How Do You Calculate Tension in a Cable and Hinge Force in a Beam System?

[shippage]

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The mass (M) is 430kg and is hanging from a rope (second vertical line). The distance between the anchor points is 1.9m (the length of the first vertical line) and the horizontal support cable (horizontal line) is 2.5m long. The uniform beam has a mass of 85kg (the diagonal line) and if the mass of the cable can be ignored, find T the tension force in the cable and the force of the hinge on the beam. (dots are insignificant just typed them to avoid the reformatting of the diagram once submitted).

Can anyone show me how to solve this, I have no idea where to start since we only had three classes on torques and I missed two.

 

[dumdum000333]

torque is equal to rXF where r is the distance from the point and f is the force acting at that distance. it should be known that only the force acting perpendicular to the distance affects torque hence the cross product. in the problem you have you have 2 downward masses(one at the center of mass for the beam and the other at the end of the beam) so you find the torques of these and counteract them with the torque created by the tension of the cable

 

[Moetasim]

First, let's define the problem. We have a system consisting of a mass (M) hanging from a rope, a horizontal support cable, and a uniform beam. The mass of the hanging object is 430kg, the distance between the anchor points is 1.9m, and the horizontal support cable is 2.5m long. The uniform beam has a mass of 85kg. We need to find the tension force in the cable and the force of the hinge on the beam.

To solve this problem, we can use the formula for torque: T = rFsinθ, where T is the torque, r is the distance from the pivot point, F is the force, and θ is the angle between the force and the lever arm.

First, let's find the torque caused by the hanging mass (M). The distance from the pivot point to the point where the rope is attached is 1.9m, and the force of gravity acting on the mass is 430kg x 9.8m/s^2 = 4214N. The angle between the force and the lever arm is 90 degrees, so sinθ = 1. Plugging these values into the formula, we get T = (1.9m)(4214N)(1) = 8017.6 Nm.

Next, let's find the torque caused by the weight of the beam. The distance from the pivot point to the center of the beam is 1.25m (calculated using Pythagorean theorem: √(2.5m)^2 - (1.9m)^2 = 1.25m). The weight of the beam is 85kg x 9.8m/s^2 = 833N. The angle between the force and the lever arm is 45 degrees, so sinθ = √2/2. Plugging these values into the formula, we get T = (1.25m)(833N)(√2/2) = 1467.8 Nm.

Now, we can set up an equation to find the tension force in the cable. Since the beam is in equilibrium, the net torque on it must be zero. This means that the torque caused by the hanging mass and the torque caused by the weight of the beam must be equal and opposite. So, we can set up the following equation:

8017.6 Nm +