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

[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.