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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.

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2 Answers 2

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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 $$\alpha =\frac{ mg. cos^2(\alpha)-F_{f- dynamic}}{m_{you}+m_{board}}$$

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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 $~F_T~,F_N$

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

Remarks

  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.

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