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michael7 · Oct 9, 2010, 11:18 PM

A horizontal force F is applied to a block of mass M1. Another block of mass M2 is adjacent to M1 as shown below. The two blocks move to the right with constant velocity. The coefficient of kinetic (static) friction between the floor and each block is μk (μs).

What is the f, the magnitude of the frictional force that the floor exerts on M2?

f = μk (M1 + M2) g
f = μk M2 g
f = F

What is F21, the magnitude of the force that M2 exerts on M1?

F21 = 0
F21 = μk M2 g
F21 = F

michael7 · Oct 9, 2010, 11:34 PM

Following the previous idea

Suppose M1 = 2kg, M2 = 5kg, μS = 0.4, and μk = 0.25. When the force F = 30N, the blocks are observed to accelerate to the right. What is F12, the magnitude of the force that M1 exerts on M2 in this case?

F12 = 0
F12 = 3.67N
F12 = 9.18N
F12 = 21.4N
F12 = 30.0N

Unknown008 · Oct 10, 2010, 12:25 AM

1. No. The normal force acting on M2 is only the mass M2, not (M1 + M2).

2. Since the two move with constant speed, the net force on M2 is 0.

$(F_{2\ on\ 1} - F_f) = 0$

$F_{2\ on\ 1} = \mu_kR$

$F_{2\ on\ 1} = \mu_kM_2g$

3. What have you tried?

michael7 · Oct 11, 2010, 07:58 PM

so I think for number 3, the acceleration for the whole system block is 30/7 m/s^2

and I think I should use the equation: F(total)=F(12)-F(friction)

so for the F(friction) I just plug in what is come up in number 1, for F(total)=mass*acceleration

am I right? But when I do it this way, I can't get the right value

Unknown008 · Oct 12, 2010, 12:32 AM

Not quite right from the start. You didn't consider the frictional forces at all.

$F_{net} = (M_1 + M_2)a$

$F - F_{f1} - F_{f2} = 7a$

$30 - (0.25)(2)(9.8) - (0.25)(5)(9.8) = 7a$

$a = 1.84 ms^{-2}$

Then, considering M2 only;

$F_{net} = M_2a$

$F_{1\ on\ 2} - F_f = 5(1.84)$

$F_{1\ on\ 2} = 9.18 + (0.25)(5)(9.8) = 21.4 N$