Question

A box of books is initially at rest a distance D = 0.600 m from the end of a wooden board. The coefficient of static friction between the box and the board is μs = 0.324, and the coefficient of kinetic friction is μk = 0.248. The angle of the board is increased slowly, until the box just begins to slide; then the board is held at this angle. Find the speed of the box as it reaches the end of the board.

Answer 1

Let m be the mass of the box, and suppose that the angle of the board is
alpha.
Notice that there are two forces acting on the box:
1)
gravitation force F=mg
2) the board reaction force N = mg cos(alpha),

3) the force of static friction Fs
see figure in the attached
file.

Then the sum F + N is directed along the board, and is equal to

T=mg sin(alpha)

The friction force is opposite to T whenever

T < μs N

Substituing the values of T and N we get:
mg
sin(alpha) < μs mg cos(alpha)
whence
tan(alpha) < μs

Thus
the will box start to move when
tan(alpha) = μs

Whence

cos(alpha) = 1/square_root(1+μs^2)
sin(alpha) =
μs/square_root(1+μs^2)





Suppose now that the box is
moved.
Then instead of the force of static friction Fs, there is a force of
kinetic friction equal to
Fk = μk N = μk mg cos(alpha)

So the
resulting force acting on the box is
F = mg sin (alpha) - μk mg
cos(alpha),
where
tan(alpha) = μs
cos(alpha) =
1/square_root(1+μs^2)
sin(alpha) = μs/square_root(1+μs^2)

Whence

F = mg μs/square_root(1+μs^2) - mg μk/square_root(1+μs^2)=
=
mg(μs-μk)/square_root(1+μs^2)

Thus the box moves with acceleration

a = F/m = g(μs-μk)/square_root(1+μs^2)
with initial zero velocity.
We need
to find its velocity at the distance d = 0.6 m
We have that
v =
at,
and
d = at^2 /2, where t is the corresponding time,
whence

t = square_root(2d/a)
and so
v = at = a square_root(2d/a)
=
square_root(2da) =
= square_root( 2dg(μs-μk)/square_root(1+μs^2)
)

Substituting the values:
d=0.6
g=9.8
μs = 0.324
μk =
0.248

we get

v = square_root( 2* 0.6 * 9.8 (0.324 -
0.248)/square_root(1+0.324^2) )=
= 0.92209 m/s