- #1
cnoa
- 5
- 0
How can they be equal at that point? And where is the force that the string applies.
By the way Fk = Buoyancy Force
A Question About Buoyancy Force (I Think My Teacher Is Wrong)
- Thread starter cnoa
- Start date
How can they be equal at that point? And where is the force that the string applies.
By the way Fk = Buoyancy Force
- #2
ehild
Science Advisor
Homework Helper
- 15,536
- 1,917
It comes from the net torque with respect to the right end of the rod. It has to be zero in equilibrium.
ehild
ehild
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question.
Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point?
Lets call the point which connects the string and rod as P.
Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
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