Two Boxes Friction Problem


by CandyApples
Tags: friction
CandyApples
CandyApples is offline
#1
Sep22-09, 10:57 PM
P: 28
1. The problem statement, all variables and given/known data
A man is pulling two suitcases, one on top of the other. Bottom case mass=25kg and top mass = 15kg. The coefficient of friction between the two cases and between the bottom case and the floor can both be represented by Uk = .4 and Us =.5. Find the maximum acceleration of the cases so they move with respect to the floor, but not with respect to one another.


2. Relevant equations
F =un


3. The attempt at a solution
Fs = un for the two boxes.
Fs = .5*15
Fs = 7.5N.
This means exceeding 7.5N of force will cause the boxes to accelerate with one another. Now I need to plug this value into another equation such that I find the maximum acceleration of the two boxes. Using F=ma setting F equal to 7.5 does not work, as this gives an extremely tiny acceleration. I think another F=un is needed, but im not sure how to set the second one up in a way that allows me to use the 7.5N I figured out before.
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rootX
rootX is offline
#2
Sep23-09, 12:56 AM
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P: 1,295
Quote Quote by CandyApples View Post
1. The problem statement, all variables and given/known data
A man is pulling two suitcases, one on top of the other. Bottom case mass=25kg and top mass = 15kg. The coefficient of friction between the two cases and between the bottom case and the floor can both be represented by Uk = .4 and Us =.5. Find the maximum acceleration of the cases so they move with respect to the floor, but not with respect to one another.


2. Relevant equations
F =un


3. The attempt at a solution
Fs = un for the two boxes.
Fs = .5*15
Fs = 7.5N.
This means exceeding 7.5N of force will cause the boxes to accelerate with one another. Now I need to plug this value into another equation such that I find the maximum acceleration of the two boxes. Using F=ma setting F equal to 7.5 does not work, as this gives an extremely tiny acceleration. I think another F=un is needed, but im not sure how to set the second one up in a way that allows me to use the 7.5N I figured out before.
Is Fs the static friction in Fs = .5*15? What are the units of friction coefficient? Do dimension analysis for the equation: Fs = .5*15
CandyApples
CandyApples is offline
#3
Sep23-09, 06:35 PM
P: 28
Ok so now there is a part two to this problem.
I figured out Part I as follows:
Fs = .5 (147)
73.5 = 15*a
a = 4.9m/s/s

Fk = 392*.4
Fk = 156.8

F-156.8 = ma
F = 40(4.9)+157
F = 353N

Now it asks if F determined in Part I is doubled (706N) what will be the acceleration of each block. This will certainly result in the to suitcase flying off.

Plugging in 706 alone does not provide the correct answer. I think I need to account for the added information that the suitcase flies off, but I am not sure how. I forget exactly how to regard the top suitcase, as the course has not covered how to handle an object that is initially at rest then is put in motion by another object below it. If I recall correctly from a statics course taken some time ago, is the mass of the top box disregarded once it is put into motion? Also, hoe could I represent this? Does knowing that it takes 353N of force to put the top box into motion with respect to the bottom box allow me to derive an equation to figure out this problem?


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