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Show that F is minimized when tan(θ) = μ

karush

Well-known member
Jan 31, 2012
2,727
An object wth 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 $\theta$ with the plane, then the magnitude of the force is

$
\displaystyle
F=\frac{\mu W}{\mu\sin{\theta}+\cos{\theta}}
$

where $\mu$ is a positive constant called the coefficient of friction
and where $$0<\theta\le \pi/2$$ Show that $F$ is minimized when $\tan\theta=\mu$

this was a problem under min/max values. I was going to find F' or try to graph this
in W|F but got a 3d graph which I didn't understand.
Am sure this is a common problem in Physics but it was put in with exercises in Calculus

Anyway curious how this is solved...
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: Show that F is mininized when tan \theta = \mu

An object wth 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 $\theta$ with the plane, then the magnitude of the force is

$
\displaystyle
F=\frac{\mu W}{\mu\sin{\theta}+\cos{\theta}}
$

where $\mu$ is a positive constant called the coefficient of friction
and where $$0<\theta\le \pi/2$$ Show that $F$ is minimized when $\tan\theta=\mu$

this was a problem under min/max values. I was going to find F' or try to graph this
in W|F but got a 3d graph which I didn't understand.
Am sure this is a common problem in Physics but it was put in with exercises in Calculus

Anyway curious how this is solved...
Hi karush, :)

I don't think you would get a 3d graph because there's only one independent variable. \(W\) and \(\mu\) are constants. So \(F\) only depends on \(\theta\). All you got to do is find the derivative of \(F\) with respect to \(\theta\) and use the first derivative test or the second derivative test. Hope you can continue. :)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Show that F is mininized when tan \theta = \mu

To see where this equation comes from, consider the following free-body diagram:

karush2.jpg

Using the conditions of equilibrium (there are no unbalanced forces), we may use Newton's second law of motion along the two components (horizontal and vertical):

(1) \(\displaystyle \sum F_x=F_x-f_k=0\)

(2) \(\displaystyle \sum F_y=n+F_y-W=0\)

Resolving the components of the applied force $F$, we find:

\(\displaystyle F_x=F\cos(\theta),\,F_y=F\sin(\theta)\)

The coefficient of kinetic friction is defined as:

\(\displaystyle \mu_k=\frac{f_k}{n}\implies f_k=n\mu_k\)

And thus, (1) and (2) become:

\(\displaystyle F\cos(\theta)-n\mu_k=0\)

\(\displaystyle n+F\sin(\theta)-W=0\)

Solving the first of these for $n$, we obtain:

\(\displaystyle n=\frac{F\cos(\theta)}{\mu_k}\)

Substituting for $n$ into the second of these, we find:

\(\displaystyle \frac{F\cos(\theta)}{\mu_k}+F\sin(\theta)-W=0\)

Multiply through by $\mu_k$:

\(\displaystyle F\cos(\theta)+F\mu_k\sin(\theta)-\mu_kW=0\)

Add \(\displaystyle \mu_kW\) to both sides:

\(\displaystyle F\cos(\theta)+F\mu_k\sin(\theta)=\mu_kW\)

Factor the left side:

\(\displaystyle F\left(\cos(\theta)+\mu_k\sin(\theta) \right)=\mu_kW\)

Divide through by \(\displaystyle \mu_k\sin(\theta)+\cos(\theta)\):

\(\displaystyle F=\frac{\mu_kW}{\mu_k\sin(\theta)+\cos(\theta)}\)