Recent content by theunloved

A mass M in held in place by the applied force F and the pulley system shown below. The pulleys are massless and frictionless. Determine the tension in each section in each section of rope and the applied force F Taking upwards as positve -T1 - F = 0 T4 - T1 - T3 = 0 T2 + T3 - T5 = 0...

In Figure 6-63, a block weighing 22 N is held at rest against a vertical wall by a horizontal force of magnitude 60 N. The coefficient of static friction between the wall and the block is 0.55, and the coefficient of kinetic friction between them is 0.38. In six experiments, a second force is...

I do not understand this problem, and since this is my first 3D problem, I can't figure out how to draw a FBD either, so I can't figure out how many forces acting on the crate. It'd be nice if someone could help me out. Thanks

and this is the last one. Do these two m1 and m2 have the same acceleration ? I think they do, but I'm not so sure... Thanks

So, I have created some blocks, and draw the FBD myself, am not sure if I did it right or not, so I'm going to upload these FBD here, and hope you guys can correct me if I'm wrong. Thanks a lot. A block of m1 and m2, m2 is siting on top of m1. Assume there is no friction on the ground under m1...

Yes, now I got it, so basically if it's put on the ground, 12N is the minimum force that you need to apply in order to move it. So fs maximum will be 12 N, right ? and since we have fs = us * mg = 12N, we could find coefficient of static friction...uhm, or coefficient of kinetic friction ...

A block of mass m1 = 4 kg is put on top of a block of mass m2 = 5 kg. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless...

Alright, so for m1, it should be fk = m1a ;) Very clear explanation, wish that you were our professor college, physics would be fun then ;) Thank you very much for your help.

I should be the one to say thanks to all your hard work Doc, not the other way around ;)

Thanks collinsmark, I finally got it and understand how third Newton's law works. On the other hand, I'm still a bit confused about the sign of the frictional force acting on m1. I uploaded the Picture and my FBD, and as you can see, if m2 is moved from right to left, m1 will be sliding from...

I think it will be opposite side with the horizontal force F ? so fk2 = -m1a a = -fk2 / m1 = -uk(m2g) / m1 ? I still don't get it why frictional force fk2 will be the pulling force on m1, weird...

So you're saying that the frictional force is the only force that makes m1 move ?

I'm sorry to bring it up again, but I'm having the same problem and reading this post helps me figure out how to do it; however, I think that what you're saying is not right Doc Al. I draw a FBD of m and M, and here are all the forces that I've found: Taking downwards as positive, x+ from left...

A slab of mass m1 = 40 kg rests on a frictionless floor. a block of mass m2 = 10 kg rests on top of the slab ? Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. The block is pulled by a horizontal force with a magnitude of 100 N...

Oh I got it, since acceleration is not constant, we can't use all the kinematic equations to solve this problem...But how are we going to find acceleration if we only have one equation ?