Force pushes three blocks, find mag of force exerted on last two blocks

gap0063

1. The problem statement, all variables and given/known data
Three blocks in contact with each other are
pushed across a rough horizontal surface by a
75 N force as shown.
The acceleration of gravity is 9.8 m/s2 .


If the coefficient of kinetic friction between
each of the blocks and the surface is 0.076,
find the magnitude of the force exerted on the
3.5 kg block by the 4.1 kg block.
Answer in units of N.

2. Relevant equations
sum of Fy=N-mg=0 => N=mg
sum of Fx= -fk=max
fk=uN


3. The attempt at a solution
I have no clue how to figure out the force just between those two block...

Do I add the first two blocks together to act a one force against the third block?

Doc Al

Hint: Start by treating the three blocks as a single system. What can you deduce?

gap0063

You mean like m1+m2+m3=9.6 kg

Then the forces acting in the y direction is N1+N2+N3 and (m1+m2+m3)g
forces acting in the x direction is fk and F

which fk=u*(N1+N2+N3)

right?

Doc Al

Right. Keep going. Apply Newton's 2nd law.

gap0063

so the sum of Fy= N1+N2+N3-Mg=0, where M is m1+m2+m3
so N1+N2+N3=Mg=> where N1+N2+N3=94.08

so the sum of Fx= -fk=Ma

fk=u(N1+N2+N3)
fk=7.15008

so -7.15008=9.6a
a=-0.7448 m/s^2
this is small... is this right?

Doc Al

OK.

ΣF = ma
But ΣF ≠ -fk. Friction is not the only horizontal force acting.

OK.

No, as explained above.

gap0063

Okay so where does fk play into?

Doc Al

Friction is one of the horizontal force acting on the blocks. What direction does it act?

What other horizontal forces act? Look at your diagram.

gap0063

is it sum of Fx= fk+F=ma?

Doc Al

Yes, but make sure you have the correction directions and thus the correct signs for the forces. (Hint: Let 'right' be positive and 'left' be negative.)

gap0063

Okay, so I did -fk+F=Ma

-7.15008+75=9.6a => a=7.0677 m/s^2 (wrong)

what did I do wrong?

Doc Al

Looks fine to me. Why do you say it's wrong? Did the problem even ask you for the acceleration? (Finding the acceleration is just a step towards the required answer.)

gap0063

You're right I need the force between m2 and m3...

So can I do Sum of F= (m2+m3)*a?

Doc Al

That won't tell you anything about the forces between m2 and m3. Hint: Analyze m3 alone or m1 + m2 together.

gap0063

So the only forces acting on m3 in the x direction is the force from m1 and m2...
in the y direction it is N and m3g

So for the force in the x direction from m1 and m2 (i'll call P12)

F-P12=m3a?
I'm not sure how to calculate P12...

Doc Al

No. What about friction? Which block is m3? Your diagram isn't labeled.
OK.

Once you account for all the forces, the only unknown will be the force between m2 and m3. You'll solve for it.

gap0063

So then sum of F = F-P12-fk=ma?

and P12 is my force bwtween m2 and m3 right?