A copy of The History of the Decline and Fall of the Roman Empire by Edward Gibb

Question

A copy of The History of the Decline and Fall of the Roman Empire by Edward Gibbon lies exactly in the middle of an otherwise empty 6.1m long bookshelf in Robarts Library. The coefficient of static friction for the book on the shelf is ?S = 11/60 and the shelf has been polished (so kinetic friction may be neglected). A tall, disgruntled librarian slowly lifts up the one end of the bookshelf at a constant rate, so that it pivots upward, anchored at the opposite end (see Figure 2). After 5.5 seconds, she sees the book begin to slide down the
tilted shelf. She continues to raise her end of the shelf at the same rate, gleefully watching until the book falls onto the floor.

a. What is the speed at which the librarian raised the shelf?
b. Draw a graph of the book

Explanation / Answer

w = angular velocity by which the librarian is tilting the board
L = Lenght of board = 6.1 m
u = static coefficent of friction.
o = angle the board makes at the pivoting end

a) board starts to move when the gravitational force is equal to the frictional force

f_friction = u * f_normal = - u * mgsin(o)
f_gravity = mg cos(o)

book moves when f_gravity + f_friction = 0 or
f_gravity = - f_friction
mg cos(o) = u mg sin(o)
tan(o) = 1/u
o = tan^-1 (1/u) = 1.327 degrees

b)
force that moves the board along the surface of the board is mgcos(o(t)). So Newton's law:
ma(t) = mgcos(o(t))
where o(t) is the angle as a function of time and since the board is tilting at a constant angular velocity
o(t) = w t
so
a(t) = g * cos(wt)
integrating and using following facts:

v(t-t0) = 0 since the book has zero velocity to start. t0 = 5.5 seconds
v(t-t0) = g/w * sin(w(t-t0))

so if I take the middle of the board to be my origin so the the starting postion of the book is
s(t-t0)=0, then
s(t-t0) = -g/(w^2) * cos(w(t-t0))
so the book hits the floor when
s(t-t0) = L/2

s(t-t0) = -g/(w^2) * cos(w(t-t0)) = L/2
cos(w(t-t0)) = L*w^2/(2g)
w(t-t0) = cos^-1(L*w^2/2g)

(t-t0) = (cos^-1(L*w^2/2g))/w = time it takes to get to the floor

Note, that the book started move at t0 = 5.5 s when so the height at the end of board that is being lifted is
h = L * sin(o(t=0)) = (6.1m) * sin( 1.327 deg) = 0.141m
w = o(t0)/t0 = 1.327 deg / 5.5 s = 0.2412 deg/s or 0.00421 radian/s

plug in to get result
(t-t0) = (cos^-1(L*w^2/2g)) / w

careful with the angular units on your calculator.

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