Homework Statement
Here is the problem I am stuck on. I have checked my process multiple times, but have come up with the same wrong result. I would like to find out where by error of thinking lies.
"An infinitely long line charge of uniform linear charge density λ = -1.30 µC/m lies parallel...
So the tension on the left side of the pulley would be the tension caused by the frictional force acting on M, and the tension on the right side of the pulley would be caused by the weight of m? (They are in opposite directions)
So the pulley disk is rotating. The string wraps around the top of the disk, and it rotates in the same direction as the acceleration of the two blocks (the same way the string is moving). Would this be a torque force acting on pulley disk? Because the force on the disk by the string is...
1. Homework Statement
I am stuck on this problem involving tension, acceleration of a system, friction and a pulley. I have an idea of how to find the acceleration of the system neglecting the mass of the pulley, but it is not the solution. How does adding a frictionless pulley in the system...
Homework Statement
I am stuck on the last problem.
Homework Equations
Just the error propagation equations
The Attempt at a Solution
I initially used the multiplication error propagation formula. So the average force would be the impulse divided by the time, the same thing as the impulse...
So I used the conservation of p equation like this:
mv0i + 2m0.5v0j = mvf + 2m0.25v0i
But I know this must be wrong because the vectors here aren't adding by components.
Homework Statement
I'm stuck on this problem, and I don't really know how to approach it.
Homework Equations
Pretty much just p=mv
And the conservation of linear momentum: total initial mv = total final mv
The Attempt at a Solution
I tried just plugging in the variables into the...
With the kinematic equations, this is what I did:
vf2 = vi2 - 2a(xf-xi)
So vi2 = 7.62 = 57.76 m/s.
And vf2 = 02 = 0 m/s.
xf - xi = 0.6 m.
Solving for a:
-57.76 = -2a(0.6)
a = 48.133 m/s/s
a = -48.133 m/s/s because it is in the opposite direction of the ball's velocity.
With a, I find the time...
Homework Statement
I figured out this homework problem, after many tries, but I'm confused about the correct final answer. Here's the question:
The rebounding ball, mass of 0.06 kg, traveling horizontally at 7.6 m/s, is caught by a player who brings it to rest. During the process, her hand...
Ok, so for each case I can solve for the acceleration for the entire system.
Then I break up the system into each mass component and treat them independently.
The Fnet of m1 would be T-F21 = m1a
The Fnet of m2 would be F12-F32 = m2a
The Fnet of m3 would be F23 = m3a
I first solve for F21 by...
Ok, so for m2, the Fnet equation would be F12-F32 = m2a2
Then I could solve for a2 by dividing Fnet by m2.
But how could you calculate the individual forces, F12 and F32?
Thanks for the reply.
Treating each cart as its own system, I'm getting here: (to the right is positive force, to the left is negative force)
The net force acting on m1 is T-F21. Which is equal to 3-(m2a)?
The net force acting on m2 is F12-F32. Which is equal to (m1a)-(m3a)?
The net force...
Homework Statement
I am stuck on this problem:
So the question is asking for F21, F12, F32, and F23. (That's the force by 2 on 1).
The mass of m2 is variable.
There are four cases where I need to solve for the forces: when m2 has mass of 3kg, 0.3kg, 0.03kg, and 0kg.
The mass of m1 = 2.5kg, the...
Ooooh. That's an embarrassing mistake. I knew where the center of mass would be, because the stick is uniform. But I thought you were supposed to subtract the overlapping length. Thanks all.
Homework Statement
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I don't understand what I'm doing wrong with this problem:
An old club-ax consists of a symmetrical 8.3 kg stone attached to the end of a uniform 2.9 kg stick. The stick is 98 cm long, and the 18 cm long stone is drilled through its center and mounted on one end of the...