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.


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*}


Is this solution correct?

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

  • 3
    $\begingroup$ +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. $\endgroup$ Jan 12, 2016 at 12:57

1 Answer 1


I would solve this by conservation of power. Linear power is force times speed; rotational power is torque times angular velocity.

You're trying to find linear force, you know torque and angular velocity, so the only thing you need to find is linear velocity.

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.

So, putting it all together:

$$ F = \frac{\tau \omega}{m \cos{\omega t}} $$

  • $\begingroup$ I would just add that all the ideal assumptions apply here; frictionless linear guide, bearings, etc. $\endgroup$
    – Chuck
    Jan 17, 2016 at 12:21

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