A 1210 N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A W = 1820 N crate hangs from the far end of the beam. The angle between the beam and the horizontal is 30 degrees upwards and the angle between the horizontal and the cable attached to the wall is 50 degrees upward.
(a) Calculate the magnitude of the tension in the wire.
(b) Calculate the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam.
Fx =
Fy =
To solve this problem, we can break down the forces acting on the beam and use Newton's laws of motion to calculate the desired quantities.
Let's start by drawing a free-body diagram of the beam, which will help us visualize the forces involved.
(a) Magnitude of the tension in the wire:
From the free-body diagram, we can see that there are two vertical forces acting on the beam - the weight of the crate and the tension in the wire. Since the beam is in equilibrium, the vertical forces must balance each other out.
The weight of the crate, W = 1820 N, acts as a downward force. To find the tension in the wire, we need to calculate the vertical component of the tension.
Using trigonometry, we can determine the vertical component of the tension:
Vertical component of tension = T * sin(angle between horizontal and cable attached to the wall)
T * sin(50°) = W
T = W / sin(50°)
Substituting the given values:
T = 1820 N / sin(50°)
T ≈ 2329.7 N
Therefore, the magnitude of the tension in the wire is approximately 2329.7 N.
(b) Magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam:
From the free-body diagram, we know that the beam has two horizontal forces acting on it - the horizontal component of the tension in the wire and the horizontal component of the force exerted by the wall. Since the beam is in equilibrium, the vertical forces must balance each other out.
To find the horizontal component of the force exerted by the wall, we need to calculate the horizontal component of the tension in the wire.
Using trigonometry, we can determine the horizontal component of the tension:
Horizontal component of tension = T * cos(angle between horizontal and cable attached to the wall)
T * cos(50°) = Fx
Fx = T * cos(50°)
Substituting the given values:
Fx = 2329.7 N * cos(50°)
Fx ≈ 1495.5 N
Therefore, the magnitude of the horizontal component of the force that the wall exerts on the left end of the beam is approximately 1495.5 N.
Since the beam is in equilibrium, the sum of the vertical forces acting on it must be zero. We can use this fact to calculate the vertical component of the force exerted by the wall:
Vertical component of the force exerted by the wall = Weight of the beam + vertical component of tension
The weight of the beam can be calculated by multiplying its mass by the acceleration due to gravity, which is approximately 9.8 m/s². The mass of the beam can be found using its weight and the acceleration due to gravity:
Weight of the beam = mass of the beam * g
Substituting back to the equation:
Vertical component of the force exerted by the wall = mass of the beam * g + T * sin(50°)
Given that the weight of the beam is equal to its mass times the acceleration due to gravity (W_beam = m_beam * g = 1210 N):
Vertical component of the force exerted by the wall = 1210 N + 2329.7 N * sin(50°)
Vertical component of the force exerted by the wall ≈ 2526.6 N
Therefore, the magnitude of the vertical component of the force that the wall exerts on the left end of the beam is approximately 2526.6 N.
To summarize:
(a) The magnitude of the tension in the wire is approximately 2329.7 N.
(b) The magnitude of the horizontal component of the force that the wall exerts on the left end of the beam is approximately 1495.5 N.
The magnitude of the vertical component of the force that the wall exerts on the left end of the beam is approximately 2526.6 N.