Finding Acceleration of Masses in Elevator

AI Thread Summary
An elevator accelerating upwards affects the forces acting on two hanging masses connected by a pulley. The equations of motion for each mass include gravitational force, tension, and the elevator's acceleration. The assumption that the accelerations of the two masses are equal in magnitude but opposite in direction (a1 = -a2) is crucial for solving the problem correctly. A misapplication of sign conventions led to confusion in the calculations. Correctly applying these principles allows for the accurate determination of the acceleration of m1 as viewed from within the elevator.
yoni162
Messages
16
Reaction score
0

Homework Statement


An elevator is accelerating upwards with acceleration a0. Two masses are hanging down from a pulley, connected to each other. I need to find the acceleration of m1, as viewed from within the elevator.

Homework Equations


The Attempt at a Solution


If I look at each mass separately, each one is subjected to 3 forces: mg, T, and ma0, when T is obviously pointed upwards, and mg & ma0 are pointed downwards..
Now, if I want to find a1 --> m1's acceleration as viewed from within the elevator, I write:

m1
T-m1g-m1a0=m1a1

When I'm assuming a1 is in the positive direction of the Y axis. Now to m2:

m2
m2g+m2a0-T=m2a2

When I'm assuming that a2 will be in the opposite direction of the Y axis (since m1 will go up, m2 will go down). That's why I write forces the are in the positive direction of Y with a (-) before them, and forces which are directed in the negative direction of Y with a + before them. I know that's only a guess, but if I'm wrong and m1 will go down, a1 will turn out to be negative, no? I've used this method when solving 'inertial' problems and it works..
Also, I'm assuming the a1=-a2, again, like in 'inertial' situations..only here I'm not so sure..is this assumption true? Because when I solve the equations like they are now, I get that:

(m2-m1)(a0+g)=a1(m1-m2)

a1=-a0-g

Which isn't true according tow the answers..where am I wrong?
 

Attachments

  • TwoMassesInAnElevator.png
    TwoMassesInAnElevator.png
    806 bytes · Views: 434
Physics news on Phys.org
yoni162 said:
m2
m2g+m2a0-T=m2a2

When I'm assuming that a2 will be in the opposite direction of the Y axis (since m1 will go up, m2 will go down). That's why I write forces the are in the positive direction of Y with a (-) before them, and forces which are directed in the negative direction of Y with a + before them. I know that's only a guess, but if I'm wrong and m1 will go down, a1 will turn out to be negative, no? I've used this method when solving 'inertial' problems and it works..
OK, but...
Also, I'm assuming the a1=-a2, again, like in 'inertial' situations..only here I'm not so sure..is this assumption true?
Now you've reversed the opposite direction you just gave things above. So you've outsmarted yourself!

Write your 2nd equation (using the usual sign conventions) as:
T -m2g -m2a0 = m2a2

And then use a1 = -a2 as your constraint. That already incorporates the fact that if one goes up, the other goes down.
 
Doc Al said:
OK, but...

Now you've reversed the opposite direction you just gave things above. So you've outsmarted yourself!

Write your 2nd equation (using the usual sign conventions) as:
T -m2g -m2a0 = m2a2

And then use a1 = -a2 as your constraint. That already incorporates the fact that if one goes up, the other goes down.

Hah! you're right..thanks a lot!
 
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}...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Back
Top