Recent content by PhoniexGuy

Hmm, perhaps the diagram isn't clear, the object is moving back and forth on the pendulum, but in sort of an arc like so: http://img11.imageshack.us/img11/9210/85456919.png Does this change anything?

Homework Statement http://img32.imageshack.us/img32/1551/filevno.jpg I have to find the tension for the rope when angle is 20 degrees, the object is moving back and forth. I know the mass is 2kg, gravity is 10N/kg Homework Equations F_t * cos θ = mg. So F_t = mg/cos θ The...

Hmm, okay, that makes sense! Thanks, but is the the answer of 25.04 right? Wait, why is it plus and not minus? Fn μs + mg sin(θ) = F cos(θ)?

Hmm.. So it's keeping it static in the x direction? or parallel to the ramp?

Hmm, ah, just misplaced negative, also, what is the purpose behind the second equation, why are you balancing those out?

Hmm, I got -25N for the force of F? Is that right? Also can you explain the principle behind the second equation?

awesome, thanks!

So, does it become something like: Fn * Us + mg sin theta = F cos theta?

Wait, what? So this is okay? Fn = mg cos theta + F sin theta And how do I set up the second equation? I don't understand what you wrote for it...

So, I think I got it. Fn = mg cos theta + F sin theta and then F = mg tan theta So F = 16.9 N and Fn = 26.10N, so max Ffric = 23.49 / cos 40 deg (component) = 30.7N for Max force F?

@ap123, so the block is not moving in the start. So I just have to find the max friction can be, and then I have my component of max force F?

I found the acceleration the block would have if there was no horizontal force, so basically the force down the incline. And then to balance it i divided by cos 40. I drew the diagram, but still don't get it..

Hmm, I think I get it. So acceleration is the F|| - ukFn right? Then a = m(sin 40 - .3 * cos 40) And i get acceleration =8.39 m/s/s Then force F = 8.39/cos40 = 10.95N, so max F = 10.95 N? Or is this wrong?

Homework Statement Assume the ramp and block make the following coefficients of friction: (μs=0.9 μk=0.3) and the 2 kg mass is placed at rest on the ramp and a horizontal force F is also applied. http://img69.imageshack.us/img69/4097/filepq.jpg What is the biggest magnitude force F...

Actually, nevermind. Thank's for all the help, i understand it now! (asked teacher)