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Thread: Solving for Theta

  1. #1
    Member Chokfull's Avatar
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    Unhappy Solving for Theta


    I got the equation
    $\displaystyle \mu * (F \cos \theta + 9.8m) = F \sin \theta$
    where $\displaystyle \mu$ is the coefficient of static friction, but I don't know how to solve for $\displaystyle \theta$ from here. I have a computer program called Maple but it can't solve the equation to anything recognizable either.
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  2. #2
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    Quote Originally Posted by Chokfull View Post

    I got the equation
    $\displaystyle \mu * (F \cos \theta + 9.8m) = F \sin \theta$
    where $\displaystyle \mu$ is the coefficient of static friction, but I don't know how to solve for $\displaystyle \theta$ from here. I have a computer program called Maple but it can't solve the equation to anything recognizable either.
    $\displaystyle \frac{1}{\mu} \cdot \frac{F \sin \theta}{F \cos \theta} = 9.8\mu m$

    $\displaystyle \frac{\tan \theta}{\mu} = 9.8\mu m$
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  3. #3
    Member Chokfull's Avatar
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    I don't get it-the first step. It looks like you factored in the $\displaystyle \mu$, then divided each side of the equation by $\displaystyle \mu F \cos \theta$, but this would result in the right side being $\displaystyle \frac {9.8m} {F \cos \theta}$
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  4. #4
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    Hello, Chokfull!


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



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

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

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

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


    Divide by $\displaystyle \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]


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

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

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

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

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

    Last edited by Soroban; Apr 22nd 2010 at 10:51 AM.
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  5. #5
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    Hi Soroban,
    "I hope that "m" is just meters."

    No. It is the mass of the mop.
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  6. #6
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    understand that you're not just solving for $\displaystyle \theta$ in his question ...

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

    $\displaystyle F\sin{\theta} = f_k$

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

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

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


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

    $\displaystyle \tan{\theta} > \mu_k$

    $\displaystyle \theta > \arctan(\mu_k) = \theta_0$
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  7. #7
    Member Chokfull's Avatar
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    Quote Originally Posted by skeeter View Post

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


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

    $\displaystyle \tan{\theta} > \mu_k$

    $\displaystyle \theta > \arctan(\mu_k) = \theta_0$
    I assume that here you are supposed to assume that F cannot be negative, thus has to be true for the denominator to be positive.
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