Recent content by ace214

... Crap Yeah that did it. Thanks. So for anyone searching the formula becomes I = Fr^2/a

A 15.0-kg object is attached to a cord that is wrapped around a wheel of radius r = 9.0 cm (Fig. P8.60). The acceleration of the object down the frictionless incline is measured to be 2.00 m/s2. Assume the axle of the wheel to be frictionless. (a) Determine the tension in the rope. 58.46681...

Oh... I guess I didn't fully learn the clockwise/counterclockwise thing. I was just working off an X and Y axis and the 10N was going in the -y direction. I understand now. Thanks for the help.

It works with -10(.15) -T1cos50(.15) + T1sin50(.3) = 0 as PhantomJay said. I just don't understand why the t1cos is negative. ADD: For completion's sake, my answers were: T1 = 11.24478 T2 = 1.385996 F = 7.228

Why should it be minus? It's going out in the positive direction Is my equation right then instead of poofaces? I don't understand why the torques on the axes wouldn't be 0. ADD: you are right about the minus (I'm getting right answers now), but will you explain why?

Ok so I have: T1sin(50) + T2 - 10N = 0 T1cos(50) - F = 0 -10N(.15m) + T1cos50(.15m) + T1sin50(.3m) = 0 Using the bottom left corner of the picture as the origin which makes the torques due to T2 and F 0 because they're on the axes. This gave me: T1 = 4.59796 T2 = 6.47776 F = 2.9555 None of...

None of those are right... Seemed right to me. Any more help from anyone? This is due 8:30 AM EST on Friday (tomorrow).

A uniform 10.0-N picture frame is supported as shown in Figure P8.62. Find the tension in the cords and the magnitude of the horizontal force at P that are required to hold the frame in the position shown. T = Fxsintheta So I got what's going on in each direction in a separate equation but I...

Ok, I was an idiot and didn't use a negative velocity in the original momentum equation... Wish somebody had caught it but oh well. Also the v1 - v2 = v2 - v1 does work for objects with different masses.

Ok, I redid the KE equation and got 32 and 24 for the larger mass. I've tried 24 already as I got it from the equation with just velocities above and it wasn't right... Gaaaaaah...

Also, the book says that that equation is supposed to work.

When I did that, I got imaginary numbers from the quadratic.

No there's two equations... I already said that. I used m1v1i + m2v2i = m1v1f + m2v2f and solved for v1f in terms of v2f and plugged this into v1i - v2i = v2f - v1f

I have also tried v2i - v1i = v1f - v2f with positive numbers and v1i +v2i = v1f + v2f with properly-signed numbers... Someone please help...

A 10.0 g object moving to the right at 22.0 cm/s makes an elastic head-on collision with a 15.0 g object moving in the opposite direction at 32.0 cm/s. Find the velocity of each object after the collision. First, I converted the masses to kg and the velocities to m/s. I used m1v1i + m2v2i =...