# Force in a crank system

## Statement

In a crank system, as described in the picture, we know the values of: the width of the connecting rod ($m$), the width of the crank ($b$), the torque in the crank ($M$) and the angular velocity $\omega$ which we apply to this torque. We want to know the result force ($F_2$) of the system.

Note: the diagram is missing $\gamma$, which is the angle between $F_1$ and $F$. Also, $F_1$ and $F$ are poorly drawn; they should verify $(1)$ below. It is just an error in the diagram, not in the concept of the question.

## Solution

My attempt was the following:

By the diagram, we can see: \begin{align} F_1 &= F \cos \gamma&&(1) \\ \frac{b}{\sin \alpha} &= \frac{m}{\sin \beta} &&(2)\\ \alpha + \beta + \gamma + 90 &= 180 &&(3)\\ F_2 &= F_1 \cos \beta &&(4) \end{align}

Puting $(1)$ in $(4)$: \begin{align} F_2 = F \cos \gamma \cos \beta && (5) \end{align}

First of all, we need a relation between $\cos \beta$ and $\alpha$. The only one is $(2)$, where we can get $\cos \beta$ in function of $\alpha$ by isolating $\cos \beta$ after having applied $\sin \beta = \sqrt{1-\cos^2 \beta}$, so: \begin{align} \frac{b}{\sin \alpha} &= \frac{m}{\sin \beta} \nonumber\\ \sin \beta &= \frac{m \cdot \sin \alpha}{b}\\ \sqrt{1-\cos^2 \beta} &= \frac{m \cdot \sin \alpha}{b}\nonumber\\ 1 - \cos^2 \beta &= \frac{m^2 \cdot \sin^2 \alpha}{b^2}\nonumber\\ \cos^2 \beta &= \frac{b^2 - m^2\cdot \sin^2\alpha}{b^2}\nonumber\\ \cos\beta &= \sqrt{\frac{b^2 - m^2\cdot \sin^2\alpha}{b^2}}\nonumber\\ \cos \beta &= \frac{\sqrt{b^2 - m^2\cdot \sin^2\alpha}}{b}\nonumber \end{align}

Puting this in $(5)$: \begin{align} F_2 = F \cos \gamma \cdot \frac{\sqrt{b^2 - m^2\cdot \sin^2\alpha}}{b} && (7) \end{align}

Now, we need a relation between $\gamma$ and $\alpha$. From $(3)$, we've got: \begin{align} \gamma = 90 - \alpha - \beta && (8) \end{align}

However, We want $\gamma$ only in function of $\alpha$, without $\beta$. So we go to $(2)$ again to calculate $\beta$ in function of $\alpha$. By applying $\arcsin$, we've got: \begin{align}\beta = \arcsin \left(\frac{m \cdot \sin \alpha}{b}\right) && (9)\end{align}

So, putting it into $(8)$: \begin{align*} \gamma = 90 - \alpha - \arcsin \left(\frac{m \cdot \sin \alpha}{b}\right) \end{align*}

And putting it inside $(7)$: \begin{align*} F_2 = F \cos \left[90 - \alpha - \arcsin \left(\frac{m \cdot \sin \alpha}{b}\right)\right] \cdot \frac{\sqrt{b^2 - m^2\cdot \sin^2\alpha}}{b} \end{align*}

We can even go further. We know that $\cos(90^\circ - x) = \sin x$. So: \begin{align} F_2 = F \sin \left[\alpha + \arcsin \left(\frac{m \cdot \sin \alpha}{b}\right)\right] \cdot \frac{\sqrt{b^2 - m^2\cdot \sin^2\alpha}}{b} \end{align}

Simplifying with Wolfram Alpha: \begin{align*} F_2 &= F \left( \sin(\alpha) \sqrt{1-\frac{m^2}{b^2}\sin^2(\alpha)} + \frac{m}{b} \sin{\alpha } \cos { \alpha} \right) \cdot \frac{\sqrt{b^2 - m^2\cdot \sin^2\alpha}}{b}\\ F_2 &= F \cdot \sin\alpha \cdot \frac{ m\cos \alpha \sqrt{b^2 - m^2 \sin^2\alpha} + b^2 - m^2 \sin^2 \alpha }{b^2} \end{align*}

Finally, if we want to express this identity in function of $\omega$, $t$ and $M$ we just have to substitute $\alpha$ by $\omega t$ and $F$ by $M/m$, so:

\begin{align*} \boxed{F_2 = \frac{M}{m} \cdot \sin(\omega t) \cdot \frac{ m\cos(\omega t) \sqrt{b^2 - m^2 \sin^2(\omega t)} + b^2 - m^2 \sin^2 (\omega t) }{b^2}} \end{align*}

## Question

Is this solution correct?

PS: my level of physics studies is quite low, so, please, restrict to this level of difficulty.

• +1 for a well-written homework help question. Your approach looks fine to me, but I didn't check all of the algebra in the middle. Your substitution of $\alpha=\omega t$ at the end is dependent on the angular speed ($\omega$) being constant which is probably not true in real life but is perhaps OK for this problem. – Chris Mueller Jan 12 '16 at 12:57

Fortunately, you know angular velocity and the radius of the crank, so the linear velocity is just $m\cos{\omega t}$, where you have chosen m to be a length and not mass.
$$F = \frac{\tau \omega}{m \cos{\omega t}}$$