Section 4.2 problem 56.
An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle θ with the plane, then the magnitude of the force
is
F =
µW
µ sin θ + cos θ
where µ is a positive constant called the coefficient of friction and where 0 ≤ θ ≤ π/2. Show that
F is minimized when µ = tan θ .
Solution: We use the closed interval method:
−µW (µ cos θ − sin θ)
(µ sin θ + cos θ)2
0
So F = 0 when µ = tan θ (setting the top = 0)
and F 0 is undefined when − µ1 = tan θ . (setting the bottom = 0)
However − µ1 = tan θ cannot happen since tan θ is always positive on the interval [0, π/2].
So the only critical value of θ happens when µ = tan θ .
To apply the closed interval method, we must compare the function value at the endpoints with the
function value at this critical point:
F0 =
F θ=0 = µW
F θ=π/2 = W
sin θW
tan θW
=
= sin θW
F µ=tan θ =
2
tan θ sin θ + cos θ
sin θ + cos2 θ
To help express the last computed function value in terms of µ (instead of θ ), we draw a right
triangle with one acute angle being θ . Since µ = tan θ , the side opposite of θ may be laµ
. Hence,
beled µ, and the adjacent side 1. From this triangle, we observe that sin θ = p
µ2 + 1
µ
F µ=tan θ = sin θW = p
W.
µ2 + 1
µ
µ
Notice, since µ is a positive number, p
is smaller than both µ and 1. So, p
W
µ2 + 1
µ2 + 1
is the absolute minimum value of the function F on the interval [0, π/2], and this happens when
µ = tan θ as desired.
Count the number times θ appears in the above text and email me that number for homework extra
credit.