Calculate the tension in the string

  • Thread starter Thread starter ssj
  • Start date Start date
  • Tags Tags
    String Tension
Click For Summary
SUMMARY

The forum discussion focuses on calculating the tension in a string under various conditions using torque equations. Participants discuss the application of the formula for torque, defined as Moment = Force x Perpendicular Distance, and how to set up equations for different scenarios involving a 200 N force and angles such as 30 degrees. The correct approach involves equating clockwise and counter-clockwise torques, leading to the conclusion that the tension T can be calculated as T = 400 N for the first problem and T = 200 N for the second problem, with the length of the beam L being irrelevant to the final tension calculation.

PREREQUISITES
  • Understanding of basic physics concepts, specifically torque and equilibrium.
  • Familiarity with trigonometric functions, particularly sine and cosine.
  • Ability to manipulate algebraic equations to solve for unknowns.
  • Knowledge of how to apply the principle of moments in static systems.
NEXT STEPS
  • Study the concept of torque equilibrium in static systems.
  • Learn how to apply trigonometric functions in physics problems, focusing on sine and cosine.
  • Practice solving problems involving tension in strings and beams under various forces.
  • Explore advanced topics in mechanics, such as rotational dynamics and moment of inertia.
USEFUL FOR

Students of physics, engineering students, and anyone interested in understanding the mechanics of tension and torque in static systems.

  • #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.
 
Physics news on Phys.org
  • #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
 
Last edited by a moderator:
  • #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 ?
 
Last edited:
  • #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?
 
Last edited:

Similar threads

Tension in a string connecting two blocks having equal and opposite velocities
  • · Replies 21 ·
Replies
21
Views
889
Waves and vibrations on a string
  • · Replies 2 ·
Replies
2
Views
2K
Finding the frequency of a string based on Mass and Tension
  • · Replies 1 ·
Replies
1
Views
3K
Time it Takes for a Transverse Pulse to Travel a Distance "L" Up a Cable Against Gravity?
  • · Replies 13 ·
Replies
13
Views
2K
Tension in the string while it descends
  • · Replies 8 ·
Replies
8
Views
2K
Tension in 4 strings suspending object at 4 different points
  • · Replies 19 ·
Replies
19
Views
5K
Find the tension in the two strings
  • · Replies 7 ·
Replies
7
Views
2K
To find the modulus of elasticity of a light elastic string
  • · Replies 7 ·
Replies
7
Views
3K
Does Tension Increase with Higher Angles in a String?
  • · Replies 2 ·
Replies
2
Views
2K
Maximum tension in thread connecting loads
  • · Replies 13 ·
Replies
13
Views
1K