Hello fellow engineers,

A friend is part-way building an outside storage chest. For whatever reason, the materials and size of storage that he is going for means that the door is heavy to lift and not very practical. I've suggested that he manufacture a counterweight system to aid in opening the door.

I've made an attempt to calculate the required weight by modelling using simple moments and balancing forces, but I'm not happy with the result as I'm pretty sure I'm missing things from my model.

Please see this sketch and my 2 calculation sets. (I've modeled this as a simple statics pulley sort of problem.) Sketch Calc 2 Calc 1

Since I've modeled this as a statics problem, my issue becomes how to account for the force applied by the person lifting the door, which I have assumed would be applied at the eyelet.

Assuming my equations are correct, as theta increases, T1 should decrease since M2 is obviously constant. But isn't tension constant?

I think I'm nearly there but perhaps need a push in the right direction... It's been a while since I've done something like this!


  • $\begingroup$ I'm voting to close this question as off-topic because this belongs on Physics.SE . $\endgroup$ Jun 8 '17 at 13:07
  • $\begingroup$ Next time please learn to use Markdown /LaTex instead of hand-scrawled equations! Our poor eyes! $\endgroup$ Jun 8 '17 at 13:07
  • $\begingroup$ I posted a similar problem on P.SE and it was moved to hear for being too "real life". Can you please explain why you think it should be moved? This is application of engineering principles, no? $\endgroup$
    – Phizzy
    Jun 8 '17 at 13:08
  • $\begingroup$ Maybe you should re-post to Physics but without numeric values. Just assume a homogeneous, rigid plate for the lid, etc. $\endgroup$ Jun 8 '17 at 13:09
  • 3
    $\begingroup$ Sounds real-life enough to me to stay here. $\endgroup$
    – hazzey
    Jun 8 '17 at 14:38

Since you considered your model as static - it's correct until things start to move. You are right, as theta increases, T1 should decrease to keep things balanced (static problem).

Tension is constant. But when you apply additional force it's not $M_2g$ anymore and that is why everything is moving.

If you want to choose correct weight $M_2$,consider two following static situations.

  1. Closed hatch: If you choose $M_2$ in a way that system is balanced when hatch is closed (max theta ), then additional small force applied to open the hatch will lead to move of the system and opening of the hatch. You already done it.

  2. Opened hatch: If you choose $M_2$ in a way when hatch is opened (theta is max) and everything is balanced. Then small additional force to close the hatch will lead to movement, because additional moment due to weight of the hatch will close the hatch. (That is the minimum required mass $M_2$, if it is less, then your hatch will be closing all the time if not fixed with some additional device.)

I would choose the mass of the counterweight taking as equilibrium point somewhere in between these two values. Depending on the exact value it will be easier close/open the hatch.

Important note, if horizontal size of hatch L is equal to vertical elevation of the pulley, you have two problems:

  • Clashing when hatch is opened, definitely something will be broken at some point of time.
  • Counterweight will touch the ground before the hatch is fully opened, thus it's impossible to stabilize hatch in fully opened position only by use of the counterweight. (or make sure that it than travel lower than hatch level)

If you move pulley higher, than angle theta at the eyelet will not be equal to angle between vertical and rope at pulley (it will be 90 deg- theta).


This is an incomplete answer, so for that I am sorry. (Limited time to answer)

I believe that tension is constant, but the vector in the positive y direction is not.

Hope that helps at all!

  • 1
    $\begingroup$ PLease don't post incomplete answers. Wait until you do have time. $\endgroup$ Jun 8 '17 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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