# Mechanics Statics

1. Oct 2, 2012

### JeeebeZ

1. The problem statement, all variables and given/known data

http://s11.postimage.org/yvt7eeg35/image.png [Broken]

2. Relevant equations

3. The attempt at a solution

So I get that

ƩFy = 0 = F(rod)y - F(weight)

ƩFx = 0 = F(rod)x - F(spring)

F(spring) = 200N * (1.5m + d)
d being the unknown distance that the rod has moved by.

So I get
F(rod) = 60N / sinθ

F(rod) = (200N * (1.5m + d)) / cosθ

Which leaves me with 3 unknowns.
d & θ & F(rod)

But only 2 equations

So I have no idea where to go from here

Last edited by a moderator: May 6, 2017
2. Oct 2, 2012

### SammyS

Staff Emeritus
The equilibrium length of the spring is 1.5 m, so the force exerted by the spring is 200*d.

There is another relationship available. How is d related to θ ?

Last edited by a moderator: May 6, 2017
3. Oct 2, 2012

### JeeebeZ

So,

cosθ = (1.5 - d) / 1.5

sinθ = √(1.52 - (1.5 - d)2) / 1.5

which would leave me with

60 / √(1.52 - (1.5 - d)2) = 200d / (1.5 - d)

What seems like a really hard problem to break down for d since to get rid of the square root I have to times everything then I'll end up with a d to powers of 1 2 3 and 4... I don't know how to solve that.

I'm only 6 days in to this class.

4. Oct 2, 2012

### runningninja

Remember, you can combine trigonometric expressions. You don't necessarily have to substitute in for both sin and cos.
Also, you should have gone over solving higher degree polynomials in precalc. What's the highest math you've taken?

5. Oct 2, 2012

### SammyS

Staff Emeritus
Flip it over, square both sides.

Maybe I'm not seeing something, but it looks like a quadratic equation to me.