Find Tension Force for Cord Holding Gate

[solarcat]

Homework Statement

A gate 4 m wide and 2 m high weights 700 newtons. It is hinged on the left side at two points, A and B (A is above B). A cord CD is connected at the other end of the gate across from A. The horizontal force at A is 0. I'm supposed to calculate the torque due to to the tension in the cord two different ways, using the lever arm of the force and the horizontal and vertical components of the tension.

Relevant Equations

Torque = Force*Distance

I thought that the upward/vertical torque due to the cord might be T*sin 30 deg * sqrt (4^2 + 2^2). Then this would have to equal the weight of the gate times the distance (700 N * 2 m). This gives a tension force of 626 Newtons but it's not the right answer. I don't even know what the lever arm of the force is or how to even use that to find tension.

 

[tnich]

I thought that the upward/vertical torque due to the cord might be T*sin 30 deg * sqrt (4^2 + 2^2). Then this would have to equal the weight of the gate times the distance (700 N * 2 m). This gives a tension force of 626 Newtons but it's not the right answer. I don't even know what the lever arm of the force is or how to even use that to find tension.

I think you are forgetting something in your torque equation. In vector form the torque would be $\vec\tau = \vec r \times \vec F$ where $\times$ represents a cross-product and $\vec r$ is the lever arm vector. You lose something when you write that as a scalar equation $T=DF$.
Also, you seem to be ignoring the torque resulting from the horizontal component of T.
It think it would be a good idea to 1) to decide which point your torques act at, 2) draw the force and lever arm vectors on the diagram, and 3) write out the torque balance equation. This would help you avoid some of incorrect assumptions you seem to be making.

 

[haruspex]

calculate the torque due to to the tension in the cord

Torque about what point?

T*sin 30 deg * sqrt (4^2 + 2^2)

You need a force and a distance which are at right angles. In what directions are the force and distance in your equation?

don't even know what the lever arm of the force

The lever arm of a force about a given point is the distance from the point to the line of action of the force.

how to even use that to find tension.

If you can write a torque balance equation in which you know the torque on one side of the equation and the length of the lever arm on the other side then you can deduce the force that goes with the lever arm.

 

[solarcat]

I thought the free body diagram might look something like this

 

[haruspex]

I thought the free body diagram might look something like this

Yes.. you don't know whether there’s a vertical force at B, but it doesn’t matter.

 

[solarcat]

OK, so I don't know what to do next though. You said earlier that I wasn't taking the vector part into account, but can't you also use the equation torque = Fd sin theta? I don't understand why this doesn't work, and I don't understand why the radius isn't the square root of 20.

 

[haruspex]

You said earlier that I wasn't taking the vector part into account,

Not me.

can't you also use the equation torque = Fd sin theta?

Yes, but in your equation:

T*sin 30 deg * sqrt (4^2 + 2^2)

the angle between the tension force and the length sqrt (4^2 + 2^2) is not 30 degrees.

 

[solarcat]

Okay, so using the side lengths of the rectangle, I can see the angle between the gates bisector is tan theta = 2/4; theta = 26.6 degrees. From the geometry of the rectangle the angle between the bisector and the tension should be 52.5. 1400=T*sqrt 20 * sin 26.6 deg T = 375 N?
So in that case how would you solve the problem the same way using components?

 

[haruspex]

tan theta = 2/4; theta = 22.5 degrees

Rather inaccurate.

 

[solarcat]

OK, I see what I did there. But I still don't understand the difference between calculating torque using the radius or the horizontal and vertical components. If I look at the components, it looks like the horizontal forces Fb and Tcos 30 don't produce any torque. The sum of torques would be 4 m *Tsin30 + 0*Fa - 2 m * 700 N, but that obviously isn't right.

 

[tnich]

The sum of torques would be 4 m *Tsin30 + 0*Fa - 2 m * 700 N, but that obviously isn't right.

I think you are missing a key concept here. To write a sum of torques, the torques have to be about the same point. You seem to be calculating the torque due to tension T about point A and the torque due to weight W about point B. I suggest that you find both torques about point B. Since the net horizontal force at A is zero, the net torque at point A about point B would be zero, and you could sum up the other two torques, set them equal to zero and solve for T.
@haruspex has been telling you that you have the angle wrong for the torque due to tension T, and I agree. I think it would help to draw the lever arms about point B on your diagram and look at the angles that result.

 

[haruspex]

it looks like the horizontal forces Fb and Tcos 30 don't produce any torque.

If you are taking moments about B (which you have not confirmed) then, yes, forces at B will not produce a torque, but why do you think the horizontal component of T will not? It acts along the top of the gate.