I'm building a climbing training apparatus in my garage that both rests on the ground and hangs by a rope from a hook attached to a ceiling joist. Here are some pictures of builds similar to what I'm going for:

enter image description here enter image description here

I'm trying to model this scenario so I can avoid slip-ups in the construction process. Specifically, I want to know if I'll need any special rubber feet for the base, and how strong the ceiling joist hooks will need to be. Here's a diagram of the model I'm proposing:

enter image description here

And here's my attempt at a free body diagram:

enter image description here

Here, $m_c$ stands for the mass of the climber, $m_r$ for the mass of the rod, $L$ for the total length of the rod, and $s$ for the position of the climber's center of mass, measured along the rod from the ground, so that $0 \leq s \leq L$. Assume the angle $\theta$, in addition to all aforementioned variables, is known and constant.

I realize I am missing something, as this diagram would suggest that friction is an unbalanced force in the horizontal direction. The $\sum F_y = 0$ equation is straightforward, but I'm stuck for how to write down my $\sum F_x = 0$ equation, which I'll need to solve for the two unknowns, $F_N$ and $F_T$.

Can anyone help explain what is missing or is incorrect about my FBD here? Once I have a correct FBD, I'll be able to solve the resulting system of equations. I have a sneaking suspicion that a third unknown will pop out, which will mean adding a $\sum \tau = 0$ equation to the mix as well. Thanks for the help!

The first two pictures are from a forum post by user Chris D linked here.


2 Answers 2


We have no reaction along X-axis as your figure shows due to the rope as long as it hangs vertically.

But if you start to swing, then you would have a $F_f$, or even movement of the rod.

Let's say you're free-swinging and your body rotates at an angle of $\theta$ then you have tension in your arms hanging from the bar= T.

$$T_{arm}= mg.cos(\theta)$$

This tension has an X component acting on the rod.

$$F_x = mg. cos^2(\theta)=F_f$$

And if this $F_x\ $becomes greater than static friction on the right end of the rod you and the rod will accelarate with an accelaration $$\alpha =\frac{ mg. cos^2(\alpha)-F_{f- dynamic}}{m_{you}+m_{board}}$$


from your FBD

take the sum of the torques about point $A~$ ( position of $~F_r$)

$$\sum \tau_A=F_{{T}}L\sin \left( \theta \right) -F_{{N}}L\cos \left( \theta \right) +\frac 12\,m_{{r}}g\,L\cos \left( \theta \right) +m_{{c}}g \left( L- S \right) \cos \left( \theta \right) =0$$

and the sum of the forces towards y

$$\sum F_y=F_{{N}}+F_{{r}}-m_{{r}}g-m_{{c}}g=0$$

you have two equations for the unknowns $~F_T~,F_N$

$$F_T=\frac 12\,{\frac {\cos \left( \theta \right) \left( -2\,LF_{{r}}+m_{{r}}gL +2\,m_{{c}}gS \right) }{L\sin \left( \theta \right) }} \\ F_N=-F_{{r}}+m_{{r}}g+m_{{c}}g $$


  1. you assumed that the rubber is a joint so you get the joint constraint forces , but the rubber is not a joint!.

  2. if $~F_T~$ is a friction force then $~F_T=\mu\,F_N~$ where $~\mu~$ is the friction coefficient.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.