- #1
hang
- 7
- 1
- Homework Statement
- seesaw
- Relevant Equations
- torque
How Does Using ##m## Instead of ##M = 2m## Affect the Seesaw and Spring System?
AI Thread Summary
The discussion centers on calculating the torque about the pivot point S in a seesaw and spring system. The user initially calculates the torque from the plate's weight and the spring force but overlooks the impact of using ##m## instead of the correct mass ##M = 2m## for the disk. It is clarified that the weight of the rod does not contribute to torque since it acts directly on the pivot. The confusion arises from the incorrect mass assumption, which affects the overall torque calculation. Accurate mass representation is crucial for determining the system's behavior.
- #2
hang
- 7
- 1
The question ask to find torque about the pivot point S.
I first find the torque due to gravity of the plate and then torque due to spring force.
I understand that the weight of the rod do no torque as it is acting on the pivot S.
Where have I gone wrong?
I first find the torque due to gravity of the plate and then torque due to spring force.
I understand that the weight of the rod do no torque as it is acting on the pivot S.
Where have I gone wrong?
- #3
- 22,812
- 14,864
You used ##m## instead of ##M = 2m## for the mass of the disk.
- #4
hang
- 7
- 1
thank you so muchOrodruin said:
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}...
Here we know that if block B is going to move up or just be at the verge of moving up
##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ##
Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
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