Normal force for stacked objects in a moving elevator

[marcusesses]

Homework Statement

Say I have two objects on top of each other in an elevator (say Object 1 on the bottom and Object 2 on top), each with a given mass. The elevator is accelerating downward (meaning the elevator is moving up and slowing down). What would the normal force be on Object 1?

Homework Equations

Newtons second law!

The Attempt at a Solution

Newton's 2nd law for Object 2 is:

$$
\Sigma F_{x2} = F_{N2} - F_{g2} =- m_2 a ,
$$
since it's the object on top, it only has a normal force and gravity acting on it. Therefore,$$
F_{N2} =- m_2 a + F_{g2}
$$

For Object 1 it is:

$$
\Sigma F_{x1} = F_{N1} - F_{g1} - F_{C1} = -m_1 a
$$

This object has the normal force from the floor of the elevator pushing up, gravity of the book, *and* the additional contact force of Object 2 pushing down on it. Since $F_{N2} = - F_{C1}$ from Newton's Third Law

$$
\Sigma F_{x1} = F_{N1} - F_{g1} + m_2 a + F_{g2} = -m_1 a
$$

Solving for the normal force on Object 1 gives

$$
F_{N1} = F_{g1} + m_2 a - F_{g2} - m_1 a
$$

$$
F_{N1} = m_1 g + m_2 a - m_2 g - m_1 a
$$

$$
F_{N1} = m_1(g-A) -m_2(g-a)
$$

$$
F_{N1} = (m_1-m_2)(g-a)
$$

which is the final result. However, if the masses are equal, there's no normal force, which doesn't make sense.

The more intuitive answer comes if I assume $F_{N2} = F_{C1}$, which gives

$$
F_{N1} = (m_1+m_2)(g-a)
$$
I'm not sure how to explain why this one is right and the other is wrong though.

 

[Chestermiller]

I don't get what you did. Here are my force balances, for your consideration:

Let $F_1$ represent the upward force that the floor of the elevator exerts on object 1
Let $F_2$ represent the upward force that object 1 exerts on object 2
Let "a" represent the upward acceleration of objects 1 and 2. If the elevator is accelerating downward, then a is negative.

$$F_1-F_2-m_1g=m_1a$$

$$F_2-m_2g=m_2a$$

Do you agree with these?

 

[marcusesses]

Yes, those are the same as the 1st and 3rd equations in my comment.

 

[Chestermiller]

Yes, those are the same as the 1st and 3rd equations in my comment.

Then what’s the problem?

 

[haruspex]

$$F_{N2} = - F_{C1}$$

No. You already did the sign switch when you wrote -F_C1 in the preceding equation. You would have avoided confusing yourself if you had written -F_N2 there and not introduced F_C1.

Also, since you are defining up as positive for forces it is safer to define it that way for accelerations too. So write ma, not -ma. a will of course take a negative value, but it should appear as +a in the equation for vertical acceleration.
I appreciate that this is at odds with how g is traditionally treated. Nearly everyone writes -g in the equation so that g is positive, but really it is illogical. But in that case we know what sign it will be, whereas sometimes we do not know in advance the sign of an unknown acceleration.

 

[Chestermiller]

In force balance problems like this, where making errors with sign are very easy to do, one way to guarantee that you always get the right signs is to write the equations as vector relationships, and by expressing all vectors as having a scalar (tentatively positive) magnitude multiplied by a unit vector in the appropriate direction. For example, in the case of gravity, one would write: $$\mathbf{g}=g(-\mathbf{i_z})$$ where $\mathbf{i_z}$ is the unit vector in the z (upward) direction. In the case of the acceleration, one could write either
$$\mathbf{a}=a\mathbf{i_z}$$
or
$$\mathbf{a}=a(-\mathbf{i_z})$$
Either way, the force balance equation would automatically guarantee that you would obtain the correct result.

For the present problem, one could write:
$$F_1\mathbf{i_z}+F_2(-\mathbf{i_z})+m_1g(-\mathbf{i_z})=m_1a\mathbf{i_z}$$
and
$$F_2\mathbf{i_z}+m_2g(-\mathbf{i_z})=m_1a\mathbf{i_z}$$