Recent content by Houdini176

Solved. In case anyone is curious, the conservation of energy relation I used is incorrect. The conservation of energy is that simple if there is no friction, but in the case of this problem there is friction. Therefore, the work done by the force applied to block A gets divided into increasing...

Alright cool. It looks like you just accidentally multiplied 2*1.5 incorrectly, and used the wrong moment of inertia for the ball.

I'm not too sure if I'm right, but hopefully this is one of the answer choices. After the collision, the ball and the turntable should be turning together. Ii*Wi=If*Wf You can just add up the moments of inertia, so: If=Turntable Moment + Ball Moment The ball is small enough that we can treat...

Homework Statement "A side view of a simplified form of vertical latch is as shown. The lower member A can be pushed forward in its horizontal channel. The sides of the channels are smooth, but at the interfaces of A and B, which are at 45 degrees with the horizontal, there exists a static...

Sweet, good to hear. Good luck on your final, and it was my pleasure.

Yea, sorry about that the book provides a figure, but I did not. The tape is at the top of the cylinder.

No problem, I'll do it all out right here M=Mass of wheel m=mass of block/weight Tension/Torque Equation T1-T2=Iaangular/r T1-T2=Ialinear/r^2 T1-T2=Ma/2 T1=Ma/2+T2 (T2=10N, I'll leave it as a variable) Force Equation T1-mg=-ma T1+ma=mg Here you combine the two, substitute...

Sorry about that, I edited my post. d is the distance the cylinder travels along the ramp.

Heh okay cool, I hope I answered your question. Good luck on your final!

T1 is on the left, T2 on the right. T1 applies the counterclockwise torque to the pulley, and T2 applies the clockwise torque to the pulley. You can think of it like this, the rope transmits a particular force (What we call tension) from the mass to the pulley and vice versa. Because of...

Oh, my bad, I was a little inconsistent with our definitions of T1 and T2. For mine, I mean T1 to be the counterclockwise tension above the mass that is going to fall. In other words, I would have written: T1-m1g=-m1a, instead of T2-m2g=-m2g To make our posts consistent, I'll switch T1 with T2...

Okay, well, remember to just take it one step at a time. It looks like you've make the force and the torque the same thing, but you should leave them apart and find one equation for torque and one for force. Torquenet=T1*r-T2*r=T1*r-10*r So all I've done here is taken the two torques and...

Torque=I*aangular so (T2+(-T1))=I*alinear/r^2 Since It's T2+(-T1), not the other way around, you're taking a tension that goes in the counterclockwise direction and adding a tension that goes in the clockwise direction. When adding these tensions/torques to find the net torque, you have defined...

The counterclockwise tension equals T*r, where T=tension in the string and r is the radius of the pulley. Because the mass is accelerating downward, you cannot say that tension on the string on the "left side" of the pulley is equal to the mass. If that were true, the mass wouldn't accelerate...

Homework Statement A solid cylinder of weight 50 lb (mass=2.3 kg) and radius 3.0 in (7.6 cm) has a light thin tape wound around it. The tape passes over a light, smoother fixed pulley to a 10-lb (mass=4.5 kg) body, hanging vertically. If the plane on which the cylinder moves is inclined 30...