Recent content by Eric [Tsu]

So, would the body in my original example have linear momentum? With the external force preventing the motion being the platform of the feet?

18 views and no answer!?

As an example, if you raise your arms in front of you, your center of mass will move slightly forward. However, you would not fall down because your feet's platform won't allow it. Does this constitute linear momentum, or does the entire system need to move in order to have linear momentum...

Here's another without friction: A mass, m1 = 4.00 kg, resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 10.0 kg, as in Figure P4.30. Find the acceleration of each mass and the tension in the cable...

I see, thanks. Assuming I had to deal with friction, how would I apply that?

Got acceleration. In addition to over-thinking it, I forgot a negative sign as you pointed out. (-T) - (68.6sin43) = (m2)a + T + 98 = (m1)a _______________________ -(68.6sin43) + 98 = (m2)a + (m1)a m1 = 10 m2 = 7 Solve for a --> a = 3.01 Edit: Tension = 67.9 N T + 98 = (m1)a T + 98...

Because I'm completely lost. Edit: Does the normal force equal w2y? Edit2: My best attempt at m2s x-axis equation: T-w2x = (m2)a T-(68.6sin43) = (m2)a m1s y-axis equation: T+w1 = (m1)a T+98 = (m1)a I know I'm missing something here, I just don't know what

1. Two packing crates of masses m1 = 10.0 kg and m2 = 7.00 kg are connected by a light string that passes over a frictionless pulley as in Figure P4.26. The 7.00 kg crate lies on a smooth incline of angle 43.0°. (a) Find the acceleration of the 7.00 kg crate. (b) Find the tension in the...

X = Xi + Vi(t) + 1/2(a)(t^2) Xi = 52.5 Vi = 0 a = 4 Chaser --> X = 52.5 + 1/2(4)(t^2) Opponent --> X = v*t Lost at this point

My bad, I wrote it opposite. X = v*t is for opponents position

1. A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 15 m/s, skates by with the puck. After 3.5 s, the first player makes up his mind to chase his opponent. (a) If he accelerates uniformly at 4.0 m/s^2, how long does it take him...