Hello, my name is matt and i have a problem.

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Matt seeks help with a physics lab involving a balance of torques and forces on a horizontal stick with attached weights. He describes an equilibrium scenario with two unknown weights, M1 and M2, and provides a diagram for context. The solution involves calculating torque by multiplying each weight by its distance from a reference point and setting the total torque to zero, as the system is balanced. Additionally, the net force must also equal zero, leading to a second equation. This approach allows for solving the two unknowns, M1 and M2, effectively.
flutieflakes
howdy. I'm not too good at physics, and i have a lab due tomorrow. it isn't too complex of a problem, so help would be appreciated. if you are willing to help, say "asyemptote" and i will type the problem. thanks a bundle
 
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If you have a question, post it. If not get to work on your lab.
 
ok here it is (i just wanted to see if anyone was still in the forum

M1 M2
| |
22cm| 78cm|
----------------------------------
6.7cm| 68cm|
| |
70g 260g

The horizontal line is a 100cm yard stick weighing 120g
The verticle lines are strings attached to weights (the ones going up are attached to pulleys)
The Lengths next to the strings are the distance from the left end of the stick to the strings.
What would I do to determine the weight of M1 and M2?
The apparatus is hanging in equilibrium.
Thanks
 
that didnt work too well, but imagine the weights correspond to the lenghs above or below them. (if you still don't get it, that's ok. it's a pretty confusing diagram of it.)
 
The "torque" of a force around a point measures how much "twist" it causes about that point and is the product of the force times the distance from the point.

What you need to do here is multiply each weight time the distance of that weight from the left end (just since that is the info you are given- since this object ISN'T "twisting", it is "balanced" about any point). Be sure to take those weights hanging down to be negative and those weight over the pulleys to be positive (their pull is upward). Use "M1" and "M2" for the unknown weights. Add all of those to get the total "torque". Since the object is in equilibrium, it is not twisting and that torque must be equal to 0. That gives you one equation. Because the object is also not moving directly up or down, the net force must also be 0: ignoring the distances, add all of the weights (again, positive for those going over the pulley, negative for those hanging down) and set that equal to 0.

You now have two equations to solve for M1 and M2.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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