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Two boxes on each other

by barthayn
Tags: boxes
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barthayn
#1
Aug22-10, 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
FNET = ma = 0


3. The attempt at a solution

FK = 1.5 * 9.8 * 0.20
FK = 2.94 N

FS = 0.5 * 9.8 * 0.35
FS = 1.715

FNET = FS = FA - FFK

1.715 N = FA - 2.95 N
1.715 N + 2.95 N = FA
4.655 N = FA
4.7 N = FA

Therefore, the maximum horizontal force that can be applied is 4.7 N without the upper, 0.5 kg, block moving.

Is this correct?
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aq1q
#2
Aug23-10, 12:12 AM
P: 158
yup, assuming you did your math right.
ehild
#3
Aug23-10, 12:39 AM
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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

aq1q
#4
Aug23-10, 12:54 AM
P: 158
Two boxes on each other

Quote Quote by ehild View Post
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
I think thats why he/she added the 1.715 N, this is so that two boxes move together with same velocity AND acceleration
ehild
#5
Aug23-10, 03:49 AM
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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 F-Fk-Fs=F-2.94-1.715 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.
aq1q
#6
Aug23-10, 08:55 AM
P: 158
Quote Quote by ehild View Post
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 F-Fk-Fs=F-2.94-1.715 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.
Ah, absolutely 4.655 isn't the maximum. I should really be careful when I look at these late at night.. *angry face* :/. F needs to be maximized such that the top box doesn't fall off.

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= F-2.94-1.715 N. Careful what m is.

sorry about that.
barthayn
#7
Aug23-10, 01:20 PM
P: 87
Quote Quote by ehild View Post
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 F-Fk-Fs=F-2.94-1.715 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.
How would this be the minimum force that keeps the boxes in motion with a constant velocity? Wouldn't it be the frictional force of the lower block? The answer I got was for the upper block not to move on top of the lower block. Therefore, in order to overcome the bottom frictional force and not overcome the top I had to add both of the frictional forces together to get the maximum force. This should be the maximum force because if you subtract the applied force by the kinetic frictional force you will get the static frictional force.
ehild
#8
Aug23-10, 01:51 PM
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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
barthayn
#9
Aug23-10, 01:57 PM
P: 87
Quote Quote by ehild View Post
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
The acceleration of the whole system is 3.10 m/s/s. So, my maximum applied force is correct?
ehild
#10
Aug23-10, 02:13 PM
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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
aq1q
#11
Aug23-10, 02:23 PM
P: 158
Quote Quote by barthayn View Post
The acceleration of the whole system is 3.10 m/s/s. So, my maximum applied force is correct?
I'm guessing that should be 3.43m/s^2? Rounding error perhaps? How did you get this?

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?
barthayn
#12
Aug23-10, 10:30 PM
P: 87
Quote Quote by aq1q View Post
I'm guessing that should be 3.43m/s^2? Rounding error perhaps? How did you get this?

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?
No, it doesn't make sense. The top box cannot have a force higher than 1.715 N and the bottom box has to overcome 2.95 N of friction. Therefore, to have a maximum force that the top box will not fall will be both frictional forces added together. The top box will have a constant speed of 0 m/s while the bottom box will be accelerating at 1.14m/s/s.

I truly, so not see where you are getting at with your answer. I am completely confused.
aq1q
#13
Aug23-10, 10:51 PM
P: 158
Quote Quote by barthayn View Post
No, it doesn't make sense. The top box cannot have a force higher than 1.715 N and the bottom box has to overcome 2.95 N of friction. Therefore, to have a maximum force that the top box will not fall will be both frictional forces added together. The top box will have a constant speed of 0 m/s while the bottom box will be accelerating at 1.14m/s/s.

I truly, so not see where you are getting at with your answer. I am completely confused.
Ok, before I attempt to explain it again.. I'll try to show the flaw in your logic (bear with me here).

"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 free-body 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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