A 1220 N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A W = 1990 N crate hangs from the far end of the beamThe 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 =
please help
a) Is the beam attached to the wall by a hinge? If not, your problem is not statically determinete.
If hinged, set the moment about that hinge equal to zero and solve for the cable tension. Make sure you include the moment due to the weight of the beam, acting at its center of mass.
b) After you have the cable tension, perform vertical and horizontal force balances to compute the two force components at the hinge.
To find the magnitude of the tension in the wire, we need to consider the equilibrium of the beam. The sum of the forces in the horizontal and vertical directions must be zero.
(a) Let's start by analyzing the vertical forces. We can break down the weight of the crate into vertical and horizontal components:
Vertical component: Wv = W * sin(30°) = 1990 N * sin(30°) = 995 N
Horizontal component: Wh = W * cos(30°) = 1990 N * cos(30°) = 1721.54 N
Now, let's consider the vertical forces acting on the beam:
Upward force at the cable: T * sin(50°)
Downward force due to the weight of the beam: Wbeam = 1220 N
Since the beam is in equilibrium, the sum of the upward and downward forces must be zero:
T * sin(50°) - Wbeam = 0
Therefore, we can solve for the tension in the wire:
T * sin(50°) = Wbeam
T = Wbeam / sin(50°)
T = 1220 N / sin(50°)
T ≈ 1642 N
So, the magnitude of the tension in the wire is approximately 1642 N.
(b) To find the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam, we need to consider the horizontal and vertical forces acting on the beam.
Horizontal component:
The horizontal component of the force exerted by the wall on the beam is equal in magnitude and opposite in direction to the horizontal component of the tension in the wire, Wh = 1721.54 N.
Vertical component:
The vertical component of the force exerted by the wall on the beam is equal in magnitude and opposite in direction to the vertical component of the tension in the wire plus the weight of the beam. Since both the wire tension and the weight of the beam act downward, the vertical component of the force exerted by the wall on the beam will be upward.
Vertical component = T * cos(50°) + Wbeam
Vertical component = 1642 N * cos(50°) + 1220 N
Vertical component ≈ 782.62 N + 1220 N
Vertical component ≈ 2002.62 N
So, the magnitude of the horizontal component of the force that the wall exerts on the left end of the beam is 1721.54 N, and the magnitude of the vertical component of the force is approximately 2002.62 N.
Therefore,
Fx = 1721.54 N
Fy = 2002.62 N
To solve this problem, we can break it down into two parts:
a) Finding the tension in the wire
b) Finding the horizontal and vertical components of the force exerted by the wall on the beam
a) Finding the tension in the wire:
To calculate the tension in the wire, we need to consider the vertical equilibrium of forces on the crate. In other words, the sum of the vertical forces acting on the crate must be zero.
We can start by resolving the weight of the crate into its vertical and horizontal components. The vertical component of the weight will be acting downwards and can be calculated as follows:
Vertical component of weight = W * cos(30°)
= 1990 N * cos(30°)
≈ 1725.4 N
Since the crate is in equilibrium, the vertical component of its weight must be balanced by the tension in the wire.
∴ Tension in the wire = Vertical component of weight
= 1725.4 N
Therefore, the magnitude of the tension in the wire is approximately 1725.4 N.
b) Finding the horizontal and vertical components of the force exerted by the wall on the beam:
To find the forces exerted by the wall on the beam, we need to consider the horizontal and vertical equilibrium of forces at the left end of the beam.
Considering the horizontal equilibrium, we know that the horizontal component of the force exerted by the wall must balance the horizontal component of the weight of the crate hanging at the far end.
Horizontal component of weight = W * sin(30°)
= 1990 N * sin(30°)
≈ 995 N
Since the horizontal force is zero (as no horizontal force is mentioned or acting), the horizontal component of the force exerted by the wall is also zero.
∴ Fx = 0 N
Moving on to vertical equilibrium, the vertical component of the force exerted by the wall must balance the vertical component of the weight of the crate, along with the vertical component of the weight of the beam itself.
Vertical component of weight of the crate = W * cos(30°)
= 1990 N * cos(30°)
≈ 1725.4 N
Vertical component of weight of the beam = 1220 N * cos(50°)
≈ 792.46 N
Summing up these vertical forces, we get:
Vertical component of force exerted by the wall = Vertical component of weight of the crate + Vertical component of weight of the beam
= 1725.4 N + 792.46 N
≈ 2517.86 N
Therefore,
Fy = 2517.86 N
To summarize:
a) The magnitude of the tension in the wire is approximately 1725.4 N.
b) The magnitude of the horizontal component of the force exerted by the wall is 0 N, and the magnitude of the vertical component of the force exerted by the wall is approximately 2517.86 N.