A 1240 N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A W = 1970 N crate hangs from the far end of the beam.
(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 =
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To solve this problem, we need to consider the forces acting on the beam and crate system.
(a) To calculate the magnitude of the tension in the wire, we will use the principle of equilibrium, which states that the sum of all forces acting on an object must be zero. In this case, the beam is in equilibrium, so the sum of the forces acting on the beam must be zero.
The forces acting on the beam are the weight of the beam itself, the weight of the crate hanging from the beam, and the tension in the wire.
The weight of the beam can be calculated using the formula:
Weight of beam = mass of beam * acceleration due to gravity
Given that the beam is uniform, we can assume that the mass is proportional to the length of the beam, and we can treat the entire weight as acting at its center.
Weight of beam = 1240 N
The weight of the crate is given as 1970 N.
Let T be the tension in the wire.
Now, considering the forces in the vertical direction (y-axis), we have:
T - 1240 N - 1970 N = 0
Simplifying the equation:
T = 1240 N + 1970 N
T = 3210 N
Therefore, the magnitude of the tension in the wire is 3210 N.
(b) To calculate the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam, we can use the fact that the beam is in equilibrium.
In the horizontal direction (x-axis), the sum of the forces must be zero since the beam is not accelerating horizontally:
Fx = 0
So, there is no horizontal component of the force exerted by the wall.
In the vertical direction (y-axis), the sum of the forces must be zero to maintain equilibrium:
Fy = 0
Considering the forces acting on the beam, we have the weight of the beam, the weight of the crate, and the vertical component of the tension in the wire. The vertical component of the tension is equal to the tension in the wire since the wire is vertical.
Fy = 1240 N + 1970 N - T
Substituting the value of T (3210 N) we found earlier:
Fy = 1240 N + 1970 N - 3210 N
Fy = 0 N
Therefore, the magnitude of the vertical component of the force that the wall exerts on the left end of the beam is 0 N.