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Setting up vector equations for problem solving

  1. Apr 8, 2015 #1
    I am confused about exactly how to set up vector equations properly when it comes to the sign of the vectors. Take the following problem for example. A 70.0 kg sailor climbs an 11.5 m long rope ladder to a mast above at constant velocity. The rope ladder is at an angle of 30.0° with the mast. Assume that the ladder is frictionless. What is the work done?

    To start, I would use Newton's second law:

    ##\sum_{1}^{n}\vec{F}_{n} = \vec{F}_{\textrm{app}} + \vec{F}_{\textrm{g}} = \vec{F}_{\textrm{net}} = m\vec{a}##
    ##F_{\textrm{app||}} + F_{\textrm{g||}} = m\vec{a_{||}}##
    ##F_{\textrm{app||}} + F_{\textrm{g||}} = 0##
    ##F_{\textrm{app||}} = - F_{\textrm{g||}} = -(-mg)\sin \theta = mg\sin \theta##

    But why would this setup be correct? For example, couldn't I have bypassed all those steps and reasoned that...

    ##F_{\textrm{app||}} - F_{\textrm{g||}} = 0## (there is a negative because the component is negative)
    ##F_{\textrm{app||}} = F_{\textrm{g||}}##
    ##F_{\textrm{app||}} = -mg\sin \theta##

    However, this has the opposite sign as the last answer! What am I doing wrong? How should I be approaching these problems when it comes to setting up my vector equations? Could someone give me a concrete example, with all steps, of how they would solve this problem (with special attention to the use of signs)?
  2. jcsd
  3. Apr 8, 2015 #2


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    (Ignoring special notation)
    2nd line: F_app+F_g=ma
    3rd line: F_app+F_g=0

    Both can't be right!
  4. Apr 8, 2015 #3
    Why not? The sailor is moving at a constant velocity, so a = 0. Right?
  5. Apr 8, 2015 #4
    When dealing with forces and Newtons 2nd Law one says the Fnet = Σ+Fi Then establish a positive reference directions for each axis so when you substitute the expression for each individual force component you will at that point determine the proper sign.to use . Just be consistent.

    In the first set of equations you wrote Fapp+Fg = 0 and in the second set Fapp-Fg = 0. that account for the discrepancy . you should not have used the neg sign in the second

    Another application where it can make big difference is in calculating work. Work can be done on something or something can be doing work.. In the sailor example the sailor is doing work in climbing the ladder. When the sailor descends gravity is doing work on the sailor. Work done by something is often consider positive and work done on something is considered negative. but that could be reversed. again just be consistent. For our sailor example up is taken as positive and the sailors effort is up thus work is positive.. So in going up the ladder work is being done by the sailor. But In coming down the distance moved is negative but the force is still positive ( the sailor is holding the ladder) so work is negative and is being done on the sailor. But also note that the force of gravity is negative and the distance moved is negative so the work done is positive but that is the work done by gravity on the sailor. Does that make sense?
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