
Trigonometry Trigonometry Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 14th, 2017, 10:13 AM  #1 
Newbie Joined: Apr 2017 From: USA Posts: 8 Thanks: 0  Base vs height of triangle when pulleys are included
Hello, Please see the attached diagram. Suppose you've got a string which begins at point F and is threaded through two rings (or pulleys) located at points D & R and ends back at point F (ie FDRF) to form a triangle with a horizontal leg of length c, angled legs of length a & b, and a height of h. Also, suppose that while point F is fixed, point D can move but only in a left or right direction (ie D cannot move vertically). Because the points forming the triangle have pulleys, the string is free to glide through each (nonbeginning or nonend) point thus causing point R to move down (due to gravity) as point D moves to the right and vice versa. In other words, as length c decreases, the height h increases and vice versa. (Point R will also have a horizontal component of movement as D is moved, but this can be ignored for now.) The diagram also shows a hook attached to a second piece of string of constant length k. Questions 1) What is the relationship between length c and height h (i.e. how far will point R descend as point D is moved to the right)? 2) How would the vertical movement of R (relative to the horizontal movement of point D) be affected if the string originated at point D and was then threaded through point F, back through point D and then through point R to again terminate at F (i.e. DFDRF)? 3) The same question but for FDFDRF? Thanks. Last edited by skipjack; April 14th, 2017 at 01:48 PM. 
April 14th, 2017, 04:15 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 16,773 Thanks: 1234 
If you assume that a = b, what progress can you make?

April 14th, 2017, 09:13 PM  #3 
Newbie Joined: Apr 2017 From: USA Posts: 8 Thanks: 0 
LOL, so you are going to make me work for this, huh? It's actually a real scenario for which I was hoping to get a solution from one of you math whizes. But based on your reply, it seems that I've joined a helpyoufigureitoutonyourown forum. Is that correct? If it is, it's not a problem but I am a pretty old fart and it's been a long time since my high school trig book was opened. But even so, I will do some stabbing if you will do the (gentle) nudging. hehe OK, so here are some things I recognize or hopefully recall from the past • a = b is right but only by intuition (ie don't know how to prove that) • P (perimeter) = a + b + c • imaginary line h is perpendicular to and bisects c • A (area) of triangle FDR = ch/2 • and by Pythagoras, a^2 = h^2 + c^2/4 so b^2 = h^2 + c^2/4 • I'm also pretty sure I remember my trig identities sin = opp/hyp cos = adj/hyp tan = opp/adj Hopefully something I need is listed there but I will wait for your nudge. Many thanks! 
April 15th, 2017, 03:48 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 16,773 Thanks: 1234 
As it's a real scenario, one ought to consider friction. If, however, friction is considered negligible, it follows that a = b and that the green dashed line is a perpendicular bisector of DF. If you also assume that the string has negligible mass and is inextensible (so that it's total length is constant), you can easily calculate the value of a (and hence b) and then use Pythagoras to calculate h. You don't need to consider area or use trigonometry. Question 2 is a bit vague, as you otherwise seem to be interested in an eventual stable position (for a given position of D), not how it is reached. 
April 18th, 2017, 05:30 PM  #5  
Newbie Joined: Apr 2017 From: USA Posts: 8 Thanks: 0  Quote:
1st what I start with. 2nd substitute a for b. 3rd use Pythagorean theorem to solve for h 4th substitute for a to get h in terms of P & c If my algebra is correct then I think the boxed solution is the simplest form and by substituting actual values for P & c, I can determine h for any value of c.  
April 18th, 2017, 06:16 PM  #6 
Newbie Joined: Apr 2017 From: USA Posts: 8 Thanks: 0 
By the way, I probably should have explained the full purpose behind this question initially to have it make more sense. It's kind of an "out there" solution to an unusual situation but one that I'm hoping will work. I'm a quadriplegic and I live alone. I'm paralyzed from the neck down except that I do have pretty good shoulder and biceps function. I'm in a power wheelchair and spend almost all of my time at my computer (which I can use fairly well with the help of adaptive hardware & software combined with typing aids on my hands). I am nearsighted and the glasses I wear most of the day are prescribed for my monitor to be in focus. I have a second pair of glasses sitting on my desk which have the typical prescription to put things in focus at a distance. However, although I can lift my arms high enough to nudge my glasses this way or that, my limited arm movement and paralyzed fingers make it impossible to swap one pair for the other. I don't go out much in the winter but love to go out in the summer, but in prior years, unless I happened to have someone here or was willing to impose on a neighbor, I had to deal with everything outside being blurred. Therefore, I've been thinking a lot about how to switch my glasses myself now that the weather is getting so nice. I've considered several different approaches but the obstacle has always been having something above me to lift my glasses off my head then lower them into my lap where I can manipulate them with my typing aids enough to move them on or off my desk. The idea I finally arrived at is to use the horizontal movement of a door (which I can nudge in an opening or closing direction with my elbow) to produce vertical movement of a small hook. Thus, point D in the diagram corresponds to the movable top corner of a door and point F represents the fixed corner of the doorframe. I had a friend measure the height of my lap (28") and the top of my head (54") so it's clear that I need to produce more than a onetoone ratio of vertical vs. horizontal excursion, hence my questions about the pulleys. As I said, it's an odd scenario but hopefully this description conveys a clearer picture of what I'm trying to accomplish. I'm grateful for the help with the math but I'm also certainly open to other suggestions for the overall project. Thanks for your time and assistance. 
