Tension; 2 cables holding weight

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Homework Help Overview

The problem involves a small sphere of weight W supported by two wires, AB and CD. The objective is to determine the tension in wire CD before and immediately after wire AB is cut, with specific values provided for reference.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equilibrium equations for the forces acting on the sphere, including the vertical and horizontal components of tension in the wires. There are attempts to derive expressions for the tensions based on the given weight and angles.

Discussion Status

Some participants have provided equations and suggested methods for solving the problem, while others express confusion about their calculations and seek clarification on the setup. There is an acknowledgment of errors in inputting values into calculators, and the discussion reflects a mix of approaches being explored without a clear consensus on the correct method.

Contextual Notes

Participants note the complexity of the problem relative to other assignments and mention the potential for overthinking simpler problems. There is a sense of frustration with the calculations, indicating that the problem may have nuances that are not immediately clear.

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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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