How Can You Solve a Physics Problem Involving Spring Tension and Angles?

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Homework Help Overview

The discussion revolves around a physics problem involving spring tension, angles, and force equilibrium. Participants are exploring the relationships between tension in springs, angles with the vertical, and geometric relationships involving lengths and heights.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to establish equations based on the tension in the springs and the geometric relationships between the lengths and angles. There is a focus on deriving multiple equations to solve for the unknowns, particularly the angle and height.

Discussion Status

Some participants have provided guidance on setting up the equations needed to relate the variables, while others express difficulty in isolating the angle or height due to the presence of multiple unknowns. The discussion is ongoing, with various relationships being explored without a clear consensus on the next steps.

Contextual Notes

There is an emphasis on the need for additional independent relationships to solve for the unknowns, as well as references to specific values related to the spring constant and the weight of the object involved. Participants are also noting the constraints of the problem setup.

tzeteng86
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Expert please help me solve this question, urgent..

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My steps are stucked as below

3 T cos @ = 400, T = tension of each spring, @ = angle of spring with vertical axis
L sin @ = 0.5
L cos @ = h
tan @ = 0.5/h

with L = length of spring

T = 1000 times x where x is the stretch length

3000x cos @ = 400
x = L - 0.5
3000 (L-0.5) cos @ = 400
3000 (h) - 1500 cos @ = 400
3000 (h) = 400 + 1500 cos @
h = (400 + 1500 cos @) / 3000

and the I don't know how...
 
Last edited:
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You are using the right approach. The Tension, T, in the spring is given by the product of the spring constant, k (= 1000 N/m) and the displacement from equilibrium, x, where x = stretched length - unstreteched length = L - 0.5 m.

But if one has two variables, then one needs two equations to solve for the variables.

Edit: deleted ref to lateral forces.
 
Last edited:
I'm sorry but can I have the solution to get the final value for the question?
 
One has both a geometric relationship, e.g. L2 = (0.5 m)2 + h2, or tan @ = 0.5/h, or h = L cos @.

And one has the force equilibrium equation in the vertical direction, as you indicated. T in the spring is simply T = k (L-0.5).

See where that get's you.
 
I tried.. but I'm not able to solve the question because I couldn't have the value of @...
 
OK, let's try again.

One has a force equilibrium equation (relationship) in which there are two unknowns, h and @. So one needs another independent relationship between h and @.

So looking at the force equilibrium in the vertical direction, one has

3 Ty = mg, or the vertical force components of the springs = weight of the ball.

Now, mg = 400 N, the ball's weight (it's mass would be about 40.8 kg), and T = k (L-0.5), where k = 1000 N/m and L is a function of h and @, and Ty = T cos @.

So the force equation is:

3 [1000 N/m *(L-0.5)] cos @ = 400 N, but

cos @ = h/L, so

3 [ 1000 N/m *(L-0.5)] h/L = 400 N

I'll leave the algebra for one to finish.

But from the Pythagorean theorem - L2 = h2 + 0.52 m2, or

L\,=\,\sqrt{h^2\,+\,0.5^2}.

So one then has an equation for h in terms of the known dimension, weight of ball, and spring constant.

Using lateral forces would not help since one simply shows that Tx = Tx, or T sin@ = T sin@.
 

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