Recent content by Drew777

Sorry. I was actually referring to the quote you put into the thread. It sounded like it was for someone else. I see why you posted it as an example though.

Was this meant for another thread?

If an air hockey table is turned on and one end is elevated so the hockey puck slides to the other end freely. Would it take longer for a heavier puck to reach the end of the table before a lighter puck or vice versa?

What was done is I slid a puck down an air table. I did this 7 times, each with a different weight. I calculated the potential energy of the high point and calculated the potential energy at the low point. The KEf I measured was using 1/2(m)(Vf^2). So the KEf is the kinetic energy from the high...

Sorry I wrote it down wrong. I went back to edit what I meant. What I have is the potential energy of a low point that I will call PEf and I got the kinetic energy it took for the object to move down the incline(KEf). I added them together to get Ef. Now I have to take this number Ef and...

I understand that. When I take the the potential energy of the higher point and subtract it from the summation of energy I get from (PEf +KEf) I get a negative number. What does this mean?

So is it possible to get a negative number as the kinetic energy of the higher elevation?

What would it mean when you subtract potential energy of a higher elevation from the summation of energy(kinetic + potential of a lower elevation)?

Homework Statement Uniform rod 200cm in legth is in equilibrium. An upward force of 200 dynes is applied 15 cm from the right edge. If the rod is 50dynes. At what pint will rod rotate about? The Attempt at a Solution Just one of many -50(100)+200(185) -5000+37000 =32000...

I guess it is taking everything into consideration and making the weight the downward force. How can this rod stay in equilibrium is what I am wondering?

I was wondering if this problem could be solved as it is or if there is something missing from the problem. A uniform rod 200cm in length is in equilibrium. An upward force of 200 dynes is applied 15cm from the right edge. If the rod is 50 dynes. At what point will the rod rotate about?

Should I expect a frictional force to get bigger or smaller as an angle is increased?

I keep thinking something is missing as well. It is nerve racking. You seem to know how to do this problem. Is there nothing missing from it? E= sum of EF= 200-50 (This should equal zero) ET= should equal zero (200)(20)-(50)(x)=0 Best I can come up with is this. -50x=(200)(20)...

LOL. Thanks everybody. The problem is hard to understand. My teacher is of very little help and won't clear up anything for me. We just now got into dealing with torque. I say we have spent one hour on it so far. What we went over was nothing like this problem, I think that is why I am having...

I agree with you. I am taking this course to learn after all. If you just gave me the solution I would actually be a little upset. The point of me posting the problem is so I can better understand it. I have done many different kinds of equations that are pretty much unrelated. If you need to...