Uniform Beam & Force

Question

A 12.0 m uniform beam is hinged to a vertical wall and held horizontally by a 5.00 m cable attached to the wall 4.00 m above the hinge, as shown in the figure below (Figure 1). The metal of this cable has a test strength of 1.20 kN , which means that it will break if the tension in it exceeds that amount.

a. What is the heaviest beam that the cable can support with the given configuration?

b. Find the horizontal component of the force the hinge exerts on the beam.

c. Find the vertical component of the force the hinge exerts on the beam.

here is the image for the figure: https://freeimage.host/i/mW4Whb

Answer 1

This is a statics problem that involves a rigid beam, a hinge, and a cable. The key to this problem is realizing that the beam is at rest. This means that the net x-force, net y-force, and net moment must be equal to zero.

Step 0: Definitions, Variables, and Assumptions

Let's define some variables. Let 'W' be the weight of the beam in N. Let 'L' be the length of the beam. Let 'T' be the tension of the cable. Let 'r' be the distance between the hinge and the attachment point. Let 'θ' be the angle between the beam and the cable. Let 'H_x' represent the hinge support force in the x-direction. Let 'H_y' represent the hinge support force in the y-direction.

First, the beam is 12.0 meters long and uniform, so its weight acts at its center, which is 6.0 meters from the hinge. The cable is 5.00 meters long and is attached to the wall 4.00 meters above the hinge. Using the Pythagorean Theorem, we can solve for 'r'.

r² + (4.00 m)² = (5.00 m)²

r = 3.00 m

Now that we have a complete triangle, we can find the sin and cos of 'θ'.

sin(θ) = opposite / hypotenuse = 4.00 m / 5.00 m = 0.8

cos(θ) = adjacent / hypotenuse = 3.00 m / 5.00 m = 0.6

Let's explicitly write out all of our known and unknown variables.

L = 12.0 m

T = 1.20 kN

r = 3.00 m

sin(θ) = 0.8

cos(θ) = 0.6

W = ?

H_x = ?

H_y = ?

Part A: Heaviest Beam the Cable Can Support

To find the heaviest beam the cable can support, we should set up an equation saying that the net torque on the beam must be equal to zero. This is because the beam is not rotating. Therefore, the torque caused by the vertical component of the cable’s tension (counterclockwise) must balance the torque from the beam’s weight (clockwise).

M = F*d

∑M = 0

(+)(Tsinθ)(r) + (-)(W)(L/2) = 0

W = 2Trsinθ / L

W = 2(1200 N)(0.8)(3.00 m) / (12.0 m)

W = 480 N (downward direction)

Part B: Horizontal Component of the Hinge Force

Next, to find the horizontal component of the hinge force, we should set up an equation saying that the net force on the beam in the x direction must be equal to zero. This is because the beam is not translating horizontally. Therefore, the force caused by the horizontal component of the cable’s tension (left) must balance the horizontal component of the hinge support force (right).

∑F_x = 0

(-)(Tcosθ) + (+)(H_x) = 0

H_x = Tcosθ

H_x = (1200 N)(0.6)

H_x = 720 N (right direction)

Part C: Vertical Component of the Hinge Force

Similarly, to find the vertical component of the hinge force, we should set up an equation saying that the net force on the beam in the y direction must be equal to zero. This is because the beam is not translating vertically. Therefore, the force caused by the vertical component of the cable’s tension (up) must balance the vertical component of the hinge support force (down) and the weight of the beam (down).

∑F_y = 0

(+)(Tsinθ) + (-)(H_y) + (-)(W) = 0

H_y = Tsinθ - W

H_y = (1200 N)(0.8) - 480 N

H_y = 480 N (downward direction)

This problem demonstrates how to apply the principles of static equilibrium (balancing forces and torques) to determine the maximum load a beam can support and the reaction forces at its hinge.