Calculating the Force Needed to Move Stacked Boxes with Friction

[oreosama]

I'm asking purely in general,

say you have two boxes stacked, if you push on the bottom one, is the top being moved forward by a friction? If this is true it implies an opposite friction on the top of the bottom box?
http://i.imgur.com/7Dya1.png
Is this "balanced" properly? If not help me fix my understanding!
Assuming the above is all good, I have a question from my homework here. same situation as above but with a rope tied to wall on top block

http://i.imgur.com/v1CgQ.png

I applied the forces the same I did before, is this correct?(it seems to make sense, T cancels out top friction meaning it doesn't move, more resistance to pull top block due to two friction forces acting on it.)

Thanks for any confirmation and help


And assuming I've gotten all of that, here's a problem

block a of mass m sits atop another block of mass 3m. block a is tied to a rope with the other end tied to a wall the coeefcient of kinetic friction between each surface is uk. the bottom block is pulled to the left with force F

given m, uk determine force F necessary to move the bottom block at constant speed

F = 4*uk*mg

 

[azizlwl]

When a force applied to the bottom box, it moves to the left.
The relative motion of the top is to the right.Friction opposes this motion to the right and and pulls the top box to the left. Maximum force is Mgμ_s

 

[ehild]

I'm asking purely in general,

say you have two boxes stacked, if you push on the bottom one, is the top being moved forward by a friction? If this is true it implies an opposite friction on the top of the bottom box?
http://i.imgur.com/7Dya1.png
Is this "balanced" properly? If not help me fix my understanding!

The top and bottom boxes interact trough their normal force and through friction. The force of friction acts against the relative motion of the boxes. The top box exerts a backward force on the bottom one and the bottom box exerts a force of same magnitude but forward to the top one. If Ff≤μN is enough to accelerate both boxes with the same acceleration the boxes are stationary with respect to each others. Otherwise the top box will lag behind the bottom one.

Assuming the above is all good, I have a question from my homework here. same situation as above but with a rope tied to wall on top block

http://i.imgur.com/v1CgQ.png


I applied the forces the same I did before, is this correct?(it seems to make sense, T cancels out top friction meaning it doesn't move, more resistance to pull top block due to two friction forces acting on it.)

Thanks for any confirmation and help

that is correct.

And assuming I've gotten all of that, here's a problem

block a of mass m sits atop another block of mass 3m. block a is tied to a rope with the other end tied to a wall the coeefcient of kinetic friction between each surface is uk. the bottom block is pulled to the left with force F

given m, uk determine force F necessary to move the bottom block at constant speed

F = 4*uk*mg

There are two forces of friction exerted to the big box, one from the top box, the other from the ground.
You need to multiple the normal force with μ to get the force of friction. The normal force between the boxes is mg, but what is the normal force between the big box and the ground? How much force is needed to keep both boxes on the surface?

ehild

 

[oreosama]

so I just need to account for force down from top box and account for it on the bottom one?(in this case just add mg to 3mg)?

normal between box and ground 3*uk*mg

to keep both on surface 4*uk*mg

F = 5*uk*mg ?

 

[ehild]

so I just need to account for force down from top box and account for it on the bottom one?(in this case just add mg to 3mg)?

normal between box and ground 3*[STRIKE]uk[/STRIKE]*mg

to keep both on surface 4*[STRIKE]uk[/STRIKE]*mg

F = 5*uk*mg ?

Considering the bottom box, force mg acts downward from the top box, and 3mg acts from gravity also downward, so the ground exerts N=4mg force upward. That is the force required to keep the whole thing on the surface.
There is a (horizontal) force of friction from the top box and the other from the ground, both in opposite direction as the applied force.

Your end result is correct, however.

ehild