
#1
Aug2210, 10:06 PM

P: 87

1. The problem statement, all variables and given/known data
A 0.5kg wooden block is placed on top of a 1.0kg block. The coefficient of static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.20. What is the maximum horizontal force that can be applied to the lower block without the upper block slipping? 2. Relevant equations F_{NET} = ma = 0 3. The attempt at a solution F_{K} = 1.5 * 9.8 * 0.20 F_{K} = 2.94 N F_{S} = 0.5 * 9.8 * 0.35 F_{S} = 1.715 F_{NET} = F_{S} = F_{A}  F_{FK} 1.715 N = F_{A}  2.95 N 1.715 N + 2.95 N = F_{A} 4.655 N = F_{A} 4.7 N = F_{A} Therefore, the maximum horizontal force that can be applied is 4.7 N without the upper, 0.5 kg, block moving. Is this correct? 



#2
Aug2310, 12:12 AM

P: 158

yup, assuming you did your math right.




#3
Aug2310, 12:39 AM

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P: 9,819

If I understand well, you calculated the force which would move the boxes with constant velocity. The question was to find the maximum force applied so as the two boxes move together, which means the same velocity and acceleration.
ehild 



#4
Aug2310, 12:54 AM

P: 158

two boxes on each other 



#5
Aug2310, 03:49 AM

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P: 9,819

The horizontal forces acting at the lower box are:
F, to the right; The kinetic friction between the lower bock and the ground Fk, on the left; The force of static friction between the boxes, Fs , on the left. The resultant force is FFkFs=F2.941.715 N. If F= 4.655, the resultant force on the lower box is zero, so it does not accelerate. This applied force is the minimum that keeps the boxes in motion with constant velocity. 



#6
Aug2310, 08:55 AM

P: 158

The maximum acceleration so that the top box doesn't fall of is: 1.715 N = 0.5 kg * a, then use that a for Fnet= ma= F2.941.715 N. Careful what m is. sorry about that. 



#7
Aug2310, 01:20 PM

P: 87





#8
Aug2310, 01:51 PM

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P: 9,819

Yes, that force, 4.665 N is equal to the net frictional force on the lower box. What is the acceleration of that box then?
ehild 



#9
Aug2310, 01:57 PM

P: 87





#10
Aug2310, 02:13 PM

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P: 9,819

You said that the maximum static friction on the 0.5 kg box is 1.715 N. This allows 3.43 m/s^2 acceleration. How did you get that 3.1 m/s^2?
ehild 



#11
Aug2310, 02:23 PM

P: 158

Now [tex] ma_{net}= F  F_k F_s [/tex], remember you are applying the force on the lower box, so you should know what the value of m is from that. **edit: Ok i see how you got 3.10. You did 4.655/1.5; however, you are forgetting the two frictional force on the lower box thats in the opposite direction of F, which also has a total magnitude of 4.655. Hence, the net acceleration is 0 if the applied force is 4.655N. Does that make sense? 



#12
Aug2310, 10:30 PM

P: 87

I truly, so not see where you are getting at with your answer. I am completely confused. 



#13
Aug2310, 10:51 PM

P: 158

"The top box cannot have a force higher than 1.715"  this is true "the bottom box has to overcome 2.95 N"  This is not true. Draw your freebody diagram for the lower box, and you will see that there is not only the kinetic frictional force of 2.95, but an additional frictional force due to the top box in the opposite direction to F < This is key! "The top box will have a constant speed of 0 m/s while the bottom box will be accelerating at 1.14m/s/s. " This is obviously not true. How could the top box have constant (or 0 m/s) speed while the bottom one is accelerating? They need to be moving(accelerating) in unison. Edit: Here are a couple of things that might help you understand the problem better: 1. Draw out the free body diagrams carefully (paste it here if you want) 2. Consider what is actually moving the top box (The frictional force is moving the box and its pointing in the direction of its motion. This might seem contradictory at first at first) 


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