# What forces are exerted on a clothespin in space?

Let's say a clothespin is modeled as a simple torsion spring as follows.

Given:

• $p_1,\ p_2,\ p_3$: point-like objects of equal mass in 2-D space.
• All objects float in space, i.e. the center of mass will not change.
• At time $t_0$ a torsion spring is inserted at $p_1$, such that it exerts torque on $p_2$ and $p_3$, with:
• $\theta$: the angle of twist from its equilibrium position in radians
• $\kappa$: the spring's torsion coefficient
• $\tau = -\kappa \theta$ is the torque exerted by the spring

Question: what are the resulting forces on $p_1$, $p_2$ and $p_3$?

My answer: Because all objects have equal mass, we can leave mass out of the equation. F2 is a force perpendicular to $p_1,\ p_2$ of magnitude $\dfrac{\tau }{ |p_1-p_2|}$ .

By Newton’s 3rd law, F2’ is a force of equal magnitude and opposite direction as F2 Similar for F3 and F3’

• I would just like to add that I do not have a physics background. I do not know whether my answer is correct, whether I am missing some input variables, or whether there is some physics law that I am missing.
– Matt
Commented Nov 3, 2013 at 13:08
• Hi, try to use math syntax in your posts for better readability. See the [?] help button when you edit the post. Commented Nov 3, 2013 at 17:07
• Nice problem. Remember the clothespin bars carry forces as well as moments. They are essentially beams (without mass). Commented Nov 3, 2013 at 17:22
• After talking to a friend, I understand it's important to note this question is time-independent: I only care about the forces at time t0.
– Matt
Commented Nov 3, 2013 at 17:35

If $\theta$ is the angle between the arms, displaced from the equilibrium $\theta_0$ by $\Delta \theta$ and the torque applied is $\tau =-\kappa \Delta \theta$, assuming equal masses of $m$ with initially motionless parts.

The first step is the kinematics, whereas the acceleration of 2 and 3 is related to the acceleration of 1 and the common angle. For simplification we have that 1 is not accelerating in the horizontal direction $\ddot{x}_1=0$ (as seen in figure below).

\begin{aligned} \ddot{x}_2 &= \ddot{x}_1 - \ell \cos \left( \frac{\theta}{2} \right) \frac{ \ddot{\theta}}{2} & \ddot{x}_3 &= \ddot{x}_1 + \ell \cos \left( \frac{\theta}{2} \right) \frac{ \ddot{\theta}}{2} \\ \ddot{y}_2 &= \ddot{y}_1 + \ell \sin \left( \frac{\theta}{2} \right) \frac{\ddot{\theta}}{2} & \ddot{y}_3 &= \ddot{y}_1 + \ell \sin \left( \frac{\theta}{2} \right) \frac{\ddot{\theta}}{2} \end{aligned}

Now for the equations of motion of each part. We start with free body diagrams in order to sum up the forces on each part.

\begin{aligned} -Fr_2 \sin \left( \frac{\theta}{2} \right) + Fr_3 \sin \left( \frac{\theta}{2} \right) + Fn_2 \cos \left( \frac{\theta}{2} \right) + Fn_3 \cos \left( \frac{\theta}{2} \right) & = m \ddot{x}_1 = 0 \\ -Fr_2 \cos \left( \frac{\theta}{2} \right) - Fr_3 \cos \left( \frac{\theta}{2} \right) + Fn_2 \sin \left( \frac{\theta}{2} \right) + Fn_3 \sin \left( \frac{\theta}{2} \right) &= m \ddot{y}_1 \end{aligned}

The EOM are done in a direction along the arm

\begin{aligned} m \ddot{x}_2 \cos \left( \frac{\theta}{2} \right) - m \ddot{y}_2 \sin \left( \frac{\theta}{2} \right) &= -Fn_2 \\ m \ddot{y}_2 \cos \left( \frac{\theta}{2} \right) + m \ddot{x}_2 \sin \left( \frac{\theta}{2} \right) & = Fr_2 \\ 0 & =\ell Fn_2 + \tau \end{aligned} with $\Rightarrow Fn_2 =-\frac{\tau}{\ell}$ \begin{aligned} m \ddot{x}_3 \cos \left( \frac{\theta}{2} \right) + m \ddot{y}_3 \sin \left( \frac{\theta}{2} \right) &= -Fn_3 \\ m \ddot{y}_3 \cos \left( \frac{\theta}{2} \right) - m \ddot{x}_3 \sin \left( \frac{\theta}{2} \right) & = Fr_3 \\ 0 & =\ell Fn_3 - \tau \end{aligned} with $\Rightarrow Fn_3 =\frac{\tau}{\ell}$

Combined the all of the above equations substituted into the kinematics are

\begin{aligned} -Fn_2 \cos \left(\theta \right) + Fr_2 \sin \left(\theta \right) - 2 Fn3 &= m \ell \frac{\ddot{\theta}}{2} \\ Fr_2 \cos \left(\theta \right) + Fn_2 \sin \left(\theta \right) + 2 Fr_3 &= 0 \\ - Fn_3 \cos \left(\theta\right) - Fr_3 \sin \left(\theta \right) + 2 Fn_2 &= - m \ell \frac{\ddot{\theta}}{2} \\ Fr_3 \cos \left(\theta \right) - Fn_3 \sin \left(\theta \right) + 2 Fr_2 &= 0 \end{aligned}

The above is solved with $$\boxed{\frac{3 \tau (\cos\theta-2)}{\ell (\sin^2\theta+3)} = m \ell \frac{\ddot{\theta}}{2}}$$

and

\begin{aligned} Fr_2 & = \frac{\tau \sin\theta ( 2-\cos\theta)}{\ell (\sin^2\theta+3)} \\ Fn_2 &= -\frac{\tau}{\ell} \\ Fr_3 &= \frac{\tau \sin\theta ( 2-\cos\theta)}{\ell (\sin^2\theta+3)} \\ Fn_3 &= \frac{\tau}{\ell} \end{aligned}