Calculating Tension in Suspended Loudspeaker Cables

[aligass2004]

Homework Statement

A 14 kg loudspeaker is suspended 2.0m below the ceiling by two 3.0m long cables that angle outward at equal angles. What is the tension in the cables?

Homework Equations

F=ma

The Attempt at a Solution

I drew the free body diagram, and I broke the tension of the two wires into their components. I'm unsure about how to tackle the problem.

 

[learningphysics]

Where are you getting stuck? What are the forces acting on the suspended mass vertically?

 

[aligass2004]

The forces acting are the tension of the two cables and the weight of the object. I just don't know how to find the angle so I can break up the tensions into components.

 

[learningphysics]

The forces acting are the tension of the two cables and the weight of the object. I just don't know how to find the angle so I can break up the tensions into components.

Find the angle... you know the y displacement is 2.0m. You know the length of the cable is 3.0m. You can find the angle.

 

[aligass2004]

Would the angle be 48.19 degrees? And if that's the case, where do I go from there?

 

[learningphysics]

Would the angle be 48.19 degrees? And if that's the case, where do I go from there?

I'm getting 41.81 degrees. Well... if T is the magnitude of the tension... what's the vertical component of tension? What are the forces in the vertical direction?

 

[aligass2004]

The forces in the vertical direction are T1x and T2x

 

[learningphysics]

Would the angle be 48.19 degrees? And if that's the case, where do I go from there?

Oops... sorry, this angle is fine... I just got the other angle of the right triangle... 90-48.19 = 41.81

 

[learningphysics]

The forces in the vertical direction are T1x and T2x

The tensions are equal... So what is Tx in terms of T?

 

[aligass2004]

Tx = Tcos(theta)

 

[learningphysics]

Tx = Tcos(theta)

So what is the F=ma equation in the vertical direction?

 

[aligass2004]

Is it Tcos(theta) = m(ax)?

 

[learningphysics]

Is it Tcos(theta) = m(ax)?

accleretation is 0... you have 2 tensions... and also gravity.

 

[aligass2004]

I put horizontal anyway. In the vertical direction it would be Fnety = T1y + T2y - mg = m(ay).

 

[learningphysics]

I put horizontal anyway. In the vertical direction it would be Fnety = T1y + T2y - mg = m(ay).

Can you put in the angles, sin/cos etc... ay = 0... and T1=T2... so just use T.

 

[aligass2004]

I don't understand.

 

[learningphysics]

I don't understand.

Did you find T?

 

[aligass2004]

No I didn't. I don't understand how you can find T.

 

[learningphysics]

I put horizontal anyway. In the vertical direction it would be Fnety = T1y + T2y - mg = m(ay).

Why are you using two different variables T1 and T2?

 

[aligass2004]

Am I not breaking the tension into components?

 

[learningphysics]

Am I not breaking the tension into components?

I don't understand... is T1 and T2 different components? I thought they were for each rope on either side of the mass...

Both of those ropes have exactly the same tension (magnitude)... As written in the question, "two 3.0m long cables"... both of those cables have the exact same tension...

 

[aligass2004]

Are the ropes straight up and down or something?

 

[aligass2004]

No, cause it says outward.

 

[learningphysics]

No, cause it says outward.

Actually you don't have to assume they're equal tensions...

\Sigma{F_x} = 0

T2cos(41.8) - T1cos(41.8) = 0

T1 = T2

so from the sum of forces in the x-direction, we get that the two tensions in the two ropes are the same.

Then \Sigma{F_y} = 0

T1sin(41.8) + T2sin(41.8) - mg = 0

sub in T1=T2

2T1sin(41.8) = mg

and you can solve for T1.

 

[aligass2004]

I think I got it now. Since the acceleration in the x direction is zero, the tensions in the x direction are zero. Since you know they're equal in the x direction you can substitute them in for the y direction. I just don't think I knew that you could substitute.