HW4 #12

#12 seems to be giving some trouble too. I'll put up this post, but if it doesn't help please ask for more specific suggestions.

Peter Pan is the mass that moves downward vertically. The problem says she starts from vo=0, and you are given the distance and the time, so you can use the x= equation to find her acceleration. That is also the acceleration of the counterweight, which is the block that accelerates upwards on the incline.

Next look at Peter Pan to find the tension in the rope: Fg acts downward, T (the tension in the rope) acts upward, and Peter Pan's acceleration is downward. If you choose downward to be the positive direction (because Peter Pan is accelerating downward) then when you use Newton's 2nd law you get ∑F=Fg-T=ma. (Note that if you had chosen up to be positive then your right hand side would be negative: =-ma.) Relace Fg with m*g to get mg-T=ma, or rearrange to get T=mg-ma. Note that the mass is Peter Pan's mass, which is known, so you can calculate the tension since you already figured out the acceleration.

Now draw a force diagram (free body diagram) for the counterweight on the incline. The force due to gravity is Fg and points downward. The normal force points up-and-to-the-left, perpendicular to the incline. The tension in the rope pulls up-and-to-the-right, along the incline.

Since the counterweight will be accelerating along the incline, make a chart with forces parallel to the incline (it will have acceleration "a"), perpendicular to the incline (will have no acceleration), and "?" (idk) which is in some other direction. The normal force is perpendicular to the incline. The tension in the rope is parallel to the incline. The force due to gravity, Fg, is some other direction, so belongs in the "?" column. Figure out the components of Fg in the parallel and perpendicular directions: Fgsinθ in the parallel direction and Fgcosθ in the perpendicular direction.

Now you can use Newton's 2nd law to look at forces in the direction parallel to the incline (use the direction up-and-to-the right which is the direction of the acceleration as the positive direction): ∑F=T-Fgsinθ=ma And this is NOT equal to zero since there IS an acceleration. Replace Fg with m*g to get T-mgsinθ=ma and solve for the mass, which is the mass of the counterweight which is what you are looking for. "The rest is just algebra."

Hope that helps. Good luck.