Maximum force on stacked boxes, when we don't want the boxes to slip.


by tomas123
Tags: boxes, force, maximum, slip, stacked
tomas123
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#1
Sep26-11, 07:29 AM
P: 5
1. The problem statement, all variables and given/known data

There are two blocks stacked on top of each other. M2 is on top of M1. M1 = 3 kg. m2 = 4 kg. μs between the two boxes = 0.6. μk between m2 and the floor = 0.2.

Using a rope, we exert a force T on m2, dragging both boxes 5m. What is the maximum force we can exert on m2, if we don't want m2 to slip on m1.



2. Relevant equations

Fk = μk * n
Fs(max) = μs * n


3. The attempt at a solution

I found Fs(max) between m2 and m1.
Fs(max) = n * μs =
mg * 0.6 =
4kg * 9.8 * 0.6 =
23.52 N.
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tomas123
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#2
Sep26-11, 09:20 AM
P: 5
1 problem! the solution is 36.6 N
gneill
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#3
Sep26-11, 10:17 AM
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P: 11,414
Sometimes it's helpful to recast the problem into a form that you've seen before, or at least one that makes the interrelationships between the forces more obvious.

Presumably you've calculated the frictional force that the floor exerts upon the bottom box when that box is moving. Call it F1. What's its derivation and value?

Also presumably you've calculated the maximum static frictional force that can be withstood between the boxes before they will slide. Call that F2. What's its derivation and value?

Now redraw the problem placing the boxes separately on the "floor". Connect them with the proverbial "light inextensible string". The tension in that string will represent the frictional force operating between the boxes; If it ever exceeds F2 then the string will snap and the boxes will become unconnected (In the original configuration, they would start to slide).

The first box, M1 has frictional force F1 retarding it and some tension T in the string pulling it forward. The second box M2 has no friction with the floor, but has tension T retarding it and the force of the rope pulling it forward. Can you find an expression for the tension in the connecting string in terms of the force applied by the rope?

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tomas123
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#4
Sep26-11, 02:59 PM
P: 5

Maximum force on stacked boxes, when we don't want the boxes to slip.


I get:

max tension = F2 = 23.52 N
Friction F1 = μk * n = 0.2 * 9.8 *7 = 13.72

tension on string = sum of forces on string = F2 - F1 = 37.24 N

The answer is supposed to be 36.6 N. What did I do wrong?

I really appreciate the help!
gneill
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#5
Sep26-11, 03:26 PM
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P: 11,414
Quote Quote by tomas123 View Post
I get:

max tension = F2 = 23.52 N
Friction F1 = μk * n = 0.2 * 9.8 *7 = 13.72
Okay, good so far!
tension on string = sum of forces on string = F2 - F1 = 37.24 N
Oops. The frictional forces are not the only ones acting. Unless the applied force is specifically metered to keep the blocks moving at a constant velocity, there will be acceleration and thus inertial forces (m*a) involved for each block. The expression for the tension will be just a bit more complicated because of this.

For the moment, concentrate on box M1 which is being pulled by the string and retarded by friction. If the acceleration of the block is a, what must the tension in the string be?
tomas123
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#6
Sep26-11, 04:49 PM
P: 5
aah I get it!

the sum of F on M1 is 9.8N. Divided by 3kg = 3.267 m/s2.

Sum of F on whole system = force by rope - friction
(7kg * 3.267 m/s2) +13.72 = force by rope
36.589 = force by rope

Thank you!!


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