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Find the tension of the system (picture included)

  • Thread starter fern518
  • Start date
6
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1. Homework Statement

In order to simplify this process and avoid possible errors I have included the scanned image of the problem from my textbook. (Let me know if the picture is too small and I will upload a larger one)

ta4p6u.jpg



2. Homework Equations

Mostly just free body diagrams..


3. The Attempt at a Solution

For this problem I split the tensions on the right and left side of the pulley as T1(from A to B) and T2(from B to C) and once I found those out I would add them together to find the total tension of the system.
For this I know I have to find the forces in the x and y directions and I have come up with the necessary equations to solve the problem:

Fx: T2cos(phi) - T1sin(theta) (equation 1)

Fy: T2sin(phi) + T1sin(theta) - 50 (equation 2)

Using equation 1, I solved for T2 in terms of T1 and got :

T2 = T1sin(theta)/cos(phi)

Then I plugged T2 into equation 2:

T1tan(phi)sin(theta) + T1sin(theta) = 50

So my problem is basically this, I need at least one of the angles (phi or theta) for me to solve this problem, however I have very little clue as to how to obtain either one. I know I am supposed to use the fact that the length of the cable is 10feet and the distance between the walls supporting the cable is 8 feet but I really don't know how to approach this.

Thank you for any and all assistance you can provide.
 

PhanthomJay

Science Advisor
Homework Helper
Gold Member
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1. Homework Statement

In order to simplify this process and avoid possible errors I have included the scanned image of the problem from my textbook. (Let me know if the picture is too small and I will upload a larger one)

ta4p6u.jpg



2. Homework Equations

Mostly just free body diagrams..


3. The Attempt at a Solution

For this problem I split the tensions on the right and left side of the pulley as T1(from A to B) and T2(from B to C) and once I found those out I would add them together to find the total tension of the system.
If the pulley is frictionless, which I assume it is, the tensions on each side are equal. In the x direction ,their x components must add algebraically to zero, and their y components must add up to the weight.
Fx: T2cos(phi) - T1sin why sin?(theta) (equation 1) equals___?

Fy: T2sin(phi) + T1sin(theta) - 50 (equation 2)equals__?

Using equation 1, I solved for T2 in terms of T1 and got :

T2 = T1sin(theta)/cos(phi)

Then I plugged T2 into equation 2:

T1tan(phi)sin(theta) + T1sin(theta) = 50

So my problem is basically this, I need at least one of the angles (phi or theta) for me to solve this problem, however I have very little clue as to how to obtain either one. I know I am supposed to use the fact that the length of the cable is 10feet and the distance between the walls supporting the cable is 8 feet but I really don't know how to approach this.

Thank you for any and all assistance you can provide.
If T1 =T2, what must the angles be?
 
6
0
Oh sorry about that:

Fx: T2cos(phi) - T1cos(theta) = 0

Fy: T2sin(phi) + T1sin(theta) = 50


I knew they had to equal 0 and 50 respectively, but the sin part was a careless error on my part.

But anyways, sp as you were saying, if t1=t2 should the angles be equal to each other?
 

jgens

Gold Member
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I think that's what PhanthomJay is hinting at, however, I thought that would only be true if the rope was hung at equivalent heights? Perhaps a misunderstanding on my part?
 
6
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That was my issue too. The fact that the cable is not at the same height led me to think that the angles were not the same. I hope I am wrong because if they do happen to be the same this would make the solution to the problem much easier.
 

PhanthomJay

Science Advisor
Homework Helper
Gold Member
7,113
456
o sorry I missed that; so is the elevation given ?
 
6
0
Nope, all that is given is what is shown in the picture, the length of the cable and the distance between the 2 walls.

I know some kind of trigonometry is involved here, but I really don't know how to approach it..

At first I was considering making a triangle that goes from one wall to another so one side is 8 ft, then from point A to point C as the hypotenuse, and then from part C to the top of part like this:

152gar5.jpg


Would it be wrong to write the distance of the hypotenuse as 10 ft then use pythagorean theorem to determine the difference in height between A and C as 6 ft?

This is about as far as I can go..
 
288
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If you use a hypotenuse of 10 ft then the sag is not taken into consideration. If the weight was fixed at a point then two tensions would exist. So, the tension in this case is expessed as

T = W/(sin(theta) + sin(phi))

The problem is to solve for sin(theta) and sin(phi). To solve, draw a vertical line through the pulley center creating two triangles. Let the horizontal distance from the left wall to the vertical line be D, then the remaining distance is 8 - D. Similarly, let the cable length from the left wall to the pulley center (the pulley is small) be L and the remaining length be 10 - L. Express the sines and cosines of theta and phi in terms of D, L, 8 - D, 10 - L. i.e.

cos(theta) = D/L, cos(phi) = (8 - D)/(10 - L), sin(theta) = sqrt(L2 - D2)/L, and
sin(phi) = sqrt((10 - L)2 - (8 - D)2)/(10 - L). Then use cos2 + sin2 =1. This will give two equations with two unknows, L and D. Solve for L and D then you can determine sin(theta) and sin(phi).
 
6
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Man thanks a lot chrisk. I still have to solve for L and D but I now know exactly how to approach the problem and solve it.

thanks again
 

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