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# M1 Forces help watch

1. A light rod in equilibrium has two forces acting on it.

If I use the horizontal component of F2 and resolve horizontally:

F2 sin a = F1

If I use the component of F1 perpendicular to the rod and resolve perpendicular to the rod:

F1 sin a = F2

Have I made a mistake? Why are these equations different?
2. Looks fine.

What do you mean 'different'?
3. (Original post by WhisperingTide)
Looks fine.

What do you mean 'different'?
F1 = F2 sin a (1)
F1 sin a = F2 (2)

Plug (1) into (2):

F2 sin^2 a = F2

sin^2 a = 1

But 'a' could be any acute angle?
4. (Original post by 0-))
F1 = F2 sin a (1)
F1 sin a = F2 (2)

Plug (1) into (2):

F2 sin^2 a = F2

sin^2 a = 1

But 'a' could be any acute angle?
Yeah, I came to the same conclusion.
I don't see how what you've done could be wrong, though, unless I'm having a brain fart.

I rationalised it by basically saying that if that rod is supposed to be in equilibrium, you'll have moments and therefor distances to be concerned with. (As it is that thing is going to spin counter-clockwise)
5. (Original post by WhisperingTide)
Yeah, I came to the same conclusion.
I don't see how what you've done could be wrong, though, unless I'm having a brain fart.

I rationalised it by basically saying that if that rod is supposed to be in equilibrium, you'll have moments and therefor distances to be concerned with. (As it is that thing is going to spin counter-clockwise)
Maybe it's impossible for this rod to be in equilibrium for an acute angle 'a'?
6. (Original post by 0-))
Maybe it's impossible for this rod to be in equilibrium for an acute angle 'a'?
If it's not fixed at a point then it will be, yeah.
7. (Original post by WhisperingTide)
If it's not fixed at a point then it will be, yeah.
Also in reality the rod would have weight so that would be included when resolving and it would make the equations work.

Thanks for helping.

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