1. Aug 5, 2006

tzeteng86

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: Aug 5, 2006
2. Aug 5, 2006

Staff: Mentor

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: Aug 5, 2006
3. Aug 5, 2006

tzeteng86

I'm sorry but can I have the solution to get the final value for the question?

4. Aug 5, 2006

Staff: Mentor

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.

5. Aug 5, 2006

tzeteng86

I tried.. but I'm not able to solve the question because I couldn't have the value of @...

6. Aug 5, 2006

Staff: Mentor

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@.