Solving Mechanics Statics Homework Problem

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The discussion revolves around solving a mechanics statics homework problem involving forces and angles. The user has established two equilibrium equations but is left with three unknowns: the distance d, angle θ, and force F(rod). A relationship between d and θ is introduced, leading to a complex equation that appears to be a quadratic. Suggestions include squaring both sides of the equation to simplify the problem. The user expresses difficulty in solving the resulting equation, highlighting their limited experience in the subject.
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Homework Statement



http://s11.postimage.org/yvt7eeg35/image.png

Homework Equations



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
 
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JeeebeZ said:

Homework Statement



http://s11.postimage.org/yvt7eeg35/image.png

Homework Equations



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
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 ?
 
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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 into this class.
 
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?
 
JeeebeZ said:
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 into this class.
Flip it over, square both sides.

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