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

#### karush

##### Well-known member
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
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

Staff member
Re: Show that F is mininized when tan \theta = \mu

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

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 $$\displaystyle \mu_k\sin(\theta)+\cos(\theta)$$:

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