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

[FruitNinja]

Homework Statement

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

Homework Equations

f=ma
kinematics
relative acceleration

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.

 

[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?

 

[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?

 

[Suraj M]

What does Aeg mean? Is that relative acceleration?

 

[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

 

[Student100]

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

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

 

[FruitNinja]

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

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

 

[Student100]

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

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?

 

[FruitNinja]

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?

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

 

[Student100]

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

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?

 

[FruitNinja]

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?

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

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

 

[Student100]

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

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

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?

 

[FruitNinja]

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?

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

 

[Student100]

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

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.