"Determine a formula for the magnitude of the force F exerted on the large block (m3) in Fig. 4-51 so that the mass m1 does not move relative to m3. Ignore all friction. Assume m2 does not make contact with m3."

Problem:


http://www.nfxclan.net/images/physics/Scan10018.png

What I came up with:


http://www.nfxclan.net/images/physics/Scan10020.png

summination of all forces = 3 normals - weight of block 2 sin angle - weight of block 1 - weight of block 3 - weight of block 2 cos angle = force needed to make block 1 stay?

Do you need to take into consideration that the angles there, since you are to ignore friction, and you are looking for an answer at which m1 does not move wrt m3?

(m1 + m2)a = m2g
a = m2g/(m1+m2)
F = m3(m2g/(m1+m2)))

Yes, I'm pretty sure for the most part since it's acts upon the tension (my guess). The tension is the same on both sides.

There's also normal forces and other things. It's frictionless but not sure.

Well okay...
First, since the system is frictionless, the force of gravity on m1 and m2 is irrelavent.

Hmm, what would be the right to left force on m2? the acceleration left to right of the pully? Sooo.... if T1 is tension from m1 to pully, and T2 is pully to m2, T2 would be... sqrt( (m2*a)^2 + (m2*g)^2)? (A^2 + B^2 = C^2)

If you solve for a, it would seem that if m2 > m1 you would have an imaginary number acceration?! It's late, that can't be right...

no, i'm betting the trig is irrelavent, m2 is moving straight up. The pully must be frictionless, one of them ideal frictionless pullies that translate force at right angles, so T1 = m1a (with respect to m3) and T2 = m2g.

yes yes, that's sounding more betterer to me right now.