# Finding the Normal Force between two Crates

pech0706

## Homework Statement

Your moving company runs out of rope and hand trucks, so you are forced to push two crates along the floor. The crates are moveing at constant velocity, their masses are m1=45kg and m2=22kg, and the coefficient of kinetic friction between both crates and the floor are 0.35. Find the normal force between the 2 crates.
(picture attatched)

m1=45kg
m2=22kg
mk=0.35

## Homework Equations

Fgrav=-mg
Ffriction=coefficient of friction*normal force

## The Attempt at a Solution

I know the answer is 75N. I attempted to solve the forces acting on each individual crate then combined the results. I got way to large of an answer. I don't know how to go about solving this another way.

#### Attachments

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Homework Helper
Gold Member
I attempted to solve the forces acting on each individual crate then combined the results. I got way to large of an answer. I don't know how to go about solving this another way.
I am not sure what you mean by combining results, but please show your work for finding the forces acting on each crate, so we can better asssist.

pech0706
Fgrav of m1=(45kg)(9.81m/s2)=441.45N
Fgrav of m2=(22kg)(9.81m/s2)=215.82N

Normal Force on both grates=441.45N+215.82=657.27N

That was my original/best guess on how to do it, but I am WAY off.

Homework Helper
Gold Member
Fgrav of m1=(45kg)(9.81m/s2)=441.45N
Fgrav of m2=(22kg)(9.81m/s2)=215.82N

Normal Force on both grates=441.45N+215.82=657.27N

That was my original/best guess on how to do it, but I am WAY off.
You are calculating the normal force of the ground on the crates. But the problem is asking you to find the horizontal normal force between the crates (the normal force of one crate acting on the other). So try looking at the the small mass and identify the forces acting on it in the horizontal direction, and then use newton's laws.

pech0706
Alright, I did the math based on the small great, and I got the right answer (basically)
Ffriction=the coefficient of friction*the normal force.
Ffriction=(0.35)(215.82)
Ffriction=75.537

That's a lot better. I don't understand why you use the small crate to solve this problem though.

Homework Helper
Gold Member
Alright, I did the math based on the small great, and I got the right answer (basically)
Ffriction=the coefficient of friction*the normal force.
Ffriction=(0.35)(215.82)
Ffriction=75.537

That's a lot better. I don't understand why you use the small crate to solve this problem though.
What you calculated was the force of friction acting on the small block. To calculate the horizontal normal force acting on the small block, you need to use one of newton's laws in the x direction, noting that the blocks are moving at constant velocity, which implies that the net force in the horizontal direction is _____?, and thus the normal horizontal force is _____?

pech0706
Okay so the forces acting on the small crate are a frictional force in the negative x direction wich is what I calculated out to be be 75.537. There is also a force from the larger mass pushing into the smaller crate... and is there also the force of the crates' velocities?

Homework Helper
Gold Member
Okay so the forces acting on the small crate are a frictional force in the negative x direction wich is what I calculated out to be be 75.537.
yes.
There is also a force from the larger mass pushing into the smaller crate...
yes, the horizontal normal contact force between the 2 crates
and is there also the force of the crates' velocities?
No. Velocity is not a force, it is a rate of change of displacement with respect to time. It is a bit misleading to show it on your diagram. Put it to the side somewhere (noted as v = constant), or just get rid of it altogether.

Homework Helper
hi pech0706! … and is there also the force of the crates' velocities?

your diagram is terrible it is essential to draw a good free body diagram to solve these problems

as PhanthomJay says, velocity is not a force …

(and you said the same thing in another thread)

it should never be on a free body diagram

(acceleration maybe, so long as it's not marked as a force, but velocity no!)

your should have two or even three free body diagrams for this case …

one showing the forces on the back crate only, one showing the forces on the front crate only, and one showing the (external) forces on the pair of crates considered as one body

in particular, your diagrams should show the reaction force between the two crates, also a separate normal force for each crate and a separate friction force for each crate …

when you've done all that, do an F = ma equation for each diagram 