# Time it takes for block to slide down an incline in elevator

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1. Jan 31, 2016

### FruitNinja

1. The problem statement, all variables and given/known data

MECHANICS:

Given Theta, L, M, and acceleration of elevator relative to ground. Find the time it takes for the block to reach the end of the incline.

Here is a diagram: http://k-elahian.com/tmp/nip.PNG [Broken]

2. Relevant equations
f=ma
kinematics
relative acceleration

3. The attempt at a solution
I've gotten 3 different answers so far. Basically I do f=ma for the m mass and then split it into x and y (slanted system) and then I relate the accelerations of the the bodes Amg=Ame+Aeg (ground is g and elevator is e).
After finding Ame I use that in my kinematics equation.

Last edited by a moderator: May 7, 2017
2. Jan 31, 2016

### Student100

Your notation is a bit odd, what does Amg=Ame+Aeg mean exactly? What is the A? Can you explain what exactly you mean?

3. Jan 31, 2016

### Suraj M

I'm not sure if I understand what you've tried to do
But start by calculating the pseudo force on the block due to the elevators Motion and also the other forces on it. I think you can proceed from there.
You say you've got 3 answers
Unless you state what were your answers, how should we judge if it's right or wrong?

4. Jan 31, 2016

### Suraj M

What does Aeg mean? Is that relative acceleration?

5. Jan 31, 2016

### FruitNinja

Yes,
Amg is acceleration of m relative to g
Aeg is acceleration of elevator relative to ground.
Since I cannot use fictitious forces I need to use relative accelerations.

Amg=Ame+Aeg just shows that the a of the mass relative to the ground is the sum of the 2 relative accelerations vectors

6. Jan 31, 2016

### Student100

Okay, so what equation did you find pre-substitution?

7. Jan 31, 2016

### FruitNinja

-mgsinΘ = mamgx for the x and -mgcosΘ+Fn=mamgy for the y
are the x and y components of f=ma for the block, but I am using a slanted coordinate frame

8. Jan 31, 2016

### Student100

Okay, so I don't understand what you're trying to write.

Maybe it will help if we just look at an incline plane first without the added acceleration, If we rotate the coordinate axis $\theta$, what's the acceleration down the plane in this case?

9. Jan 31, 2016

### FruitNinja

yeah I did that in the equation above. for the y, -mgcosΘ+Fn are the 2 forces acting on the block. (slanted frame). amgy just means the acceleration of the mass relative to the ground in the y direction

10. Jan 31, 2016

### Student100

Why is there an acceleration in the y direction in the rotated coordinate system?

Hang with me for a second I'll get you to the answer, ignore the elevator for a second, what's the sum of the forces in the x and y looking only at an inclined plane?

11. Jan 31, 2016

### FruitNinja

Ok, It's just Fg and Fn. Those are the only 2 forces acting on the block.

for X: -mgsinΘ , no fn here and for Y: -mgcosΘ+Fn

12. Jan 31, 2016

### Student100

So $mgsin(\theta)=ma$ and $N - mgcos(\theta) = 0$, correct?

You know why the y direction is in equilibrium when we rotate the system, correct?

13. Jan 31, 2016

### FruitNinja

Yes, if we were not in a elevator yes. But that would be a non-inertial frame so we can't do that, according to my teacher

14. Jan 31, 2016

### Student100

Yes, so if you now look back at the accelerating system relative to the plane, this time don't rotate your coordinate system and solve for $a_{rel}$ relative down the plane.