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

  • Thread starter Thread starter tzeteng86
  • Start date Start date
AI Thread Summary
The discussion focuses on solving a physics problem involving spring tension and angles, with the primary equations derived from force equilibrium and geometric relationships. The tension in the spring is expressed as T = k(L - 0.5), where k is the spring constant. The vertical force equilibrium equation is established as 3T cos(θ) = mg, with mg representing the weight of the ball. Participants emphasize the need for two independent equations to solve for the two unknowns, h and θ, using relationships from geometry and force equilibrium. The conversation concludes with a suggestion to use the Pythagorean theorem to relate L, h, and the known dimensions to derive a solution.
tzeteng86
Messages
3
Reaction score
0
Expert please help me solve this question, urgent..

help11222222.jpg


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:
Physics news on Phys.org
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@.
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top