Tension of two ropes given mass and angle

[alexas]

Homework Statement

A 1400 kg steel beam is supported by two ropes.

Homework Equations

Tx = mgsin(ANGLE) ?
Ty = mgcos(ANGLE) ??

g = 9.8?

Tension of one wire = [((Tx)^2) + ((Ty)^2)]^(1/2) ?

The Attempt at a Solution

I wasn't too sure whether or not i should use the angle: 20 or 70.
So i did it both ways they were both wrong, but i did it both ways...

[((1400(9.8)sin(20))^2) + ((1400(9.8)cos(20))^2)]&(1/2)

And then i did:

[((1400(9.8)sin(70))^2) + ((1400(9.8)cos(70))^2)]&(1/2)

Am i approaching this problem correctly? I have a test Thursday and i feel pretty F***** right now. Any help would be appreciated.

 

[LowlyPion]

Homework Statement

A 1400 kg steel beam is supported by two ropes.

Homework Equations

Tx = mgsin(ANGLE) ?
Ty = mgcos(ANGLE) ??

g = 9.8?

Tension of one wire = [((Tx)^2) + ((Ty)^2)]^(1/2) ?

The Attempt at a Solution

I wasn't too sure whether or not i should use the angle: 20 or 70.
So i did it both ways they were both wrong, but i did it both ways...

[((1400(9.8)sin(20))^2) + ((1400(9.8)cos(20))^2)]&(1/2)

And then i did:

[((1400(9.8)sin(70))^2) + ((1400(9.8)cos(70))^2)]&(1/2)

I can't see your picture as yet.

But resolving angles is no sweat. You always want the component in the direction of the axes, and however your angle is defined - relative to the horizontal or the vertical - you want the angle function determined by your drawing. Side opposite is sine, and side adjacent is cosine.

As to your method I think you need to think about what you are doing.

Most importantly remember for statics that the ∑ F in the x are 0 and the ∑ F in the y are 0.

Merely take the components of your tensions in the x and they add to 0. Now the components in the Y are supporting the beam. so those components of tension must add to the weight.

∑ Ty = T1y + T2y = m*g

For the x ∑ Tx = 0 then T1x = - T2x

 

[alexas]

So the the total Ty should equal 13720N correct?

m*g
1400*9.8 = 13720N ??

How would i go about finding the tensions of the individual ropes?

I was thinking that:

13720(sin20)) = gives you the tension of rope 1 with an angle of 20 degrees, and 13720(sin30)) = gives you the tension of rope 2 with an angle of 30 degrees.

Am i approaching this correctly?

 

[LowlyPion]

So the the total Ty should equal 13720N correct?

m*g
1400*9.8 = 13720N ??

How would i go about finding the tensions of the individual ropes?

I was thinking that:

13720(sin20)) = gives you the tension of rope 1 with an angle of 20 degrees, and 13720(sin30)) = gives you the tension of rope 2 with an angle of 30 degrees.

Am i going in the right direction?

If the angles are with the horizontal then yes, the sum of T1*Sin(20) and T2*Sin(30) will yield the m*g.

Now write an equation for the x components.

Then all you need to do is solve for T1 and T2 directly.

2 equations and 2 unknowns means the end of the voyage is in sight.

 

[alexas]

T1sin(20) + T2cos(30) = 0 for the x-components?

I'm a little confused on how to go further from here...

The angles i believe are with the vertical axis, they are in the picture as rope 1 being 20 degrees to the left of the vertical y-axis and rope 2, 30 degrees to the right of the vertical y axis.

 

[LowlyPion]

T1sin(20) + T2cos(30) = 0 for the x-components?

I'm a little confused on how to go further from here...

The angles i believe are with the vertical axis, they are in the picture as rope 1 being 20 degrees to the left of the vertical y-axis and rope 2, 30 degrees to the right of the vertical y axis.

Sorry I can't see your picture. I previously said if the angles were with the horizontal. So if they are with the vertical then simply swap sin and cos. Sin is the x components then and cos is vertical.

 

[alexas]

http://tinypic.com/view.php?pic=1hpmav&s=5

 

[LowlyPion]

Got it. So ...

T1*Cos(20) and T2*Cos(30) = m*g.

And

T1*Sin(20) = -T2*Sin(30)

 

[alexas]

Shouldn't it be:

T1*Sin(20) - T2*Sin(30) = 0

then...
T1*Sin(20) = T2*Sin(30)

then i did...

T1 = [(T2*Sin(30))/(Sin(20))]

using...
T1*Cos(20) + T2*Cos(30) = m*g.

Replace T1, with what i found earlier and...
[(T2*Sin(30))/(Sin(20))]*Cos(20) + T2*Cos(30) = m*g.

Pull out T2...

T2([(Sin(30))/(Sin(20))]*[Cos(20)] + Cos(30)) = 13720 N

T2 = 13720 N / {([(Sin(30))/(Sin(20))]*[Cos(20)] + Cos(30))}

T2 = 6125.64 N ?I know its wrong, i am just not sure where i am messing up.

 

[LowlyPion]

Shouldn't it be:

T1*Sin(20) - T2*Sin(30) = 0

I think you are already anticipating the direction of the Tension.

It's certainly true that |T1*Sin(20)| - |T2*Sin(30)| = 0

But the x components are opposed and hence not interchangeable if you take their directionality into account as

T1*Sin(20) = T2*Sin(30) might imply.

Ultimately you are really only interested here in |T1| = |T2|*Sin30/Sin20 for solving for T1 and T2, so I suppose it doesn't matter here.

 

[alexas]

I think you are already anticipating the direction of the Tension.

It's certainly true that |T1*Sin(20)| - |T2*Sin(30)| = 0

But the x components are opposed and hence not interchangeable if you take their directionality into account as

T1*Sin(20) = T2*Sin(30) might imply.

Ultimately you are really only interested here in |T1| = |T2|*Sin30/Sin20 for solving for T1 and T2, so I suppose it doesn't matter here.

I'm a little confused. Because i think what your saying is that i am doing it correctly? But when i solved out my answer for T2 the online system says I am wrong.

 

[LowlyPion]

As for your calculation

I'd put it in numbers. It's easier for me to follow.

T1 = 1.462 T2

Then

1.462 T2 * Cos20 + T2 * Cos 30 = 1.374T2 + .866 T2 = 1400*9.8 = 13720

T2 = 6125 (Which is what you got.)
T1 = 1.462*6125 = 8955

 

[LowlyPion]

I'm a little confused. Because i think what your saying is that i am doing it correctly? But when i solved out my answer for T2 the online system says I am wrong.

Yes it looks like you are ok. Maybe you need to reverse the values for T1 and T2 in the answering system?

 

[alexas]

I'm an idiot. I was putting the T2 answer in the T1 box. LOLLL

 

[alexas]

I really appreciate your help. Thank You.

 

[LowlyPion]

Sure.

Cheers.