# Solving for $\theta$: $\frac{F - ma}{w} = M\cos\theta + \sin\theta$

I am doing a physics problem, and I am stuck at one formula. I need to find a specific angle of inclination ($\theta$), and have come up with the formula

$$\frac{F - ma}{w} = M\cos\theta + \sin\theta$$

I need get $\theta$, and the only thing I can think of is setting $\sin\theta = \sqrt{1-\cos^2\theta}$, but that only made solving for a single trig function (in this case $\cos$) worse. Is there a better way of eliminating one of the two different trig functions?

For this specific problem the left equation is equal to 0.395, and M on the right is 0.07, if that helps.

• The RHS seems to be dimensionally incorrect . ( I only say this since it is a physics problem where dimensions matter a lot ). – user369582 Sep 27 '16 at 16:47
• @Avi if $w$ has units of force and $M$ is dimensionless then it's fine ... – John Sep 27 '16 at 16:51
• Yes , you are right. I take back my comment. I assumed that M had dimensions of mass. – user369582 Sep 27 '16 at 16:53
• F is applied force, ma = mass*acceleration (net force), w = weight. The M is actually the coefficient of friction. That may be were the confusion is coming from. – FyreeW Sep 27 '16 at 16:53

Use the Weierstrass substitution $r=\tan\left(\frac{\theta}{2}\right)$, then we get $\sin\left(\theta\right)=\frac{2r}{1+r^2}$ and $\cos\left(\theta\right)=\frac{1-r^2}{1+r^2}$:

$$\frac{\text{F}-\text{m}\cdot\text{a}}{\text{w}}=\text{M}\cdot\cos\left(\theta\right)+\sin\left(\theta\right)\Longleftrightarrow\frac{\text{M}}{1+r^2}+\frac{2r}{1+r^2}-\frac{\text{M}r^2}{1+r^2}-\frac{\text{F}-\text{m}\cdot\text{a}}{\text{w}}=0$$

So, we get that:

$$r-\frac{1}{\text{M}+\frac{\text{F}-\text{m}\cdot\text{a}}{\text{w}}}=\pm\sqrt{\frac{1-\left(-\text{M}-\frac{\text{F}-\text{m}\cdot\text{a}}{\text{w}}\right)\left(\text{M}-\frac{\text{F}-\text{m}\cdot\text{a}}{\text{w}}\right)}{\left(-\text{M}-\frac{\text{F}-\text{m}\cdot\text{a}}{\text{w}}\right)^2}}$$

Solving this for $r$ and setting $r=\tan\left(\frac{\theta}{2}\right)$ back:

$$\tan\left(\frac{\theta}{2}\right)=\frac{1}{\text{M}+\frac{\text{F}-\text{m}\cdot\text{a}}{\text{w}}}\pm\sqrt{\frac{1-\left(\frac{\text{m}\cdot\text{a}-\text{F}}{\text{w}}-\text{M}\right)\left(\text{M}+\frac{\text{m}\cdot\text{a}-\text{F}}{\text{w}}\right)}{\left(\frac{\text{m}\cdot\text{a}-\text{F}}{\text{w}}-\text{M}\right)^2}}$$

When $\frac{\text{F}-\text{m}\cdot\text{a}}{\text{w}}=\frac{79}{200}$ and $\text{M}=\frac{7}{100}$. Then we find for $\theta$ (in radians):

$$\theta=2\pi n+\arctan\left(\frac{200\pm\sqrt{33955}}{93}\right)$$

Where $n\in\mathbb{Z}$

Since the equation depends only on $\cos\theta$ and $\sin\theta$ you can use the Weierstrass substitution: $$t = \tan\left(\frac\theta2\right)$$ giving $$\cos\theta = \frac{1 - t^2}{1 + t^2},\qquad\sin\theta = \frac{2t}{1 + t^2}.$$

After substituting, you will get a quadratic equation which you can solve with the formula.