The physics of a dropweight tensioner

[Irvin]

Originally Posted by swfh

Quick question for you. I'm taking ap physics b (ninth grade) and I learned torque to be F x d x sin fee. Why did you use cos?

Just remember SOH-CAH-TOA. Here is a short video that explains.

http://www.youtube.com/watch?v=_S35Ht4imhs

[swfh]

Originally Posted by Koz

You pick Cos or Sin depending on what angle you know (or what angle you want to use). In the original post we use cos because the angle we measure is the angle Q, which is the angle between the horizontal direction and the arm. Cos gives the correct properties for this setup. It has a max value (1) at zero degrees (when the bar is horizontal), and a minimum value (0) at 90 degrees (when the bar is vertical). In comparison sin has a max value at 90 degrees. So if we used Sin in the equation we'd be saying you get max tension when the weight arm is pointing straight up or straight down! We know this isn't true, because if you're really good at balancing, you could balance the weight arm vertically without ANY tension on the string! You definitely can't hold the arm horizontal without tension on the string though!




However, you could also look at the problem from a different direction. If you defined the angle Z as the angle between the lever arm and the vertical direction (instead of the horizontal direction), your tension would be max when Z=90 degrees. Z=90 degrees is the same position as Q=0 degrees.

The original tension equation in the 1st post was:
f = (M g / r) (cos Q) R

If you use Z instead of Q you'd just get:
f = (M g / r) (sin Z) R

Note that the physics didn't change at all, you just looked at it differently the 2nd time. So you pick Sin or Cos depending on how you look at the problem.

It's good you ask the question here, because if you can make sense of this now it will be really helpful for you in your AP class. A lot of times you'll learn an equation one way (with sin), but you'll be given a problem where the angle you're given means you need to use Cos instead. Try to do a quick "does this make sense" check when looking at these problems. Ask yourself "what happens when the angle is zero?" or "what happens when the angle is 90?" If the value is biggest at zero, you know to use cos. If it's biggest at 90, you know to use sin.

So in this case, you ask "what happens when Q is zero?" well the bar is horizontal and you get your max tension. Which trig function is maxed at zero? Cos.

If you asked "what happens when Q is 90?" you'd have to think about it...if Q is 90 then the bar would be standing up vertically. If you had it perfectly balanced so that it was standing up all by itself, how much tension would you have to apply to keep it vertical? Zero lbs! In other words, your tension is a minimum (zero) when the arm is vertical. What trig function is a minimum at zero? Cos. So you use Cos for this problem when given angle Q.

Note, if you are dead set on using sin in your equation, you still can't use Sin(Q), but you can use Sin(90-Q), since that's the same thing as Cos (Q). It's just whatever way you're most comfortable with, or what makes the most sense to you.

I got it. Thanks for the response. My teacher wants it to be perpendicular to the force, so i was a little confused. It makes perfect sense.

[Speed Kat]

Originally Posted by JackSkellington

lovely write up--I'm wondering, though, if ever to let the weight "freefall" even for only an inch or two. The correct method is to keep the weight in hand and move it back to horizontal without it ever leaving the hand? Since you say that putting a 1lb force on the top of the weight can translate to a 30 lb difference, I'm a bit curious because sometimes I'll press downward on the lever at the last string to compensate for slack, a bad habit I realize after reading this

I only let it "freefall an inch or two sometimes on the crosses to help get rid of some of the excess slack on days where due to particular strings and string pattern combos, things get a bit jammed and tight. Also, instead of pressing down on the lever at the last string to compensate for slack, just increase the tension by a couple of pounds although this is a controversial practice too.

[Ray_cn]

It's a great post. Recently i'm interesting in the stringing machine too. And read some docs about that.

For the dropweight machine.

If considered the string tension lose eg. the string will be elongated. For example sofer strings. The dropweight will not be a accurated system. In my pic, left, the string is pulled at a constant tension, and continuously whether the string is elongated. Right, if the string is elongated eg. tension lost, the lever will not still holding the horizontal position. For example, extreme situation if the string losed too much, when the Q=-90degree, the string tension is 0. Stringer have to adjust the tension for the tension lose. Increase the Q between the arm and the string gripper. One hand hold the 1-2kg lever, and one hand hold the string gripper, and the eye have to watch the indictor. It's complicate operation. I saw some video, if the stringer need to adjust the Q, at least adjust 2-3times.

To avoid the string elongated, many stringer using Pre-stretch method. It's easy to control.

I agree with your opinion about the F=ma, the speed should not be fast. and your example catapult, I always using a manual Manual crane for example, it is easy to find in a car repair shop. And easy to use to elevate a car. I think that is why someone adjust the Q many times then the tension is much higher than correct.

If the string is hard to be elongated, the tension is excatly accuracate, if the string is softer, the tension maybe need more experience to adjust.



The Stringway machine invients a method to make "The tension is the same for every angle of the lever" It's great, but I don't have that machine, wish to learn how they do that.
http://www.stringway-nl.com/USA/

Learned much from this topic and reply, it's great.

[dyldore]

I didn't look through the entire thread so I apologize if this has been answered, but what if the drop weight goes passed horizontal? I'm curious because I ordered the klippermate and am not sure how much of a pain the non-ratcheting tensioner is going to be.

[Lakers4Life]

Originally Posted by dyldore

I didn't look through the entire thread so I apologize if this has been answered, but what if the drop weight goes passed horizontal? I'm curious because I ordered the klippermate and am not sure how much of a pain the non-ratcheting tensioner is going to be.

I think he answered that in the first page. If the bar were to dip slightly below horizonal, 1.2 deg can be as high as 15 lbs.

[enderx1x]

Originally Posted by Lakers4Life

I think he answered that in the first page. If the bar were to dip slightly below horizonal, 1.2 deg can be as high as 15 lbs.

If I understand correctly, extra tension only occurs when external force is added to the bar, such as dropping it or pushing on it. If the drop weight ends up past horizontal, but was lowered slowly then it is still fine. 5-10 degrees past horizontal would result in about 1 pound less tension.

[Lakers4Life]

I just realized that too, but what about torque? I've never really used a DW other than to mess around with. I've even had a ML200TT that I hated with a vengence, but that's another story.

[enderx1x]

The torque for 5 degrees over horizontal and 5 degrees under horizontal should be the same, as for any other number of degrees.

[Technatic]

To get the total picture it might be good to add the theory of the “automatic” dropweight to this discussion.
It shows why the tension is independent of the angel of the lever.