Recent content by Impathy

Actually, it would make sense to use the "new" position and the "old" force, I suppose, right?

Alright ... do I just do that to the position vector, or do I do it to the force vector as well and then cross product the new position and new force? Or do I just use the new position and the old force?

I have a two-part problem. The first part gve me a position vector r and a Force vector F acting upon r and I needed to find the magnitude of the torque about the origin. I took rxF and got the correct answer. Part two asks me to find the torque about the a new point (a,b). So instead of r...

Awesome, got it. Thanks a million for your help! :smile:

The problem: A projectile of mass m=1.8 kg moves to the right with speed v=24.8 m/s. THe projectile strikes and sticks to the end of a stationary rod of mass M=6.75 kg and length d=1.71 m that is pivoted about a frictionless axle through its center. Find the angular speed of the system right...

Thanks for your help -- I got it. :) I just needed 2l/t instead of l/t and it worked out. :)

Disregard this post, I will reply when I can get the equations to work for me! Sorry ... Here's a problem I'm struggling with. It says I have a can of soup with a given mass (m), height (h), and diameter (2r). It's placed at rest on the top of an incline with a given length (l) and angle...

Well (somebody correct me if I'm wrong!) ... when water falls from a high place, its potential energy decreases. For a waterfall, the change in PE per second is equal to the rate times acceleration due to gravity times the height. When converted into electrical energy, the power is equal to the...

Ahh, I see what you mean. Does this seem reasonable then? a = \frac{F cos \left(\theta-\phi\right) - mg sin \phi - \mu mgcos\phi}{m} ... Wait, I think I get part of what you're saying ... maybe not all of it. Hm. If I use the above equation I get 2.9518 m/s², which seems physically...

I don't know the answer ... I get one more chance to submit, so I want to be sure! And I want to understand it ... how did you come about that answer? Did you use a formula similar to one above?

Alright, I see what you mean. I guess I'm confused, because I thought I was rotating the axis already (hence the mgsin). I tried just solving it algabraeically like you suggested ... when I rotate my axis, would my equation look like this: a = \frac{F cos \theta - mg sin \phi - \mu mg}{m}...

Hello ... looking for some assistance! I have one two-part question I'm struggling with. I got the first part of the first question: A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 160 N at an angle of 35...