T) acts upward.
- The vertical component of the force exerted by the person (5.99 * 10^2 N) acts downward.
- The vertical component of the force exerted by the wall (R) acts upward.

Now, let's write the equations of equilibrium:

In the horizontal direction:
ΣF_horizontal = 0
0 = T*cos(53.0°) - R*cos(180°) -- (1)

In the vertical direction:
ΣF_vertical = 0
0 = T*sin(53.0°) + R*sin(180°) - 2.88 * 10^2 N - 5.99 * 10^2 N*sin(θ) -- (2)

Using these equations, we can solve for the unknowns T and R.

Substituting the values, we have:
- For equation (1): 0 = T*cos(53.0°) - R*(-1)

Simplifying equation (1), we get:
T*cos(53.0°) = R -- (3)

- For equation (2): 0 = T*sin(53.0°) + R*sin(180°) - 2.88 * 10^2 N - 5.99 * 10^2 N*sin(θ)

Since sin(180°) = 0, equation (2) becomes:
0 = T*sin(53.0°) - 2.88 * 10^2 N - 5.99 * 10^2 N*sin(θ)

Now, let's substitute equation (3) into equation (2):
0 = T*sin(53.0°) - 2.88 * 10^2 N - 5.99 * 10^2 N*sin(θ)
0 = T*sin(53.0°) - 2.88 * 10^2 N - 5.99 * 10^2 N*T*cos(53.0°)

Simplifying further, we have:
2.88 * 10^2 N = T*(sin(53.0°) - 5.99 * 10^2 N*cos(53.0°))

Now, we can solve for T:
T = (2.88 * 10^2 N) / (sin(53.0°) - 5.99 * 10^2 N*cos(53.0°))

Calculating this will give you the magnitude of the tension vector T in the cable.

To find the force vector R exerted by the wall on the beam, we can substitute the value of T we just found into equation (3):
R = T*cos(53.0°)

Simplifying this equation will give you the magnitude of the force vector R exerted by the wall on the beam.