How long will it take for the small cube to fall?

In summary: The size of A makes no difference to the distance covered. It's only a question of whether the mass centre has to traverse 1m of the lower cube or 0.5m (or anything else up to 1m).
  • #1
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Homework Statement



A cube with 1 kg mass is put on top of a bigger cube with 3kg mass and 1 meter edges.
If 10 N force is exerted on the bigger cube, while the maximum friction between the cube surface is 2 N, and g=10m/s^2 , then at one time the small cube will fall to the floor.
The time needed for the small cube to reach the floor since the force is exerted is about ...
(Assume there is no friction between the floor and the big cube)
A. 1 s
B. 1.4 s
C. 1.7 s
D. 2.4 s
E. 2.7 s

Homework Equations


Newton's law of motions
Kinematics

The Attempt at a Solution



Free-body diagram for block A

A ----> F (friction between A and B) = 2N

Free-body diagram for block BF(friction between A and B) = 2N <-------B --------> F = 10 NThen, I get stuck..
Please help

 
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  • #2
No need to get stuck. apply the kinematic equations and see what comes out !
You calculate the acceleration of block A and B and find out when A doesn't stay on top of B any more.
The size of A isn't given, so it'll be educated guessing which of the answers is right (there's also some time needed to fall down 1 m)

Must be hollow styrofoam blocks or something: 3 kg for 1 m3 is very light :smile: !
 
  • #3
No need to get stuck. apply the kinematic equations and see what comes out !
You calculate the acceleration of block A and B and find out when A doesn't stay on top of B any more.
The size of A isn't given, so it'll be educated guessing which of the answers is right (there's also some time needed to fall down 1 m)

Must be hollow styrofoam blocks or something: 3 kg for 1 m3 is very light :smile: !
 
  • #4
+1

From the FBDs you can calculate the net force on each block and hence it's acceleration. The accelerations are different so the displacements vs time are different. Figure out the condition under which the displacements are different enough for the top one to fall off. Write the kinematic equations and solve for time.
 
  • #5
BvU said:
No need to get stuck. apply the kinematic equations and see what comes out !
You calculate the acceleration of block A and B and find out when A doesn't stay on top of B any more.
The size of A isn't given, so it'll be educated guessing which of the answers is right (there's also some time needed to fall down 1 m)

Must be hollow styrofoam blocks or something: 3 kg for 1 m3 is very light :smile: !

Okay,

For the block A

Ffriction = ma aa
aa = 2/1 = 2 m/s^2

For the block B

F - Ffriction = mb ab
ab = (10-2)/3 = 8/3 m/s^2

The time when the displacement is equal

0.5 * aa * t^2 = 0.5 * ab * t^2
aa * t^2 = ab * t^2
2 t^2 = 8/3 t^2

which means there are no solution for t

I don't know how to determine the time from initial position to the position where A is about to fall from B.
 
  • #6
terryds said:
The time when the displacement is equal
That is the case at t=0. After t = 0 block A lags behind B (it accelerates slower). It can't stay behind too far or it will fall off ! So you are interested in the difference 0.5 * ab * t^2 - 0.5 * aa * t^2
 
  • #7
Does the exercise really tell you nothing at all about the initial position of A ? Or about its size ?
 
  • #8
BvU said:
Does the exercise really tell you nothing at all about the initial position of A ? Or about its size ?
No...
Maybe it's treated as a particle :|

BvU said:
That is the case at t=0. After t = 0 block A lags behind B (it accelerates slower). It can't stay behind too far or it will fall off ! So you are interested in the difference 0.5 * ab * t^2 - 0.5 * aa * t^2

Alright,

0.5 * ab * t^2 - 0.5 * aa * t^2 = 1
8/3 t^2 - 2 t^2 = 2
2/3 t^2 = 2
t^2 = 3
t = √3 s ≈ 1.7 s

Fall duration :
t = √(2h/g) = √(2/10) = √(1/5) = 0.4 s

So, it's about 2.1 second, right ??
 
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  • #9
That's correct -- if A is small and its starting position is on the right edge of B. What if it starts in the center (also a reasonable assumption, I would say) ?
 
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  • #10
BvU said:
That's correct -- if A is small and its starting position is on the right edge of B. What if it starts in the center (also a reasonable assumption, I would say) ?
The size of A makes no difference to the distance covered. It's only a question of whether the mass centre has to traverse 1m of the lower cube or 0.5m (or anything else up to 1m).
Where it does matter is in the process of falling off. If not a particle, it has to rotate, slowing its descent.
 
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  • #11
BvU said:
That's correct -- if A is small and its starting position is on the right edge of B. What if it starts in the center (also a reasonable assumption, I would say) ?

0.5 * ab * t^2 - 0.5 * aa * t^2 = 0.5
8/3 t^2 - 2 t^2 = 1
2/3 t^2 = 1
t^2 = 3/2
t = √(3/2) s ≈ 1.2 s

Total time = 1.2 + 0.4 = 1.6 s
Thanks
 
  • #12
well done - let's hope it's the preferred answer :rolleyes:
 
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1. What is friction?

Friction is a force that opposes the motion of two surfaces that are in contact with each other.

2. How does friction affect two blocks?

Friction between two blocks will cause a resistance to the motion of the blocks, making it harder for them to slide past each other.

3. What factors affect the amount of friction between two blocks?

The amount of friction between two blocks is affected by the type of surfaces in contact, the force pressing the blocks together, and the roughness of the surfaces.

4. How can friction be reduced between two blocks?

Friction can be reduced by using a lubricant, such as oil or grease, to create a slippery layer between the two surfaces.

5. Why is friction important in the study of two blocks?

Friction plays a crucial role in understanding the behavior of objects in contact with each other, and it allows us to control the movement of objects and prevent them from sliding uncontrollably.

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