   Chapter 3.7, Problem 44E

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 a plane, then the magnitude of the force is F = μ W μ sin θ + cos θ where μ is a constant called the coefficient of friction. For what value of θ is F smallest?

To determine

To find:

The value of θ for which F is smallest.

Explanation

1) Concept:

i. A critical number of a function f   is a number c in the domain of f  such that either  f'c=0 or f'c does not exist.

ii. First derivative test for absolute extreme values- suppose that c is a critical number of a continuous function f defined on an interval.

If f'x>0 for all x<c and f'x<0 for all x>c, then fc is the absolute maximum value of f.

If f'x<0 for all x<c and f'x>0 for all >c, then fc is the absolute Minimum value of f.

2) Given:

F=μWμsinθ+cosθ

3) Calculation:

If the rope is making an angle θ with the plane, then magnitude of force is F=μWμsinθ+cosθ

For finding θ  forsmallest value of force first find critical points.

Therefore, differentiating F with respect to θ by using rules of differentiation,

By using rule of constant multiplication.

F'θ=ddθμWμsinθ+cosθ=μWddθ1μsinθ+cosθ

By using power rule,

=μW- 1μsinθ+cosθ2·ddθ(μsinθ+cosθ)

=μW- 1μsinθ+cosθ2(μcosθ-sinθ)

=-μW(μcosθ-sinθ)μsinθ+cosθ2

So for F'θ=0, μcosθ-sinθ=0

Therefore, μcosθ=sinθ;

Dividing by cosθ on both sides,

μ=tan

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