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.

 

[nasu]

Both g and a are negative (or they are both positive) as long as they are in the same direction - downwards. You need a minus in front of the "ma" term because the friction force is opposite to the weight, not because the acceleration is negative. This is so as long as the acceleration is less than g.
If the acceleration becomes larger than g, the friction force has to acts downwards, otherwise the block will move towards the ceiling of the elevator. The formula for F will change for this case becaue the friction and weight act in the same direction.

 

[MatinSAR]

Both g and a are negative (or they are both positive) as long as they are in the same direction - downwards. You need a minus in front of the "ma" term because the friction force is opposite to the weight, not because the acceleration is negative. This is so as long as the acceleration is less than g.
If the acceleration becomes larger than g, the friction force has to acts downwards, otherwise the block will move towards the ceiling of the elevator. The formula for F will change for this case becaue the friction and weight act in the same direction.

So...
You are disagree with my answer in post 1? (for condition that "a" is less than g)

I only need to find answer for first condition. (a < g)

 

[nasu]

For a<g, the force (F) should be less than for the case with zero acceleration. And both accelerations have the same sign. Obviously, your guess is not right. Why don't you write Newton's second law rather than guessing? It's just one line.

 

[MatinSAR]

Both g and a are negative (or they are both positive) as long as they are in the same direction - downwards. You need a minus in front of the "ma" term because the friction force is opposite to the weight, not because the acceleration is negative. This is so as long as the acceleration is less than g.
If the acceleration becomes larger than g, the friction force has to acts downwards, otherwise the block will move towards the ceiling of the elevator. The formula for F will change for this case becaue the friction and weight act in the same direction.

If |a|<|g| : (I consider down as negative.)

$u_{s}F - mg = ma$​

So we have :

$F = \dfrac {m(g+a)} {u_s}$ (a<0 so F<$\dfrac {mg} {u_s}$)

​

 

[FinBurger]

So, the formula that holds for all a is:
F = |mg-ma|/us - the absolute value of the resultant force is what matters to figure out the friction.

As an aside, since most elevators are suspended with cables, you might find that it is very difficult to make the elevator accelerate downward with an acceleration greater than g. Have you ever pushed a car up a hill with a rope?

 

[Frabjous]

If |a|<|g| : (I consider down as negative.)

$u_{s}F - mg = ma$​

So we have :

$F = \dfrac {m(g+a)} {u_s}$ (a<0 so F<$\dfrac {mg} {u_s}$)

​

Let's define down as negative. The elevator has an acceleration of -a_el. For an elevator accelerating downward, this implies a_el>0.

In the frame of the elevator, there are three forces on the brick
1) gravity = -mg
2) friction = μF_normal
3) a pseudo force due to the accelerating reference frame f=ma_el
If the brick is to remain stationary in this frame, the sum of the forces must be zero
-mg+μF_normal+ma_el=0
giving
F_normal=(m/μ)(g-a_el)

If you define a = -a_el, this becomes what you have in your OP. This is a confusing definition, which is why I told you that you needed to define "a" clearly in an earlier post.

In an external non-moving frame, the acceleration of the brick (-a_br) and of the elevator (-a_el) must be identical for the brick to remain "stationary".
In this frame, there are two forces on the brick
1) gravity = -mg
2) friction = μF_normal
The force on the brick is
-ma_br=-mg+μF_normal
giving
-a_br=-g+μF_normal/m=-a_el
which gives
F_normal=(m/μ)(g-a_el)

 

[nasu]

If |a|<|g| : (I consider down as negative.)

$u_{s}F - mg = ma$​

So we have :

$F = \dfrac {m(g+a)} {u_s}$ (a<0 so F<$\dfrac {mg} {u_s}$)

​

If you consider down as negative, both g and a should be negative. You are not consistent in your signs.
You should have $u_{s}F - mg = - ma$

 

[MatinSAR]

@Frabjous @nasu @FinBurger I understand now. Thanks for your help and time.