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.
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
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
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)
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) )
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