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need your help. I realized today, if you do not practice it becomes hard to understand simple concepts even you were once a science student :(

I have attached a picture to help make the problem statement clear. Hopefully, I can convey my message. enter image description here

I am using this simple weight bar mechanism to exert horizontal pull on a string on the let side. Goal is to mark the scale on the right in kgs. Once the weight is on the right is moved to a certain mark(10kg), it should exert the tension of 10kg on the string on left. This I can measure with any hanging luggage scale or fish scale.

I need your help as to which physics law/equations helps to achieve that? I was thinking of Torque, but not sure how to then convert it back to equivalent mass units.

The small object with mass M2 is a fixed length object, but it can end up in any of that range when the right hand bar is horizontal. need also help to add that variability in the equation to see how that makes the same 10kg mark on the bar to be practically different, based on the angle at which the object M2 stops.

e.g. if I put a mark on the right bar at 10kg. I move the weight M1 to that point. To keep it horizontal I should apply the force of 10kg on the left side (which lets say is attached with a string). Once I have that, I expect there should be a small variation in the result, if the object M2 is at 60deg to vertical vs 30 degrees to vertical.

I hope I was able to state my problem clearly. If there is already existing question then happy to get a reference. I searched and still searching. maybe I am not using the correct terminology.

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  • $\begingroup$ wouldn't it be simpler if the string were vertical and attached to M1 at the bottom? $\endgroup$ – JMLCarter Jul 6 '18 at 17:36
  • $\begingroup$ Well, yes in a simple one string way it would be. But this is a setup for stringing machine. Where the string roll around the center. The stretched string is fixed by a clamp and the next section from the loop will be stretched again. This continues till the whol string is used till the end of the racket. $\endgroup$ – arundeep78 Jul 7 '18 at 20:43
  • $\begingroup$ but can' you use a vertical string to mark the scale, then return to the orginal configuration? Otherwise tension is dependent upon the deviation of the string from straigtht. $\endgroup$ – JMLCarter Jul 7 '18 at 21:30
  • $\begingroup$ Sorry, I could not understand your statement " Otherwise tension is dependent upon the deviation of the string from straigtht. " What "straight" is referring to here? If the point is to mark the scale by putting the setup vertically. But then would it not give different results, when the actual tension is done horizontally? Here is a video of setup, showing how it is done in practice. youtube.com/watch?v=adMW3mARDV4 . Hopefully this helps explains the problem statement. My final goal is to calculate the variation added by the final position of M2. $\endgroup$ – arundeep78 Jul 8 '18 at 6:13
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I watched your video, M2 is relatively small, and seems mainly to have the function of grasping the string. Anyway, what you need to do is calibrate the device by measuring the tension in the string and by accurately marking the position of all relevant variable elements of the machine that provide that tension.

However to provide control over tension you only need one element of the machine to change. It will be much easier, for example if you always position m2 in the same place, and always set the rod holding m1 to horizontal. Only use the position of M1 on the rod to alter the tension.

To measure the tension accurately for calibration, instead of attaching a racket either attach a spring balance (such as for suitcase weight measurement, perhaps) or a set of vertically hanging weights.

Modify the position of M1 along the rod, and mark the rod appropriatly so the positions can be recovered when a racket is connected


I wouldn't recommend, as the M2 is probably not significant, but here anyway

$$T=rg( M_1x_1 - M_2x_2cos\theta)$$

$x_1$ is the distance of the center of mass of $M_1$ from the pivot

$x_2$ is the distance of the center of mass of $M_2$ from the pivot (L1/2?)

$\theta$ is the angle from the horizontal anticlock-wise/positive or downwards to the direction of $M_2$

$r$ is the radius of the spool = distance from string to pivot.

$g$ is acceleration due to gravity.

$T$ is the tension force.

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  • $\begingroup$ Well, that would work if I could fix M2 at the same place. But if you watch the video, it varies based on how much stretch is added while putting the string in the "string gripper" M2. It also depends on the M1 being high or low. Measuring by scale would not have caused the need to come to physics stackexchange, right?. My main pain point was to able to capture the variation that M2 can cause, because of the inherit nature of the design. This is where I thought to seek physics help to calculate it. Is there no equation that can really help to calculate it? $\endgroup$ – arundeep78 Jul 9 '18 at 10:46
  • $\begingroup$ It's small compared to M1 and close to the pivot, I suspect its position is not significant. Which you could test, otherwise it's going to be a bit more complex. $\endgroup$ – JMLCarter Jul 9 '18 at 16:54
  • $\begingroup$ Thanks for the equation. I understand than M2X2cos(theta) is the variation added to M1X1; assuming r and g are constant. In my case M1 is 1.17kg and M2 is .36kg. At the reference mark of 10kg X1 is 21cm and L1 is 11cm, thus X2 5.5cm. Using above formula M2 contribution(M2X2cos(theta) in %age of M1 (M1X1), varies from .14% to 8%!. On the reference mark of 12kg X1 is 27cm thus reducing the max %age to 6%. In worse case when I assume 10kg, it would actually be 9.2 kg and for 12kg would be 11.3kg, right? If I have used the formula correctly, this does not seem like negligible. Am I right? $\endgroup$ – arundeep78 Jul 11 '18 at 10:18
  • $\begingroup$ That's right, but 8% is your worst case scenario. If you can arrange for $\theta$ to be near 90 you can neglect M2. I expect a tool that is sold with calibration marks to be reasonably well designed to for those marks to be accurate; which leads me to believe it is possible to arrange for $\theta$ to be near 90degress. Perhaps it even describes this in the instructions. $\endgroup$ – JMLCarter Jul 11 '18 at 16:41
  • $\begingroup$ I agree. If it can be managed between 85-90 then error is max .6%. Well manual does not say anything about it. Maybe it is just for this brand. I did not see anyone ensuring that in other instructional videos as well . But sure this discussion helped to reach a valuable conclusion. Thanks a lot for your help. $\endgroup$ – arundeep78 Jul 12 '18 at 5:25

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