Recent content by BSCS

Ok, this makes sense. So, I can have a separate Ug = 0 point for each element in the system?

1.24 m/s is correct.

"0.65m diamter", you are correct.

I can't comment on your sanity :rolleyes: , but that goes back to my problem... How can it be 20 cm for the hanging block if Ug is not 0 at the point we would expect?

\Delta U_{g} = (m_{2}sin\theta-m_{1})gx \Delta U_{g} = [20.0 kg sin40 - 30.0 kg](9.80)(0.200) = -33.6 N

(2.53 x 60) / (2 * pi * .65) = 37.1688006 I think you're forgetting r.

A 20.0 kg block is connected to a 30.0 kg block by a string that passes over a light frictionless pulley. The 30.0 kg block is connected to a spring that has negligible mass and a force constant of 250 N/m, as shown in the figure. The spring is unstretched when the system is as shown...

Yes, and it/they were getting complex, so I wanted to get some feedback. I came up with the force exerted down on the chair as having an x component and a y component (different from what I posted). From there I was thinking I could apply that vector to the FBD of the chair and come up with my...

I'm trying to think up situations that will help me understand more complicated situations... Let's say I have: a box on a slanted slope on a chair on the floor The floor exerts a normal force on the chair. But, equal to what? is it: a box m_{1}gcos\theta on a slanted slope m_{2}g on a...

I have a question: If R were to drop to zero, the coaster would fall, right? So, for v=9.9 m/s, it would actually lose contact with the track and fall. But for any value even minutely larger than R=0 (and thus v=9.9 m/s) the coaster would just barely "stick" to the track? Even for a super...

I didn't mean to give the impression that I thought it was easy. I felt stupid because I made algebraic mistakes that I shouldn't have made. After spending so much time on the problem, I guess I was getting used to the "situation" in the problem... Thanks again.

Thanks! I got it. Boy, do I feel stupid. I think it looks ok, now, right? (this isn't homework, btw, it's just extra practice) I made 2 algebraic errors: 1) when I pulled out the N, I didn't change the - to a + 2) when I divided by m, I didn't divide Q by m I found them while...

Thanks for the help. I tried the problem over again using ma=Q/2 and was able to easily solve both (a) and (b). FYI The correct answer is listed as having the cos and sin reversed from what you said... I'm a bit concerned though... I was shown how to do part (a) as follows: Find...

Dorothy and Doc Al, Thanks for the replies. My original reply was lost... Question: For UCM problems like this in general, is it generally safe to ignore forces that would act in the tangential direction? (in this case mg*cos(theta)) Thanks again!

Given: 2 wedge shaped blocks arranged as a rectangle. The bottom block is on a frictionless horizontal surface. Horizontal force Q is applied to the vertical side of the bottom block. Find Q max, the maximum value of Q where there is no slipping, in terms of u[SIZE="1"]s, m, and theta...