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Homework Help: Motion (newton's 2nd law + weight question)

  1. Apr 9, 2010 #1
    [SOLVED] Motion (newton's 2nd law + weight question)

    1. The problem statement, all variables and given/known data

    A waterskier of mass 70 kg is towed in a northerly direction by a speedboat with a mass of 350 kg. The frictional forces opposing the forward motion of the waterskier total 240 N.

    (a) If the waterskier has an acceleration of 2.0 m due north, what is the tension in the rope towing the waterskier?

    (b) If the frictional forces opposing the forward motion of the speedboat total 600 N, what is the thrust force applied to the boat due to the action of the motor?

    2. The attempt at a solution

    I answered a) correctly getting the tension as 380N north. For part b) though I got the answer wrong. According to the solutions it says that the mass used in for Fnet = ma is 350kg. Why? Doesn't the boat also pull the skier? So the net weight it 420kg? I'm confused as to why they used 350kg rather than 420kg. Same with part a), although I got it correct, looking back I'm not sure why 70kg is used rather than the combined mass of 420kg.

    So, in problems with two objects connected by a string, why is the mass of an object ONLY it's mass rather than the added mass of both, because isn't the mass pulled by the object the overall mass of the pair?

    Sorry if I'm not clear :(

    Last edited: Apr 9, 2010
  2. jcsd
  3. Apr 9, 2010 #2
    Have you drawn out your FBDs? For a) you should only have 1 FBD, and for b) you should again only have 1 FBD.

    If you haven't, that should answer your question concerning which mass to choose and why its not both.

    EDIT: It depends on how you analyze the problem, and to find the tension in the rope, you need to separate it into two systems and analyze it from one.
    Last edited: Apr 9, 2010
  4. Apr 9, 2010 #3
    Sorry, but what's an FBD?
  5. Apr 9, 2010 #4
    free body diagram.
    If you are unsure of what that is, it is a diagram that shows the system you are analyzing as a point particle (a dot or circle or what not on your paper) and you draw in the forces.

    So for a, you could draw a small circle labeled Ws for waterskier and draw a vector upwards labeled T for tension and a vector downwards labeled f for friction as those are the only two forces acting on the skier. Sum of F=ma. Since your only mass in your system is the skier, it is only 70kg. The tension accounts for the mass/forces on the boat.
  6. Apr 9, 2010 #5
    Ah yes. Solved my problem. So they are essentially seperate objects (in terms of mass), except for the tension between them?
  7. Apr 9, 2010 #6
    I think it all has to do with how you analyed it. Say the motorboat dies. And a tug boat comes by and offers to tow the motorboat and the skier too, just how they were.

    And you wanted to find the Tension in THAT rope (between the motorboat and the tug, so you still have the skier being towed by the motorboat).. Well you could analyze it as the tug boat of mass 1000kg and motorboat/waterskier 420kg with tension T between them..

    So it just depends on how you look at it and need to analyze it. The motor boat is definitly pulling the weight of the boat and skier, against frictional forces on the boat and the skier, but to analyze it, we split them up, but as a whole system, the mass is 420kg.
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