Question

To determine the magnitude of force\(F\)is minimized if\(\tan \theta = \mu \).[B1]

[B1]Question must be exactly same as in book

Short answer

When\(\mu = \tan \theta \)the value of\(F\)is minimum.

The given statement is proven.

Step-by-step solution

Given information

The magnitude of the force, \(F = \frac{{\mu W}}{{\mu \sin \theta + \cos \theta }},0 \le \theta \le \frac{\pi }{2}\) where \(\mu \) is a positive constant called the coefficient of friction and \(W\) is the weight of the object.

Concept of Quotient Rule

Quotient Rule:\({\left( {\frac{f}{g}} \right)^\prime } = \frac{{{f^\prime }g - {g^\prime }f}}{{{g^2}}}\)

Differentiate the function and apply Quotient rule

Consider the given function,\(F = \frac{{\mu W}}{{\mu \sin \theta + \cos \theta }}\).

Differentiate\(F\)with respect to\(\theta \).

\({F^\prime }(\theta ) = \frac{d}{{d\theta }}\left( {\frac{{\mu W}}{{\mu \sin \theta + \cos \theta }}} \right)\)

Apply the Quotient Rule and simplify further.

\(\begin{aligned}{c}{F^\prime }(\theta ) &= \frac{{(\mu \sin \theta + \cos \theta )(0) - \mu W(\mu \cos \theta - \sin \theta )}}{{{{(\mu \sin \theta + \cos \theta )}^2}}}\\{F^\prime }(\theta ) &= \frac{{ - \mu W(\mu \cos \theta - \sin \theta )}}{{{{(\mu \sin \theta + \cos \theta )}^2}}}\end{aligned}\)

Set\({F^\prime }(\theta ) = 0\)and obtain the critical numbers.

\(\begin{aligned}{c}\frac{{ - \mu W(\mu \cos \theta - \sin \theta )}}{{{{(\mu \sin \theta + \cos \theta )}^2}}} &= 0\\ - \mu W(\mu \cos \theta - \sin \theta ) &= 0\\\mu \cos \theta - \sin \theta &= 0\\\mu &= \frac{{\sin \theta }}{{\cos \theta }}\end{aligned}\)

So, \( = \tan \theta \).

 Step 4: Put the Critical value and simplify

Substitute the obtained critical number\(\mu = \tan \theta \to F\).

\(\begin{aligned}{c}F(\tan \theta ) &= \frac{{(\tan \theta )W}}{{(\tan \theta )\sin \theta + \cos \theta }}\\ &= \frac{{(\tan \theta )W}}{{\left( {\frac{{\sin \theta }}{{\cos \theta }}} \right)\sin \theta + \cos \theta }}\\ &= \frac{{(\tan \theta )W}}{{\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta }}}}\\ &= \frac{{W\left( {\frac{{\sin \theta }}{{{{\cos }^2}\theta }}} \right)}}{{\frac{{{{\sin }^2}\theta {{\cos }^2}\theta }}{{(\cos \theta )}}}}\end{aligned}\)

Simplify further as follows:

\(\begin{aligned}{c}F(\tan \theta ) &= W\left( {\frac{{\sin \theta }}{{{{\sin }^2}\theta + {{\cos }^2}\theta }}} \right)\\ &= Wn\left( {\frac{{\sin \theta }}{1}} \right)\;\;\;\\{\sin ^2}\theta + {\cos ^2}\theta &= 1\\ &= W\sin \theta \end{aligned}\)

Express\(\sin \theta \)in terms of\(\mu \).

If\(\mu = \tan \theta \), use the Pythagoras formula to find the value for\(\sin \theta \).

\(\begin{aligned}{c}\tan \theta &= \frac{{{\rm{ opposite side }}}}{{{\rm{ adjacent side }}}}\\ &= \frac{\mu }{1}\end{aligned}\)

From this, it is observed that opposite side is\(\mu \)and adjacent side is\(1\).

By Pythagoras formula, hypotenuse will be\(\sqrt {{\mu ^2} + 1} \).

Thus, the value of\(\sin \theta \)becomes:

\(\begin{aligned}{c}\sin \theta = \frac{{{\rm{ opposite }}}}{{{\rm{ hypotenous }}}}\\ = \frac{\mu }{{\sqrt {{\mu ^2} + 1} }}\end{aligned}\)

Therefore, \(F(\tan \theta ) = \frac{{W\mu }}{{\sqrt {{\mu ^2} + 1} }}\). …... (1)

 Step 5: Put the Critical value and simplify

Find the values of\(F\)for the end points of the given interval\(\left( {0,\frac{\pi }{2}} \right)\).

Substitute,\(\theta = 0\)in\(F\).

\(\begin{aligned}{c}F(0) &= \frac{{\mu W}}{I}\\F(0) &= \frac{{\mu W}}{{\mu \sin (0) + \cos (0)}}\backslash \\\sin 0 &= 0,\cos 0 &= 1\\ &= \mu W\end{aligned}\)

When,\(\theta = \frac{\pi }{2}\).

\(\begin{aligned}{c}F\left( {\frac{\pi }{2}} \right) &= \frac{{\mu W}}{{\mu \sin \left( {\frac{\pi }{2}} \right) + \cos \left( {\frac{\pi }{2}} \right)}}\\ &= \frac{{\mu W}}{\mu }\;\;\;\\\sin \frac{\pi }{2} = 1,\cos \frac{\pi }{2} &= 0\\ &= W\end{aligned}\)

Since,\(\sin \theta \le 1,\frac{\mu }{{\sqrt {{\mu ^2} + 1} }} \le 1\). …... (2)

Multiply by\(W\)on both the sides of the equation (2).

\(\frac{{\mu W}}{{\sqrt {{\mu ^2} + 1} }} \le W\) …... (3)

And,\(\frac{\mu }{{\sqrt {{\mu ^2} + 1} }} \le \mu \). …... (4)

Multiply by\(W\)on both the sides of the equation (4),\(\frac{{\mu W}}{{\sqrt {{\mu ^2} + 1} }} \le \mu W\). …... (5)

Now compare the equations (1) with (3) and (5).

When\(\mu = \tan \theta \)the value of\(F\)is minimum.

Hence, the given statement is showed.