Pushing a block against the wall of an elevator that is accelerating

[MatinSAR]

Homework Statement

Imagine you push a block against wall of an elevator that is accelerating downward. What is the minimum force you should apply to prevent block's free fall?

Relevant Equations

N's laws.

Easier case: Elevator is at rest.
We need to prevent box from free fall so friction should be bigger than "mg".(And they can be equal)
When we push with force F we know that the maximum static friction is $u_sF$.
"mg" should be smaller than $u_sF$ or should be equal to it so the minimum value for $F$ is : $F=\frac {mg} {u_s}$. Am I right?!

But here the elevator is accelerating downward ... I can guess that here F is equal to $\frac {mg+ma} {u_s}$ ... Is my guess correct?! Is there any good reason for it?!

 

[Hill]

Homework Statement: Imagine you push a block against wall of an elevator that is accelerating downward. What is the minimum force you should apply to prevent block's free fall?
Relevant Equations: N's laws.

Is my guess correct?

No, it is not. Consider what happens if $a=g$.

 

[MatinSAR]

No, it is not. Consider what happens if $a=g$.

Good point. According to me F should be 0 in this case ...
What about 1st part (Whe elevator was at rest)? Was it correct?

 

[Hill]

What about 1st part? Was it correct?

I don't see any problem there.

 

[FinBurger]

Your reasoning is substantially correct. However, let me ask you this: When you drew your diagram, you made "mg" point down (ok!) and you made curly-f=ma point up (also ok!). Since they point in opposite directions, what does that say to you about your equation? How would you correct it?

 

[MatinSAR]

I don't see any problem there.

Thanks.

and you made curly-f=ma point up (also ok!).

Actually I can't understand what do you mean in this part. Can you explain more? Thanks.

 

[MatinSAR]

@Hill @FinBurger
My friend's reasoning for the 2nd part.

The elevator is accelerating downward. "mg" and friction act on the box. If the box is supposed to be stationary with respect to the elevator, its acceleration must be negative of the elevator so F is equal to :
(mg-ma)/u_s

 

[Hill]

@Hill @FinBurger
My friend's reasoning for the 2nd part.

The elevator is accelerating downward. "mg" and friction act on the box. If the box is supposed to be stationary with respect to the elevator, its acceleration must be negative of the elevator so F is equal to :
(mg-ma)/u_s

What if $a \gt g$?

 

[Frabjous]

What is the force when a=0?

 

[MatinSAR]

What is the force when a=0?

$mg/u_s$

What if $a \gt g$?

I don't see any problem.

 

[Hill]

I don't see any problem.

Then your $F$ is negative. Is it OK?

 

[Frabjous]

$mg/u_s$

The elevator is accelerating under gravity.
The brick is acclerating under gravity.
So the frictional force needs to be …

 

[MatinSAR]

Then your $F$ is negative. Is it OK?

How is that possible?
I said F= $\frac {mg-ma} {u_s}$ and $a$ is negative so $mg-ma$ is always positive.

 

[MatinSAR]

The elevator is accelerating under gravity.
The brick is acclerating under gravity.
So the frictional force needs to be …

Bigger than gravitational force?
So the net force acts upward...

 

[Frabjous]

Bigger than gravitational force?
So the net force acts upward...

You are guessing.
Try this.
Write down the position of the brick as a function of time.
Write down the position of the wall of the elevator where the brick touches as a function of time.
Compare.

 

[Hill]

How is that possible?
I said F= $\frac {mg-ma} {u_s}$ and $a$ is negative so $mg-ma$ is always positive.

If $g$ is positive and $a \gt g$ then $a$ is positive.

 

[MatinSAR]

If $g$ is positive and $a \gt g$ then $a$ is positive.

In this case the elevator's speed should decrease. Is it true?

You are guessing.
Try this.
Write down the position of the brick as a function of time.
Write down the position of the wall of the elevator where the brick touches as a function of time.
Compare.

They are similar to each other.

 

[Frabjous]

They are similar to each other.

You want to calculate a quantitative value. “Similar” is not useful.

 

[Hill]

In this case the elevator's speed should decrease. Is it true?

No. I consider $a$ having the same direction as $g$. As the elevator accelerates downwards, $a$ is positive. When $a=g$, the elevator free falls - both $a$ and $g$ are positive. When $a \gt g$, the elevator accelerates downwards more than the free fall acceleration $g$.

