(I attempted on my own already) 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 = 1860 N crate hangs from the far end of the beam.

a) Calculate the magnitude of the horizontal component of the force that the wall exerts on the left end of the beam if the angle between the cable and horizontal is θ = 49°. My answer: 1429.630896 N

b) Calculate the magnitude of the vertical component of the force that the wall exerts on the left end of the beam. My answer: 1435.397793 N

Figure: imgur(dot)com/a/y6aJbOo

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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 can use trigonometry and the principle of equilibrium.

Let's start by understanding the forces acting on the beam. There are three forces involved: the weight of the beam, the weight of the crate hanging from the far end of the beam, and the tension in the cable supporting the beam.

The weight of the beam is given as 1220 N, and we can assume it acts at the center of the beam. The weight of the crate is given as 1860 N, acting downward at the far end of the beam.

Now, let's break down the forces acting on the beam. Since the beam is in equilibrium (not accelerating), the net force on it must be zero. Considering the forces acting along the horizontal (x-axis) and vertical (y-axis) directions separately, we can set up the following equations:

Sum of forces in the x-direction: T * cos(θ) = 0 (Equation 1)
Sum of forces in the y-direction: T * sin(θ) + 1860 N - 1220 N = 0 (Equation 2)

In Equation 1, T represents the tension in the cable, and cos(θ) represents the horizontal component of the tension. In Equation 2, sin(θ) represents the vertical component of the tension.

a) To find the magnitude of the horizontal component of the force that the wall exerts on the left end of the beam, we need to solve Equation 1 for T * cos(θ). Rearranging the equation, we get:

T * cos(θ) = 0
T = 0 / cos(θ)
T = 0

According to Equation 1, the horizontal component of the tension is zero. Therefore, the wall exerts no horizontal force on the left end of the beam.

b) To find the magnitude of the vertical component of the force that the wall exerts on the left end of the beam, we need to solve Equation 2 for T * sin(θ). Rearranging the equation, we get:

T * sin(θ) + 1860 N - 1220 N = 0
T * sin(θ) = 1220 N - 1860 N
T * sin(θ) = -640 N

Using the given angle θ = 49°, we can substitute the values into the equation:

T * sin(49°) = -640 N
T = -640 N / sin(49°)
T ≈ -1435.397793 N

Taking the absolute value of T, we find that the magnitude of the vertical component of the force that the wall exerts on the left end of the beam is approximately 1435.397793 N.

Please note that the negative sign indicates that the force is acting in the opposite direction to the positive y-axis convention. Magnitudes of forces are always positive.

Hence, based on the given information and calculations, the magnitude of the horizontal component of the force exerted by the wall on the left end of the beam is 0 N (no horizontal force), and the magnitude of the vertical component of the force exerted by the wall on the left end of the beam is approximately 1435.397793 N.