Calculate the tension in the string

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    String Tension
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Homework Help Overview

The discussion revolves around calculating the tension in a string in the context of torque and equilibrium, specifically involving a beam with forces applied at various points. Participants are analyzing the effects of these forces and how they relate to the moments generated about a pivot point.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between torque and tension, attempting to set up equations based on the moments created by forces acting on the beam. There are questions about the correct application of trigonometric functions in the context of tension and torque.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem and attempting to clarify their understanding of the concepts involved. Some guidance has been offered regarding the application of sine and cosine in the context of tension, but no consensus has been reached on the final calculations.

Contextual Notes

There are constraints regarding the information provided, such as the length of the beam and the specific points at which forces are applied. Participants are questioning how these factors influence their calculations and the assumptions they must make.

  • #61
ssj said:
So if I replace L with 2 I get T=600/sin50 meaning T=783.
Correct. But realize you don't have to "replace" L with any number--the L just drops out:
300*L=L/2*T*Sin50 becomes 300=1/2*T*Sin50 if you cancel the L on both sides. (Which means dividing both sides by L.)

Now I am stuck with the following question ...
Another equilibrium problem. What are the conditions for equilibrium? Forces must add to zero and torques about any point must add to zero. Label the vertical support forces (at each end of the bridge) F1 & F2. Hint: A good point to use as a pivot is either end of the bridge.
 
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  • #62
Doc Al said:
Another equilibrium problem. What are the conditions for equilibrium? Forces must add to zero and torques about any point must add to zero. Label the vertical support forces (at each end of the bridge) F1 & F2. Hint: A good point to use as a pivot is either end of the bridge.

Alright so if I lable the force between 4kN and 10kN F1 and from 10kN to 8kN F2 and place an pivot at the left hand side of the bridge would this be correct and what would the equation be for the clockwise rotation= counter clockwise rotation ?

As seen below
http://img211.imageshack.us/img211/5756/klcw7.th.jpg
 
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  • #63
ssj said:
Alright so if I lable the force between 4kN and 10kN F1 and from 10kN to 8kN F2 and place an pivot at the left hand side of the bridge would this be correct and what would the equation be for the clockwise rotation= counter clockwise rotation ?
No. F1 is the upward force of the support on the left end of the bridge, F2 is the upward force on the right end of the bridge.

If you pick the left side of the bridge as your pivot, you will have the three axial loads producing the clockwise torques and the force F2 producing the counter-clockwise torques. Find those torques just like in all the previous problems.
 
  • #64
Would it be *F1 14kN*10.5M=8kN*5.25M*F2 ? am I correct ?
 
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  • #65
ssj said:
Would it be *F1 14kN*10.5M=8kN*5.25M*F2 ? am I correct ?
No. There are three forces (each at a different distance from the pivot) that create clockwise torques. And the force F2, which creates the counter-clockwise torque, is how far from the pivot?

I encourage you to study the examples discussed on the webpage I linked in post #37.
 
  • #66
Doc Al said:
No. There are three forces (each at a different distance from the pivot) that create clockwise torques. And the force F2, which creates the counter-clockwise torque, is how far from the pivot?

I encourage you to study the examples discussed on the webpage I linked in post #37.

Alright so is this correct ?
kl.jpg

Meaning that ths distance from the pivot is 5.25M+5.25M?
 
  • #67
No, the distance of F2 from the left pivot point is: 7m + 3.5m + 5.25m + 5.25m = 21m. (You didn't even have to do that addition: the problem states that the bridge supports are 21m apart.)
 
  • #68
Doc Al said:
No, the distance of F2 from the left pivot point is: 7m + 3.5m + 5.25m + 5.25m = 21m. (You didn't even have to do that addition: the problem states that the bridge supports are 21m apart.)

I think the total Clockwise force F1 would be equal to 22kN*21m=462 correct but what would the counter clockwise force be then?
 
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