 

[MatinSAR]

You want to calculate a quantitative value. “Similar” is not useful.

 

[MatinSAR]

No. I consider $a$ having the same direction as $g$. As the elevator accelerates downwards, $a$ is positive. When $a=g$, the elevator free falls - both $a$ and $g$ are positive. When $a \gt g$, the elevator accelerates downwards more than the free fall acceleration $g$.

According to above picture... My friend was wrong and elevator's acceleration is same as block's acceleration...

 

[Frabjous]

View attachment 336576

You did not go correctly from your second to the third line, but you got the idea. The accelerations need to be identical. Plug in the accelerations from the problem.

 

[Frabjous]

According to above picture... My friend was wrong and elevator's acceleration is same as block's acceleration...

Yes!

 

[MatinSAR]

You did not go correctly from your second to the third line, but you got the idea. The accelerations need to be identical. Plug in the accelerations from the problem.

Yes I see. The acceleration is "a".

Yes!

But then if I use N's 2nd law I get :
f - mg = ma
uF - mg = ma

F = (mg + ma) / u
And this is wrong.

 

[Frabjous]

Yes I see. The acceleration is "a".

But then if I use N's 2nd law I get :
f - mg = ma
uF - mg = ma

F = (mg + ma) / u
And this is wrong.

You are mixing up the variables because the names are too generic. Try using subscripts (variable_br and variable_el).

 

[MatinSAR]

You are mixing up the variables because the names are too generic. Try using subscripts (variable_br and variable_el).

I have only m witch is mass of the object. And a which is acceleration of the box. F is the force that pushes the box. What variables am I mixing up?

 

[Frabjous]

Define f_external= M_elevatora_external.

What is the acceleration of the elevator, a_el?

Let F_friction=μF_{applied to brick}.

What is the acceleration of the brick, a_br, as a function of F_friction?

How are a_el and a_br related?

 

[MatinSAR]

What is the acceleration of the elevator, ael?

$a_{el} =a_{external}$

Let Ffriction=μFapplied to brick.

What is the acceleration of the brick, abr, as a function of Ffriction?

I cannot understand what do you mean by this.

How are ael and abr related?

They were equal.

 

[Frabjous]

I cannot understand what do you mean by this.

Let acceleration downward be negative, and acceleration upward be positive.

ma_br = F_friction-mg
a_br=F_friction/m - g

I over-complicated things slightly. Just assume that the elevator has an acceleration, -a_el (a_el is positive).

 

[MatinSAR]

mabr = Ffriction-mg

But there is no difference between this equation and my answer!
Why do I need that $- a_{el}$?

I think I am completely lost.

 

[Frabjous]

But there is no difference between this equation and my answer!
Why do I need that $- a_{el}$?

I think I am lost completely...

Your original answer was correct. I apologize.

You are doing the problem in the frame of the brick, so you had a pseudo force f. You were essentially solving from your post #20, $\ddot x_2=0$.

I misinterpreted what you did.

My derivation was in a frame external to the elevator. I was essentially solving from your post #20, $\ddot x_1=\ddot x_3$.

They are different ways of solving the same problem, so they come up with the same answer.

Once again my apologies.

 

[MatinSAR]

Your original answer was correct. I apologize.

You are doing the problem in the frame of the brick, so you had a pseudo force f. You were essentially solving from your post #20, $\ddot x_2=0$.

I misinterpreted what you did.

My derivation was in a frame external to the elevator. I was essentially solving from your post #20, $\ddot x_1=\ddot x_3$.

They are different ways of solving the same problem, so they come up with the same answer.

Once again my apologies.

Thank you for your time @Frabjous . So you are agree with my answer in post one. Yes?
$F= \dfrac {mg+ma_{el} } {u_s}$

 

[Frabjous]

Thank you for your time @Frabjous . So you are agree with my answer in post one. Yes?
$F= \dfrac {mg+ma} {u_s}$

I would more clearly define "a" so that it is clear that it is a negative number for a downward accelerating elevator. Other than that , yes.

 

[MatinSAR]

I would more clearly define "a" so that it is clear that it is a negative number for a downward accelerating elevator. Other than that , yes.

I completely agree. Thanks.

 

[Frabjous]

Good point. According to me F should be 0 in this case ...

Notice that if the elevator is in free fall, F should be zero. Your comment is a little ambiguous.