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Maximum force on stacked boxes, when we don't want the boxes to slip. 
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#1
Sep2611, 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. 


#2
Sep2611, 09:20 AM

P: 5

1 problem! the solution is 36.6 N



#3
Sep2611, 10:17 AM

Mentor
P: 11,678

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? 


#4
Sep2611, 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! 


#5
Sep2611, 03:26 PM

Mentor
P: 11,678

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? 


#6
Sep2611, 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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