Friction: Two Blocks Sliding

[Naeem]

Q. Two blocks are sliding across a table pushed by a force F from the left equal to 23 N. The force of friction from the table on m1 is 4 N and the force of friction on m2 is 12 N. m1 = 2 kg and m2 = 6 kg.

a) What is the magnitude acceleration of the entire system?

I figured this part out. Turns out to be 0.88 m/s2

But need help with the following parts.

b) What is the magnitude of force from m1 on m2?
c) What is the magnitude of acceleration of only m1?
d) What is the magnitude of force from m2 on m1?

Thanks,

Naeem

[Doc Al]

Start by identifying (labeling) the forces on each block. Then apply Newton's laws to one of the blocks and see what you can deduce.

[Naeem]

I found out that the acc, for part c is 0.88 m/s2, but please help me with setting up
part b and d.

Thanks,

[Doc Al]

Start by doing what I suggested in my last post. Pick a block (m1, say) and identify all the forces on it.

[Naeem]

I tried setting up but no use.

all I know is F= ma

and

Ffric - Fapp = m * a

found combined acc. for part a which is correct.


Please help me in setting up eqns,

Thanks,

[Chronos]

You need to solve for the vector force.

[Doc Al]

I tried setting up but no use.

all I know is F= ma

and

Ffric - Fapp = m * a

On m1 there are three forces acting (horizontally):
(1) the applied force: 23 N to the right
(2) the friction from the table: 4 N to the left
(3) the force that m2 exerts on m1: that's what you need to find out! (which direction does it act?)

Now set up your equation.

[Naeem]

May be something like this:

F21 + F fric - Fapp = max

F21 + 4 - 23 =6 ( 0.88 )

[Doc Al]

May be something like this:

F21 + F fric - Fapp = max

F21 + 4 - 23 =6 ( 0.88 )
Almost. But direction, and thus the signs, matters. (Also, you have the wrong mass.) I'll take positive to mean "to the right", thus:
F_{app} - F_f - F_{21} = m_1 a
and so:
23 - 4 - F_{21} = (2)(0.875)

[Naeem]

Thanks, very much Doc, for all you guiding and help