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

• bearhug
In summary, a 225 kg uniform square sign 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. To solve for the tension in the cable, the sum of Fx, sum of Fy, and sum of Torque equations should be used. The angle between the rod and the cable can be found using arctan 4/3 if the sign were a point load.
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.

bearhug said:
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?

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.

## 1. What is the formula for calculating tension in a cable?

The formula for calculating tension in a cable is T = (m x g) + (F x sinθ), where T is the tension, m is the mass, g is the acceleration due to gravity, F is the force acting on the cable, and θ is the angle of the cable with respect to the horizontal.

## 2. How do you determine the weight of the sign and the force acting on the cable?

The weight of the sign can be determined by multiplying the mass (225 kg) by the acceleration due to gravity (9.8 m/s^2), giving a weight of 2205 N. The force acting on the cable can be determined by considering the weight of the sign and the angle of the cable, using the formula F = W/sinθ.

## 3. Can the tension in the cable be greater than the weight of the sign?

Yes, the tension in the cable can be greater than the weight of the sign. This can happen if there is an additional force acting on the cable, such as wind or an uneven distribution of weight on the sign. In this case, the tension in the cable will be equal to the weight of the sign plus the additional force.

## 4. How does the length of the rod affect the tension in the cable?

The length of the rod does not directly affect the tension in the cable, as long as the rod is strong enough to support the weight of the sign. However, a longer rod may cause the cable to have a larger angle, which can increase the force acting on the cable and therefore increase the tension.

## 5. Is there a maximum tension that the cable can withstand?

Yes, there is a maximum tension that the cable can withstand before it breaks. This depends on factors such as the material and thickness of the cable, as well as its age and condition. It is important to ensure that the tension in the cable does not exceed the maximum limit to prevent accidents or damage.

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