Calculate Tension in Cable for 225 kg Square Sign Hanging from 3.00 m Rod

[bearhug]

A 225 kg uniform square sign, 2.00 m on a side, is hung from a 3.00 m rod of negligible mass. A cable is attached to the end of the rod and to a point on the wall 4.00 m above the point where the rod is fixed to the wall. What is the tension in the cable?
I know I need to use:
sum of Fx
sum of Fy
sum of Torque

What I'm wondering is how should I do this when I don't know any angles whatsoever? Any help on this would be useful, just knowing this should get me through the problem.

 

[PhanthomJay]

A 225 kg uniform square sign, 2.00 m on a side, is hung from a 3.00 m rod of negligible mass. A cable is attached to the end of the rod and to a point on the wall 4.00 m above the point where the rod is fixed to the wall. What is the tension in the cable?
I know I need to use:
sum of Fx
sum of Fy
sum of Torque

What I'm wondering is how should I do this when I don't know any angles whatsoever? Any help on this would be useful, just knowing this should get me through the problem.

The rod is 3 m and the cable is attached 4 m up from the wall. Now if the sign was just a point load at the end of the rod, the angle would be arctan 4/3, correct?

 

[kyle8921]

To calculate the tension in the cable, we can use the equations of equilibrium, which state that the sum of all forces and the sum of all torques acting on a body must be equal to zero in order for it to be in static equilibrium.

First, let's draw a free-body diagram of the sign and the rod. We have the weight of the sign acting downwards with a magnitude of 225 kg * 9.8 m/s^2 = 2205 N. The tension in the cable will act upwards and to the right, and we can label this as T. The weight of the rod will also act downwards, but since it is negligible compared to the weight of the sign, we can ignore it in our calculations.

Next, we can apply the equations of equilibrium. In the x-direction, we have T acting to the right and no other forces, so we can set the sum of forces in the x-direction equal to zero. This gives us the equation:

∑Fx = T = 0

In the y-direction, we have the weight of the sign acting downwards and the tension in the cable acting upwards. This gives us the equation:

∑Fy = T - 2205 N = 0

Finally, we can apply the equation for torque, which states that the sum of all torques acting on a body must be equal to zero. We can choose any point as the center of rotation, but it is convenient to choose the point where the rod is attached to the wall. This gives us the equation:

∑Torque = T * 4.00 m - 2205 N * 3.00 m = 0

Solving for T in the first two equations, we get T = 2205 N, which is the tension in the cable.

To address your concern about not knowing any angles, we do not need to know any angles in this problem because we are dealing with forces and distances that are perpendicular to each other. In cases where angles are involved, we can use trigonometry to break down the forces into their x and y components, but in this problem, that is not necessary.

I hope this helps! Let me know if you have any further questions.