# Elevator problem: considering normal force on a body inside while calculating net force on the system

I faced this problem in Fundamentals of Physics (Halliday and Resnick):

A $0.250\ \mathrm{kg}$ block of cheese lies on the floor of a $900\ \mathrm{kg}$ elevator cab that is being pulled upward by a cable through distance $d_1=2.40\ \mathrm{m}$ and then through distance $d_2= 10.5\ \mathrm{m}$. (a) Through $d_1$, if the normal force on the block from the floor has constant magnitude $F_N= 3.00\ \mathrm{N}$, how much work is done on the cab by the force from the cable? (b) Through $d_2$, if the work done on the cab by the (constant) force from the cable is $92.61\ \mathrm{kJ}$, what is the magnitude of $F_N$?

When I looked at the solution there was equation like this

$$F+F_N−(m+M)g=(m+M)a$$

(where $m = 0.250\ \mathrm{kg}$ is the mass of the cheese, $M = 900\ \mathrm{kg}$ is the mass of the elevator cab, $F$ is the force from the cable, and $F_N = 3.00\ \mathrm{N}$ is the normal force on the cheese.)

The question: Why do we take into consideration $F_N$ (normal force) while calculating net force on the system? Isn't this an internal force?

You can (and should!) always draw a Free Body Diagram for the given problem (at least at this level of physics, at higher levels the Lagrangian is more informative, but FBD are STILL useful even then), and this will tell you exactly how the forces effect an object's acceleration; this is important because the sum of those forces must satisfy Newton's Second Law for a given object

$\sum \vec{F} = m\vec{a}$

regardless of whether they are internal or not. You are correct that internal forces do no net work, but what has been written in the solution seems to be actually the sum of two equations, namely one for the m, and one for the elevator M, actually I don't think that the $F_N$ term should appear in this summation either since it should cancel out.

Explicitly we have (down is negative, up is positive):

Smaller mass:

$F_N - mg = ma$

Elevator:

$F-F_N - Mg = Ma$

The sum then yields:

$F - (m+M)g= (m+M)a$

ALSO, you should note that the question asks nothing about work being done by the normal force (which is the internal force), only about the work done by the cable, which is clearly an external force.

• Thanks for your interest. I also drawed FBD and found the same answer as yours but for the b part of the question which we don't know the normal force and have to set an equation to find it, it affects the answer. Can the answer be wrong? Jan 10, 2016 at 18:35
• You have enough information though to solve part (b) because in part (a) you derived an equation that tells you how work done by the cable relates to $F_N$; think carefully how you can use the information given. If you believe my answer was helpful, please accept and upvote it :). Jan 10, 2016 at 19:12
• We are thinking in the same way actually :) This part your sum is F - (m+M)g = (m+M)a is also same as mine the equation which I wrote above: F+FN−(m+M)g=(m+M)a is the equation in the book. Jan 10, 2016 at 19:32

You are correct that the normal force is irrelevant when calculating the overall acceleration of the system, which is equivalent to an object of mass $M + m$ being accelerated by a force $F$.

However, the story is different when calculating work, because you are calculating the work done on the cab, not the work done on the system. In this case, there is a normal force exerted downwards by the block onto the floor of the elevator that acts opposite to the direction of overall travel.

Think about it this way: if you were to ignore the normal force, that would imply that the work performed by the cable (which is equal to $Fd$) is transferred completely into a cab of mass $M + m$. This is clearly false, because only some of the energy goes into the cab. The rest goes into the block.

In short: when you draw the FBD for the cab, you should note that there is a normal force pointing downwards exerted by the block (with magnitude equal to the normal force from the floor onto the block), which performs negative work.