We can get even more exited if we consider only one rigid
body of the mass M, shaped like it is shown on the picture (of course, the
shape does not matter, but it helps to see the similarity).
Now we have just four forces acting on the body:
F is the force acting on the block m from the spring.
Fcf is the force of kinetic friction acting on
the body from the surface.
N is the normal force form the surface to the body.
If μ is the coefficient of the kinetic friction between
the body and the surface, Fcf
= μ N.
Mg is the force of gravity acting on the bodies.
a - is the accelerations of the bodies.
The 2nd NL for the object (in components to the
horizontal and vertical axes) is:
F Fcf = M
a Mg N = 0
Plus, we can add the definition of the μ; Fcf = μ N.
The solution is obvious:
F - μ Mg = M a
In the case of the motion with a constant velocity we have a
= 0 and get the solution for the coefficient of friction
But we know that we can always treat a rigid body as a
system of interacting parts (we can do it, for example, to prove the 2nd
NL for the center mass). Lets divide (mentally!) the given object into two
parts, like on the next picture.
What
would be the free body diagram and the 2nd NL for the part 1, when
the acceleration is 0? Can we say that the force interacting between the parts
is 0 (similarly to the situation of two bodies)?