Tension; 2 cables holding weight

AI Thread Summary
The discussion focuses on calculating the tension in two wires holding a weight, specifically after one wire is cut. The initial equations set up the balance of forces in both the x and y directions, leading to the determination of tensions TCD and TAB. A participant initially miscalculates TCD, obtaining 1.066W instead of the correct 0.742W, due to input errors in calculations. After revising the equations and simplifying, the correct tensions are confirmed, with TCD being 0.940W immediately after wire AB is cut. The conversation highlights the common struggle with seemingly simple problems amidst a heavy workload.
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Homework Statement



A small sphere of weight W is held as shown by two wires AB and CD. If wire AB is cut, determine the tension in the other wire (a) before AB is cut, (b) immediately after AB has been cut. (Ans: (a) TCD = 0.742 W; (b) TCD = 0.940 W).

http://img4.imageshack.us/img4/3763/picture11td.png

Homework Equations



\SigmaFy=may
Ty=Tsin\theta

The Attempt at a Solution



TCDy+TABy-W=may ; ay=0 (not moving)

TCDSin(70\circ)+TABSin(50\circ)-W=0 ; solve fot TAB and plug in

TABSin(50\circ)=W-TCDSin(70\circ) ; divide Sin(50\circ) from both sides

TAB=[W/Sin(50\circ)]-[TCDSin(70\circ)/Sin(50\circ)]

TCDSin(70\circ)+[W/Sin(50\circ)]-[TCDSin(70\circ)/Sin(50\circ)]-W=0 ; Plugging in TAB from above

TCD(Sin(70\circ)-(Sin(70\circ)/(Sin(50\circ)=W(1-1/(Sin(50\circ)

Solving for TCD I get 1.066W when its supposed to be 0.742 W

What am i doing wrong?
 
Last edited by a moderator:
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T(CD)*sin70 + T(AB)*sin50 - W = 0.
After this write
T(CD)*cos70 + T(AB)*cos50 = 0
Now solve.
 
rl.bhat said:
T(CD)*sin70 + T(AB)*sin50 - W = 0.
After this write
T(CD)*cos70 + T(AB)*cos50 = 0
Now solve.

so do the sum of the forces in the x direction?
Wouldn't it then be T(CD)*cos70 - T(AB)*cos50 = 0 ?

---
Worked out the sum of the forces in x and y direction
solved for TAB Set them equal to each other and still wrong answer :(
 
Last edited:
In my equation I have given the vector sum. Your equation is correct.
T(CD)*sin70 + T(AB)*sin50 = W ...(1)
After this write
T(CD)*cos70 = T(AB)*cos50
T(AB) = T(CD)*cos70/cos50
Put it in the eq, 1, you get
T(CD)*sin70 + [T(CD)*cos70/cos50]*sin50 = W
Now simplify.
For part (b) T(CD) = component of W along the wire.
 
thanks, turned out i just wasn't inputting in the calculator right.

funny how this one of the simplest problems out of the assignment and i was having trouble with it and easily solved the complex ones. Always the easy stuff that gets me, i guess i think about it too hard, make it more complex then it is.

been working on so much homework (thermo, dynamics, eaII, and solids) that my mind is fried!
 
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