April 20th, 2017, 02:50 PM  #7 
Newbie Joined: Apr 2017 From: USA Posts: 8 Thanks: 0 
I posted a couple of comments two days ago but they have not shown up. I will post them again. [Moderator note: linking direct to an image may avoid the problem. I've removed the redundant posts.] Last edited by skipjack; April 21st, 2017 at 03:05 AM. 
April 20th, 2017, 02:54 PM  #8  
Newbie Joined: Apr 2017 From: USA Posts: 8 Thanks: 0  Quote:
1st is what I start with. 2nd substitute a for b. 3rd use Pythagorean theorem to solve for h 4th substitute for a to get h in terms of P & c If my algebra is correct, then I think the boxed solution is the simplest form and by substituting actual values for P & c, I can determine h for any value of c. Last edited by skipjack; April 21st, 2017 at 02:58 AM.  
April 20th, 2017, 03:23 PM  #9 
Newbie Joined: Apr 2017 From: USA Posts: 8 Thanks: 0 
I probably should have explained the full purpose behind this question initially to have it make more sense. It's kind of an "out there" solution to an unusual situation but one that I'm hoping will work. I'm a quadriplegic and I live alone. I'm paralyzed from the neck down except that I do have pretty good shoulder and biceps function. I'm in a power wheelchair and spend almost all of my time at my computer (which I can use fairly well with the help of adaptive hardware & software combined with typing aids on my hands). I am nearsighted and the glasses I wear most of the day are prescribed for my monitor to be in focus. I have a second pair of glasses sitting on my desk which have the typical prescription to put things in focus at a distance. However, although I can lift my arms high enough to nudge my glasses this way or that, my limited arm movement and paralyzed fingers make it impossible to swap one pair for the other. I don't go out much in the winter but love to go out in the summer, but in prior years, unless I happened to have someone here or was willing to impose on a neighbor, I had to deal with everything outside being blurred. Therefore, I've been thinking a lot about how to switch my glasses myself now that the weather is getting so nice. I've considered several different approaches but the obstacle has always been having something above me to lift my glasses off my head then lower them into my lap where I can manipulate them with my typing aids enough to move them on or off my desk. The idea I finally arrived at is to use the horizontal movement of a door (which I can nudge in an opening or closing direction with my elbow) to produce vertical movement of a small hook. Thus, point D in the diagram corresponds to the movable top corner of a door and point F represents the fixed corner of the doorframe. I had a friend measure the height of my lap (28") and the top of my head (54") so it's clear that I need to produce more than a onetoone ratio of vertical vs. horizontal excursion, hence my questions about the pulleys. As I said, it's an odd scenario but hopefully this description conveys a clearer picture of what I'm trying to accomplish. I'm grateful for the help with the math but I'm also certainly open to other suggestions for the overall project. Thanks for your time and assistance. 
April 20th, 2017, 06:49 PM  #10 
Newbie Joined: Apr 2017 From: USA Posts: 8 Thanks: 0 
By the way, is there a way to upload an image (to be included in a post) directly from your hard drive or is the only option to capture a screenshot, upload it to a website (like imgur.com), then post the link here? The latter is what I did above although it did not like my placing it in img tags nor would it allow me to include the initial "http...".
Last edited by skipjack; April 21st, 2017 at 03:08 AM. 

Tags 
base, height, included, pulleys, triangle 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Base length of trapezoid given area, height, and nonparallel side lengths  caters  Geometry  3  March 11th, 2015 08:58 AM 
How do I find the base height of a Tetrahedron?  swm06  Algebra  4  January 29th, 2013 02:44 AM 
Triangle plane surface 2m base and 3 height  rsoy  Physics  4  April 23rd, 2012 08:27 AM 
Change of triangle height  Valar30  Calculus  1  April 22nd, 2011 11:43 AM 
Finding the Height of a Spherical Cap with it Volume Base  bhtp  Algebra  1  June 20th, 2010 11:14 AM 