   Chapter 3.1, Problem 64E

Chapter
Section
Textbook Problem

# 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 = μ W 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 tan θ = μ .

To determine

To show:

That F is minimized when tanθ=μ.

Explanation

1) Concept:

Use the Closed Interval Method to show that F is minimized when tanθ=μ

The Closed Interval Method:

To find the absolute maximum and minimum values of a continuous function f on a closed interval a, b:

i. Find the values of f at the critical numbers of f in a, b

ii. Find the values of f at the end points of the interval

iii. The largest of the values from step (i) and (ii) is the absolute maximum value; the smallest of these values is the absolute minimum value.

2) Given:

The magnitude of the force is

F=μWμsinθ + cosθ,   0  θ  π/2

3) Calculation:

Differentiate F with respect to θ by using the quotient rule of derivative.

F'θ= (μsinθ + cosθ) 0- μW  (μ cosθ-sinθ)μsinθ + cosθ2

F'θ= -μW  (μ cosθ-sinθ)μsinθ + cosθ2

To find critical points, equate F'θ=0.

-μW  (μ cosθ-sinθ)μsinθ + cosθ2=0

-μW  (μ cosθ-sinθ)=0

μ cosθ-sinθ =0

μ=tanθ ………. this is the only critical point since F'θ  exists for all values of θ in the given interval.

Substitute μ=tanθ in F

F=tanθ·Wtanθsinθ+cosθ

F=sinθcosθ·Wsinθcosθ ·sinθ+cosθ

Factor out 1cosθ from the numerator and denominator of the above step

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