Homework Help: Two blocks and friction

1. May 11, 2016

terryds

1. The problem statement, all variables and given/known data

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

2. Relevant equations
Newton's law of motions
Kinematics

3. The attempt at a solution

Free-body diagram for block A

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

Free-body diagram for block B

F(friction between A and B) = 2N <-------B --------> F = 10 N

Then, I get stuck..

2. May 11, 2016

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 !

3. May 11, 2016

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 !

4. May 11, 2016

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.

5. May 11, 2016

terryds

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. May 11, 2016

BvU

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. May 11, 2016

BvU

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

8. May 11, 2016

terryds

No...
Maybe it's treated as a particle :|

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 ??

9. May 11, 2016

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) ?

10. May 11, 2016

haruspex

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.

11. May 11, 2016

terryds

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. May 12, 2016

BvU

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