Force required to negate vertical acceleration with friction

[Obliv]

Homework Statement

Two blocks m (16kg) and M (88kg) are as shown in the figure. If the co-efficient of friction b/w the blocks is 0.38 but the surface beneath the block is frictionless, what is the minimum force required to hold m against M?

Homework Equations

F_net = ma
F_s = μ_s*F_a

The Attempt at a Solution

By using this equation for m F_nety = mg - F_s = 0
One can find the force required F_req = (16*9.8) / μ_s to be 412.63N

From there I have read that because that is the force required on m on M, it is not the required applied force on the system. That comes out to be around ~487N using proportions. How is this part solved formally? Please be as formal as possible because I want to get this concept down fundamentally.
Thank you very much.

(side question regarding the concept of force transferring through a system of mass like this one: Say there was a basketball in space weighing 5kg. If you applied a force of 10N on it alone, it would accelerate at 2m/s². If the basketball was connected, by a nearly massless rigid rod that extended millions of miles, to a rock weighing 5000kg, and you pushed on the basketball with 10N, would its acceleration ever change or would it be 10N/5005kg from beginning to end?)

 

[SammyS]

Homework Statement

Two blocks m (16kg) and M (88kg) are as shown in the figure. If the co-efficient of friction b/w the blocks is 0.38 but the surface beneath the block is frictionless, what is the minimum force required to hold m against M?

Homework Equations

F_net = ma
F_s = μ_s*F_a

The Attempt at a Solution

By using this equation for m F_nety = mg - F_s = 0
One can find the force required F_req = (16*9.8) / μ_s to be 412.63N

From there I have read that because that is the force required on m on M, it is not the required applied force on the system. That comes out to be around ~487N using proportions. How is this part solved formally? Please be as formal as possible because I want to get this concept down fundamentally.
Thank you very much.

(side question regarding the concept of force transferring through a system of mass like this one: Say there was a basketball in space weighing 5kg. If you applied a force of 10N on it alone, it would accelerate at 2m/s². If the basketball was connected, by a nearly massless rigid rod that extended millions of miles, to a rock weighing 5000kg, and you pushed on the basketball with 10N, would its acceleration ever change or would it be 10N/5005kg from beginning to end?)

It's generally recommended for problems such as this, to sketch an FBD (free body diagram) for each mass (object) involved.

What you have labeled as F_{net y} is actually the minimum normal force required, between m and M, so that m doesn't slide down.

 

[Obliv]

Well it's also the net force on m in the y-direction is it not? I thought the only forces acting on m are gravity and the static force of friction through the applied force.
FBD I drew quickly. That is what I see visually along with the image. F_a is equal in both FBD's, correct?

 

[haruspex]

What you have labeled as F_{net y} is actually the minimum normal force required, between m and M, so that m doesn't slide down.

You mean, what is labelled F_req is the minimum normal force, yes?
Obliv, what is the relationship between the normal force and F? What is the relationship between the normal force and M?

 

[Biker]

What you have missed is the contact force effect on mass M, When you push something it moves. You have to add to that force the required force to make the two objects get the same acceleration.

A great place to understand fcontact is http://www.physicsclassroom.com/class/newtlaws/lesson-3/double-trouble
When you push an objects it pushes back with the same force as an attempt to move it (Newton third law). So when you treat each object individually there should be a force acting leftward which in your case you will treat it as N ( Us * N)
So the force overall should be:
F - n = m a

Now after you figure things out you will end up with the answer.

 

[Obliv]

What you have missed is the contact force effect on mass M, When you push something it moves. You have to add to that force the required force to make the two objects get the same acceleration.

A great place to understand fcontact is http://www.physicsclassroom.com/class/newtlaws/lesson-3/double-trouble
When you push an objects it pushes back with the same force as an attempt to move it (Newton third law). So when you treat each object individually there should be a force acting leftward which in your case you will treat it as N ( Us * N)
So the force overall should be:
F - n = m a

Now after you figure things out you will end up with the answer.

Thank you for that link. For some reason, Newton's third law did not occur to me.
for m -> F_net = ma = F_A - F_C (contact force)
for M -> F_net = Ma = F_c
for m in y-direction -> F_net = ma_y = F_s - mg = 0
F_s = μ_sF_C = mg
F_C = 412.63

Plugging the value of F_C back into the net force equation for M yields a = 4.688m/s² and using this information
F_net = (16)(4.688) = F_A - 412.63
F_A = 487.65N

I finally get it :)