"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 1960-N crate hangs from the far end of the beam. Find a) the magnitude of the tension in the cable and b) the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam. (The cable makes an angle of 50 degrees with respect to the horizontal and the beam makes an angle of 30 degrees.)"

Question

According to my textbook, the answers are a)2260N and b) 1450N for both the horizontal and vertical components. I just really need some help in setting up this problem to get those answers.

Bot · Answer

Let's call the tension in the cable T₁, the force of the wall on the beam (horizontal component) Fx, and the force of the wall on the beam (vertical component) Fy. We'll also call the length of the beam L and the angle between the cable and the horizontal θ.

Here's how to set up the problem:

1. Determine the angles: The cable makes an angle of 50 degrees with the horizontal (given), and the beam makes an angle of 30 degrees with the horizontal (given).

2. Break the weight of the crate (1960 N) into its horizontal and vertical components:
- Wx = 0 (since it's only acting vertically)
- Wy = 1960 N

3. Break the tension in the cable (T₁) into its horizontal and vertical components:
- T₁x = T₁ * cos(50°)
- T₁y = T₁ * sin(50°)

4. Sum the forces in the horizontal and vertical directions:
a. Sum of forces in x-direction (horizontal): Fx + T₁x = 0
b. Sum of forces in y-direction (vertical): Fy + T₁y - Wy = 0

5. Set up a torque equation using the left end of the beam as the pivot point. Since torques are equal and opposite, the net torque is zero: T₁ * L * cos(30°) = Wy * L

6. Solve the equation for T₁:
- T₁ * cos(30°) = 1960 N
- T₁ = 2260 N

7. Now that we have the tension in the cable, we can solve for Fx and Fy using the horizontal and vertical force equations:
a. Fx + (2260 N) * cos(50°) = 0
- Fx = -2260 * cos(50°)
- Fx ~ 1450 N (opposite direction for the wall force)

b. Fy + (2260 N) * sin(50°) - 1960 N = 0
- Fy = 1960 N - 2260 * sin(50°)
- Fy ~ 1450 N

So, the magnitude of the tension in the cable is 2260 N, and the magnitudes of the horizontal and vertical components of the force that the wall exerts on the left end of the beam are both 1450 N.