Simple frictional forces and forces on boxes

  • #1

Homework Statement



First question: A force of magnitude 7.5 N pushes three boxes with masses m1= 1.3 kg, m2=3.2 kg, and m3=4.9 kg. Find the contact force between

a-boxes 1 and 2
b-boxes 2 and 3

There is a diagram, so I'll try to describe it...it's pretty simple. The boxes are just sitting next to each other. From left to right, Box 1 is next to box 2 which is next to box 3, and the force is being applied to box 3 from the right.

Second question: A 10 kg box rests on a horizontal floor. The coefficient of static friction is .4 and the coefficient of kinetic friction is .3. Determine the force of friction acting on the box if a horizontal external applied force is exerted on it of magnitude:

a) 0 N b) 10 N c) 20 N d) 38 N e) 40 N

Homework Equations



for question 1, F=ma
for question 2, f=(coefficient of static/kinetic friction) * (normal force)

The Attempt at a Solution



First question: I really just don't understand the concept behind this. At first I thought that I needed to find the acceleration of the entire system, so

7.5 = (4.9+3.2+1.3) * a
a= .798

But then I'm not sure if I'm right with where I go now. I did:

let F= contact force between boxes 1 and 2 and
1.3 * .798= 7.5 N - F
F= 6.5 N

The answer is supposedly 1 N

for part b, I did pretty much the same thing except it was the mass of box 2, or 3.2, instead of 1.3.

The answer for b is 3.59 N

my main question is how do I know which box I use as the mass, and are these answers even right. My physics teacher is horrible, and I have no confidence in her answers as she has proven to me before...

For question 2, is the force of friction just the same as the applied force up until the max static friction or 9.8 * 10 * .4 which is 39.2. If so, then how do I get the answer for when the force is 40 N? Is it just 9.8 * 10 * .3?

Thanks for any help guys...this is my first time asking for an answer, and probably not the last time since my teacher is so incompetent...
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
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First question: I really just don't understand the concept behind this. At first I thought that I needed to find the acceleration of the entire system, so

7.5 = (4.9+3.2+1.3) * a
a= .798

Good.

But then I'm not sure if I'm right with where I go now. I did:

let F= contact force between boxes 1 and 2 and
1.3 * .798= 7.5 N - F
F= 6.5 N

The answer is supposedly 1 N

The force is being applied from the right, so the contact force between boxes 1 and 2 is just enough to cause box 1's acceleration to be 0.798....does that help?

For part(b), the force is just enough to push box 2 and box 1 at an acceleration of 0.798 :wink:
 
  • #3
OOOOOOOOH, I understand now!

So force between box 1 and 2 is:

f=ma=1.3 * .798= 1

and force between box 2 and 3 is:

f= (1.3+3.2) * .798 = 3.59 N

IS THAT IT???

Also, could you answer my second question involving frictional forces?

Thanks a lot!
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
7
OOOOOOOOH, I understand now!

So force between box 1 and 2 is:

f=ma=1.3 * .798= 1

and force between box 2 and 3 is:

f= (1.3+3.2) * .798 = 3.59 N

IS THAT IT???

Also, could you answer my second question involving frictional forces?

Thanks a lot!

Looks good! :smile:

...now on to the second question:

For question 2, is the force of friction just the same as the applied force up until the max static friction or 9.8 * 10 * .4 which is 39.2. If so, then how do I get the answer for when the force is 40 N? Is it just 9.8 * 10 * .3?

seems reasonable to me!:approve:

I guess your physics teacher can't be that bad after all :wink:
 
  • #5
Oh no...my teacher didn't teach me any of that. She just gave us the equation f=uk * N. I had to guess with the rest by getting my own textbook...

Thanks so much...you made everything clear with one sentence!!!
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Your welcome; good luck with your studies! :smile:
 

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