Basic Newton's Laws and Applied Force: 3 boxes

AI Thread Summary
The discussion revolves around calculating the acceleration, net forces, and contact forces of three boxes on a frictionless surface when a 90 N force is applied to the first box. The total mass of the boxes is determined to be 60 kg, leading to an acceleration of 1.5 m/s². The net forces on each box are calculated as 15 N for the first box, 30 N for the second, and 45 N for the third. Contact forces between the boxes are derived, with 75 N between the first and second boxes, and 45 N between the second and third. The participants confirm the calculations and clarify the concept of contact forces, emphasizing the interaction between the boxes due to the applied force.
cassie123
Messages
15
Reaction score
0

Homework Statement


Three boxes are in contact with each other on a frictionless horizontal surface as shown. The masses of the boxes are m1 = 10 kg, m2 = 20 kg, and m3 = 30 kg. A horizontal force F = 90 N is applied to m1.
Calculate:
a. the acceleration of the three boxes.
b. the net force on each box.
c. the contact forces between the boxes.

Screen Shot 2015-07-16 at 3.45.49 PM.png


Homework Equations


Newtons second law
F=ma

The Attempt at a Solution


total mass = 60 kg
F = ma

1) a = F/m = 90/60 = 1.5 m/s^2

2) net force on each box
F1 = m1a = 10*1.5 = 15N
F2 = m2a = 20*1.5 = 30 N
F3 = m3*a = 30 *1.5 = 45 N

Am I missing forces in calculating the net force?

c) contact force
90-15 = 75 N btwn 1st and 2nd box
75- 30 = 45 N btwn 2nd and 3rd box

The above is my best guess although I am not confident that this is the way to calculate contact forces. (What are contact forces?) Should I be using something like f=(F*(m2+3))/(m1+(m2+3) or f=(F*m2)/(m1+m2) to find the contact force between boxes 1 and 2?
 
Physics news on Phys.org
Theres a two force which affects first two boxes (m1 and m2).But there's only one force effects third box which is contact between m2 and m3.Contant force means If you push m1 the force will affect the system and in the system m1 will push m2.But every force has a opposite force so m2 applies force on m1.
For this problem,If you want to answer question quick,start with the last box.Theres one force act it.Find that force.

Your find the acceleration right.
After find that force think this way.
Theres one box in the middle which its m2.Now there's two forces acting on it the contant force between m2 and m3 and contant force between m1 and m2.And their net force will be m2a=F which its 30N.

To this solution step by step.Youll find the answer
 
cassie123 said:
The above is my best guess although I am not confident that this is the way to calculate contact forces.

You should be more confident as you've got the solution precisely correct.
 
I did not notice you were right.What a shame for me
 
PeroK said:
You should be more confident as you've got the solution precisely correct.

So it's right as shown? Never mind the other guesses for contact force? Thanks so much!
 
RyanH42 said:
Theres a two force which affects first two boxes (m1 and m2).But there's only one force effects third box which is contact between m2 and m3.Contant force means If you push m1 the force will affect the system and in the system m1 will push m2.But every force has a opposite force so m2 applies force on m1.
For this problem,If you want to answer question quick,start with the last box.Theres one force act it.Find that force.

Your find the acceleration right.
After find that force think this way.
Theres one box in the middle which its m2.Now there's two forces acting on it the contant force between m2 and m3 and contant force between m1 and m2.And their net force will be m2a=F which its 30N.

To this solution step by step.Youll find the answer

Thanks for your explanation! :)
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top