1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Directions/signs question about force-equilibrium problem

  1. May 14, 2013 #1

    s3a

    User Avatar

    1. The problem statement, all variables and given/known data
    The problem and its solution are given in the attached TheProblemAndSolution.jpg file (as well as in this link - http://i.imgur.com/baa0G1S.jpg - since the jpg file attached here seems blurry).

    2. Relevant equations
    E1: Q##n_Q## + S##n_S## + R##n_R## = 80 lb
    and
    E2: –4Q – 4S – 4R = –(80)(5)

    3. The attempt at a solution
    Assuming ##n_Q##, ##n_S## and ##n_R## are positive, I could see how Q##n_Q## + S##n_S## + R##n_R## – 80 lb = 0 ⇒ Q##n_Q## + S##n_S## + R##n_R## = 80 lb since the tension forces are directed from the plate to the knob and the plate weighs downward however, the book says they are negative and, also, E2 just seems to be –5*E1. Why do this (where "this" is multiplying by –5)?

    Also, it seems inconsistent to assume that ##n_Q##, ##n_S## and ##n_R## are negative since that would imply that the component of the tensions along the z axis and the weight of the plate are in the same direction (adding up to a nonzero & negative value rather cancelling out).: Q##n_Q## + S##n_S## + R##n_R## – 80 lb = 0 ⇒ negative + negative + negative - 80 lb = negative

    For E1, should the book just have said that E1 is Q##(–n_Q)## + S##(–n_S)## + R##(–n_R)## = 80 lb where ##n_Q##, ##n_S## and ##n_R## are, each, negative (such that the two negatives make a positive)?

    I am being pedantic about this because I would like to know what the logic is behind these decisions if it's correct or if the book has mistakes.

    Any input would be greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. May 14, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    The vectors formed by the cables all have 5 in their denominators. Multiplying through each equation by 5 makes the coefficients of the unknowns is not necessary, but it makes for less typing in the solution (Didn't you read this? It is very clearly explained.)

    The unit vectors are negative because the origin is the point of suspension, and the lid is below this point. Once you have selected a coordinate system, it is important that you remain consistent throughout the solution to the problem. In this case, the origin could have been selected to be the center of the lid, and the signs of the tension equations could have been adjusted to suit.

    In solving equilibrium problems, a certain positive direction is assumed for forces and moments. When a solution is obtained, if the unknown force or moment happens to come out negative, it just means that the assumed positive direction should have been the opposite to what was initially assumed.
     
  4. May 15, 2013 #3

    s3a

    User Avatar

    It's not 5; It's -5.

    I don't think I explained myself correctly. As you likely know, vectors can be translated in their direction of motion such that the tail end of the tension (=force) vectors are placed on the knob with the direction arrows pointing above the knob. What I am saying is that the solution seems to want to reuse the direction cosines of the length vectors (which are opposite in direction to the tension/force vectors). Because the book wants to reuse these direction cosines, I was suggesting that ##Q####n_Q## + ##S####n_S## + ##R####n_R## = 80 be ##Q####(-n_Q)## + ##S####(-n_S)## + ##R####(-n_R)## = 80 since ##n_Q## + ##n_S## + ##n_R## are already negative. I believe I am correct in thinking this and that the solution is wrong but, just in case I am the one who is wrong, I would like to go over this with someone else (such as you).

    I know. :)
     
  5. May 15, 2013 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    If the solution is wrong, you can confirm by adding the tension vectors together and seeing if they are opposite and equal to the weight of the lid. A further check would be to calculate the moments as well.

    If you have a number like -(a/b), it doesn't matter if you assume it is (-a/b) or (a/-b). 1/5 was a common factor in the direction cosine values.
     
  6. May 15, 2013 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    s3a, the signs on the direction cosines are correct for the way the three vectors have been defined. A problem arises when these are used in conjunction with the tension variables; the signs should have been reversed at that point, since the defined vectors are opposite in sense to the tensions. The author seems to have fixed this up by arbitrarily inserting a minus sign in front of "(80)(5)". I think it all makes sense if the minus sign had been inserted one step sooner (in front of "80lb").
     
  7. May 15, 2013 #6

    s3a

    User Avatar

    I meant that the work had something wrong in it but that the final answer was still correct.

    I know they're both correct but I was questioning why one would want to multiply both sides of the equation by a negative number if all the terms are positive but, all of this no longer matters since I got a confirmation from haruspex about what I was trying to get a confirmation about.

    The way the direction cosines were defined were in terms of the length of the cables from the knob and, when thinking about lengths of stuff, directions don't matter much as long as they're consistent but, when thinking about forces/tensions, it's more logical for the tensions to be upward (in this problem) ... I'm pretty sure you agree with this but, I just want to state my logic. Basically, I was right all along that the author basically just made a mistake and that mistake was that for E1 where E1: ##Q####n_Q## + ##S####n_S## + ##R####n_R## = 80 lb should have been ##Q####n_Q## + ##S####n_S## + ##R####n_R## = -80 lb (changing the direction of the plate's weight since the "recycled" direction cosines are in the incorrect logical direction ). :) By the way, I'm not saying it's unacceptable to have a positive weight of the plate and negative direction cosines (for the z axis) since I get that as long as one is consistent with signs, everything is fine but, the author was not consistent even when doing it assuming directions I wouldn't have assumed myself.

    So, this entire thread was just about omitting that negative sign, lol (because, I wanted to make sure I wasn't actually misunderstanding something). (Yes, my question for this entire thread was that simple!)

    Thank you both! :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Directions/signs question about force-equilibrium problem
Loading...