0
$\begingroup$

So, I have a camera crane that has a pivot leaving 2 distances, 0.5mts and 1.5mts respectively. Plus a small L-shaped piece of metal and the end of d2 (1.5) which holds a camera, which I suppose I have to add it to the force applied on that side. One additional variable is that there's a piece of metal attached to the fulcrum and then connected to the second distance (d2, 1.5) at the end, which leaves me the doubt of where to add that weight (which is 580grs, let's say d2*) to the force, or add it as another variable in the equation. The current equation that i'm using (which was without the second piece of weight on distance 2) is: $$ w = \dfrac{f * d2}{d1} $$

Lever example

So i'm dubious if this is the right equation taking in count the variables I have explained above. Thanks!

$\endgroup$
6
  • $\begingroup$ Why is exactly homework a tag? I hardly disagree, i'm not a physicist like Mr. Qmechanic (who edited this question and added the tag) to figure out this by myself. If i'm asking this in SE is because I would really like a more personal explanation, so then I can understand well. $\endgroup$ Sep 4, 2013 at 2:31
  • $\begingroup$ is 870grs the weight of one link of $d2$ length, or both of them? $\endgroup$ Sep 4, 2013 at 12:58
  • $\begingroup$ A Free Body Diagram would be very helpful here. Try it! More here and here. $\endgroup$ Sep 4, 2013 at 13:00
  • $\begingroup$ ja72 i'm sorry, the 870grs are only for the second piece under $d2$, which I call $d2*$ $\endgroup$ Sep 5, 2013 at 2:23
  • $\begingroup$ @DilipRamirez: See the hw policy. It doesn't need to be a real hw question. $\endgroup$ Sep 9, 2013 at 14:04

2 Answers 2

0
$\begingroup$

You have

  1. A counter weight $W$ hanging a distance $d_1$ from the pivot on the left.
  2. A payload and bracket of combined weight $F$ hanging a distance $d_2$ from the pivot on the right.
  3. A link of weight $w_1$ on the left with its center of gravity $\frac{d_1}{2}$ from the pivot.
  4. Two links of weight $w_2$ each on the right with their combined center of gravity $\frac{d_2}{2}$ from the pivot.

All together you balance the moments with

$$ d_1 W + \frac{d_1}{2} w_1 = d_2 F + 2 \frac{d_2}{2} w_2 $$

which is solved for

$$ W = \frac{d_2 (F+w_2)}{d_1} - \frac{w_1}{2} $$

$\endgroup$
3
  • $\begingroup$ I'm sorry to be such a PITA here, but what do you mean by "links of weight"? Specially the $w1$ variable, which I can't figure out. $\endgroup$ Sep 5, 2013 at 2:23
  • $\begingroup$ Oh by links of weight you mean $d1$ and $(d2+d2*)$, right ? $\endgroup$ Sep 5, 2013 at 2:36
  • $\begingroup$ I mean the weight of link $d_1$ is $w_1$, etc. $\endgroup$ Sep 5, 2013 at 12:13
0
$\begingroup$

The equation you provided is valid for the camera holder and camera itself because their size is negligible to their distance to the fulcrum. for the two poles, $d_2$ must be calculated as distance of the pole CG to the fulcrum, i.e $d_2 = 0.5$, $d_ 2 = 0.75$

counterweight required is summation of all $w$'s.

note that this $d_2$ given implicitly shows net torque due to the long pole itself is downwards on the right of fulcrum. the long form is $w_{\mathrm{Long}} = \dfrac{f_2l_2-f_1l_1}{d_1}$, where $f2$ and $f1$ are $\dfrac{3}{4}$ and $\dfrac{1}{4}$ of the pole weight respectively, and $l2$ $l1$ are $0.75$ and $0.25$ respectively.

$\endgroup$
2
  • $\begingroup$ Thanks for the answer. Maybe you can edit it to use math formatting in order to make it more readable. $\endgroup$ Sep 4, 2013 at 13:14
  • 2
    $\begingroup$ mobile phone :/ $\endgroup$
    – gregsan
    Sep 5, 2013 at 10:12

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.