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

[terryds]

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

 

[BvU]

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 !

 

[BvU]

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 !

 

[CWatters]

+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.

 

[terryds]

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 !

Okay,

For the block A

F_friction = m_a a_a
a_a = 2/1 = 2 m/s^2

For the block B

F - F_friction = m_b a_b
a_b = (10-2)/3 = 8/3 m/s^2

The time when the displacement is equal

0.5 * a_a * t^2 = 0.5 * a_b * t^2
a_a * t^2 = a_b * 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.

 

[BvU]

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 * a_b * t^2 - 0.5 * a_a * t^2

 

[BvU]

Does the exercise really tell you nothing at all about the initial position of A ? Or about its size ?

 

[terryds]

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 :|

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 * a_b * t^2 - 0.5 * a_a * t^2

Alright,

0.5 * a_b * t^2 - 0.5 * a_a * 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 ??

 

[BvU]

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) ?

 

[haruspex]

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.

 

[terryds]

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

 

[BvU]

well done - let's hope it's the preferred answer