Clearly if ##F = 0## and ##\tan\theta > \mu_s##, then using the above equations for ##f_s## and ##n##, we get ##f_s > \mu_s n## so the block will slip. However, it seems that as long as the force ##F## is directed to the right with a certain minimum magnitude, namely ##\frac{\tan\theta -...
@PeroK thanks for the hint but can you provide some equations?
From my understanding, one should find the component of the normal force along his legs and find the component of this force along the ground. This gives ##F_s = N\cos \theta \cos \phi## where ##\theta## is the angle between the...
I get how to solve (a); my method involves finding the net torque about the man's hands and setting it to zero, which can be used to solve for the normal force acting on his feet and the normal force on his hands can be solved using Newton's first law. Then divide by 2 for each to get the normal...
I'm assuming there is a torque because that seems to be implied by the question. If there is no torque, then since the angular velocity was initially zero, it should remain that way; that is, the ruler should remain at rest and not rotate.
I think both spools will land about at the spot x (there aren't any horizontal forces causing them to land away from the x).
Also, I think student 2 might be the closest to being correct, but I'm not sure if they're entirely correct; isn't it possible that force doesn't only go to translational...
Thanks for offering that hint. I said gravitational force is one, and I think it acts about the center of mass. But what about the others? That's the whole point of asking my question.
I think the angular acceleration is counterclockwise (and thus so is the torque) in the diagram, but what would a free body diagram look like? After the system is released from rest, isn't the only force the gravitational force about the center of mass? And if so, what's causing the angular...
The formula for ##\vec{E}## is ##\vec{E} = \hat{j} E_{max} \cos(kx - \omega t)## and the formula for ##\vec{B}## is ##\vec{B} = \hat{k} B_{max} \cos(kx - \omega t)##, where ##\omega## is the angular frequency and ##k## is the wave number, equal to ##2\pi## divided by the wavelength, ##t##...
Ok, so to simplify things I'll redefine ##v## and ##v'## so that they can be negative (e.g. they are velocities). Conservation of momentum says ##mv + MV = mv' + MV' (1)## in both parts A and B, where ##v## and ##v'## are the initial and final speeds of the ball and ##V## and ##V'## are the...