- #1
pedro_crusader
- 1
- 0
- Homework Statement
- I need to understand where the very first equation comes from?
- Relevant Equations
- torque?
Torques acting on a cylinder, with friction
- Thread starter pedro_crusader
- Start date
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- Tags
- Torque
AI Thread Summary
The discussion focuses on a problem involving a rolling cylinder that appears to be dragging a rod with a plate experiencing friction. Participants express the need for a detailed description of the problem rather than assumptions. There is confusion regarding the meaning of labeled arrows in the diagram, particularly the curvy arrows labeled ##\beta## and ##\omega##. A torque equation is mentioned, but its derivation is unclear without additional context. Clarity on these points is essential for a proper understanding of the mechanics involved.
- #2
- 15,778
- 8,960
Please describe the problem in detail not just part of the proffered solution. What is going on here? It looks like you have a rolling cylinder that is dragging a rod at the end of which is a plate with friction. However, this is only a guess and we do not like guessing if it can be avoided.
Furthermore, what do all these labeled arrows represent, especially the curvy arrows labeled ##\beta## and ##\omega##? Again, we do not like guessing.
That said, yes, the first equation looks like a torque equation. Where it comes from depends on details that we do not have.
Furthermore, what do all these labeled arrows represent, especially the curvy arrows labeled ##\beta## and ##\omega##? Again, we do not like guessing.
That said, yes, the first equation looks like a torque equation. Where it comes from depends on details that we do not have.
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
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}...
I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?
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