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
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
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...