# Solving for Theta

• April 20th 2010, 08:28 AM
Chokfull
Solving for Theta
http://www.mathhelpforum.com/math-he...9&d=1271780556
I got the equation
$\mu * (F \cos \theta + 9.8m) = F \sin \theta$
where $\mu$ is the coefficient of static friction, but I don't know how to solve for $\theta$ from here. I have a computer program called Maple but it can't solve the equation to anything recognizable either.
• April 20th 2010, 08:35 AM
e^(i*pi)
Quote:

Originally Posted by Chokfull
http://www.mathhelpforum.com/math-he...9&d=1271780556
I got the equation
$\mu * (F \cos \theta + 9.8m) = F \sin \theta$
where $\mu$ is the coefficient of static friction, but I don't know how to solve for $\theta$ from here. I have a computer program called Maple but it can't solve the equation to anything recognizable either.

$\frac{1}{\mu} \cdot \frac{F \sin \theta}{F \cos \theta} = 9.8\mu m$

$\frac{\tan \theta}{\mu} = 9.8\mu m$
• April 21st 2010, 08:15 AM
Chokfull
I don't get it-the first step. It looks like you factored in the $\mu$, then divided each side of the equation by $\mu F \cos \theta$, but this would result in the right side being $\frac {9.8m} {F \cos \theta}$
• April 21st 2010, 12:28 PM
Soroban
Hello, Chokfull!

Edit: Okay, so the "m" is mass.

Quote:

I got the equation: . $\mu\cdot(F \cos \theta + 9.8{\color{red}m}) \:=\:F \sin\theta$

but I don't know how to solve for $\theta$.

We have: . $\mu F\cos\theta + 9.8m\mu \:=\:F\sin\theta \quad\Rightarrow\quad F\sin\theta - \mu F\cos\theta \:=\:9.8m\mu$

. . . . . . . . $F(\sin\theta - \mu\cos\theta) \:=\:9.8m\mu \quad\Rightarrow\quad \sin\theta - \mu\cos\theta \:=\:\frac{9.8m\mu}{F}$

Divide by $\sqrt{1+\mu^2}:\quad\frac{1}{\sqrt{1+\mu^2}}\sin\t heta - \frac{\mu}{\sqrt{1+\mu^2}}\cos\theta \:=\:\frac{9.8m\mu}{F\sqrt{1+\mu^2}}$ .[1]

$\text{Let: }\,\tan\alpha \:=\:\mu \quad\Rightarrow\quad\sin\alpha \:=\:\frac{\mu}{\sqrt{1+\mu^2}},\;\;\cos\alpha \:=\:\frac{1}{\sqrt{1+\mu^2}}$

Substitute into [1]: . $\cos\alpha\sin\theta - \sin\alpha\cos\theta \:=\:\frac{9.8m\mu}{F\sqrt{1+\mu^2}}$

$\text{And we have: }\;\sin(\theta - \alpha) \:=\:\frac{9.8m\mu}{F\sqrt{1+\mu^2}}$

. . . . . . . . . . . . $\theta-\alpha \;=\;\sin^{-1}\left(\frac{9.8m\mu}{F\sqrt{1+\mu^2}}\right)$

. . . . . . . . . . . . . . . $\theta \;=\;\sin^{-1}\left(\frac{9.8m\mu}{F\sqrt{1+mu^2}}\right) + \alpha$

. . . . . . . . . . . . . . . $\theta \;=\;\sin^{-1}\left(\frac{9.8m\mu}{F\sqrt{1+\mu^2}}\right) + \tan^{-1}(\mu)$

• April 21st 2010, 05:23 PM
sa-ri-ga-ma
Hi Soroban,
"I hope that "m" is just meters."

No. It is the mass of the mop.
• April 21st 2010, 06:17 PM
skeeter
understand that you're not just solving for $\theta$ in his question ...

for part (a) , the mop head is in dynamic equilibrium.

$F\sin{\theta} = f_k$

$F\sin{\theta} = \mu_k(F\cos{\theta} + mg)$

$F\sin{\theta} - \mu_k F\cos{\theta} = \mu_k mg$

$F = \frac{\mu_k mg}{\sin{\theta} - \mu_k \cos{\theta}}$

for part (b) , note that for the above equation to make sense, $\sin{\theta} > \mu_k \cos{\theta}$

$\tan{\theta} > \mu_k$

$\theta > \arctan(\mu_k) = \theta_0$
• April 22nd 2010, 07:23 AM
Chokfull
Quote:

Originally Posted by skeeter

$F = \frac{\mu_k mg}{\sin{\theta} - \mu_k \cos{\theta}}$

for part (b) , note that for the above equation to make sense, $\sin{\theta} > \mu_k \cos{\theta}$

$\tan{\theta} > \mu_k$

$\theta > \arctan(\mu_k) = \theta_0$

I assume that here you are supposed to assume that F cannot be negative, thus http://www.mathhelpforum.com/math-he...15825d37-1.gif has to be true for the denominator to be positive.