# magnitude of the forces

• Oct 18th 2007, 12:26 PM
rcmango
magnitude of the forces
A 32 kg crate rests on a horizontal floor, and a 69 kg person is standing on the crate.
Determine the magnitude of the normal force that the floor exerts on the crate.
N

Determine the magnitude of the normal force that the crate exerts on the person.
N

not sure why the answers would not be: 101 N and 69 N.

thanks for looking at this one.
• Oct 18th 2007, 02:15 PM
topsquark
Quote:

Originally Posted by rcmango
A 32 kg crate rests on a horizontal floor, and a 69 kg person is standing on the crate.
Determine the magnitude of the normal force that the floor exerts on the crate.
N

Determine the magnitude of the normal force that the crate exerts on the person.
N

not sure why the answers would not be: 101 N and 69 N.

thanks for looking at this one.

I'm going to use a specific notation for this one. It might seem a little confusing at first, but once you get used to it it saves a lot of time and trouble.

Define the symbol $\displaystyle F_{ab}$ to be the force on object a that is coming from object b.

So Free Body Diagram time. In each FBD I'm defining upward to be positive. We don't need a horizontal axis.

For the box, we have a normal force $\displaystyle N_{bf}$ (f represents the floor) pointing
up, a weight on the box $\displaystyle w_{bE}$ (E is the Earth) acting straight down, and a normal force $\displaystyle N_{bp}$ acting straight down.

So
$\displaystyle \sum F = N_{bf} - w_{bE} - N_{bp} = 0$ (since the box is not accelerating.)

We need $\displaystyle N_{bf}$, so
$\displaystyle N_{bf} = w_{bE} + N_{bp}$

So what is $\displaystyle N_{bp}$?

Do a FBD on the person:
We have a normal force $\displaystyle N_{pb}$ acting upward and a weight $\displaystyle w_{pE}$ acting downward.

Thus
$\displaystyle \sum F = N_{pb} - w_{pE} = 0$

So
$\displaystyle N_{pb} = w_{pE}$

So the magnitude of the normal force on the person is equal to the magnitude of the weight of the person.

Now, to finish the problem we need to find $\displaystyle N_{bp}$, not $\displaystyle N_{pb}$. The notation is meant to suggest a relationship between the two: they are Newton's 3rd Law pairs. So by Newton's 3rd these have equal magnitudes. Thus
$\displaystyle N_{bf} = w_{bE} + N_{bp} = w_{bE} + N_{pb} = w_{bE} + w_{pE}$

You can do the rest from here.

Edit: After going through all this it appears that the whole problem is that you gave your answer in terms of the mass, not the weight. Ah well, it was good for a demonstration. :)

-Dan
• Oct 19th 2007, 10:31 AM
rcmango
Okay, so in other words Fnet = mass * acceleration,

got it.