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Newtons 3rd Law

  1. Nov 7, 2006 #1
    I have to apply the 3rd law i think to this, just not sure how to.
    A 2000 kg motorboat's propeller pushes on the water toward the west with a 2000 N force. The wind blowing toward the south pushes the motorboat with a 500 N force. Determine the location of the motorboat 10 seconds from the starting point where its initial velocity is zero.

    I have like 3 different problems like this one, so i choose thise one and hope someone can help me figure it out, then i can do the other ones easily. So what do i do now? I drew a graph with the N E W S directions and have 2000N going W and 500N going south, now what?
  2. jcsd
  3. Nov 7, 2006 #2
    Draw your free body diagram and determine the net force acting on the boat. You know vi = 0. Now, determine the net acceleration on the boat (you know the mass), and plug it into a kinematic equation to find out how far it went. Alternatively, you can calculate each component separately, and then determine the net effect in the end.
  4. Nov 7, 2006 #3
    Im not exactly sure how to do all that, im new to all this Newton stuff. How would i be able to find the acceleration on the boat using just the fact that its 2000kg, and what would the kinematic equation look like?
  5. Nov 7, 2006 #4
    F = ma (You should try to memorize this one if you could.)

    [tex]x_{f} = x_{i} + v_{x,i} \Delta t + \frac{1}{2} a_{x} (\Delta t)^{2}[/tex]

    [tex]v_{f} = v_{i} + a \Delta t[/tex]

    [tex]v_{f}^{2} = v_{i}^{2} + 2a \Delta x[/tex]

    Where x is simply a component value.

    Pick the one that suits your needs. If you are curious what any of the formulae mean, they will be in your text book.
  6. Nov 7, 2006 #5
    Ok, so force = mass * acceleration. What would the force be in that equation, since there are 2 different forces, one pushing west 2000N and one pushing south 500N? Thats what i didnt understand when putting together the acceleration.
  7. Nov 7, 2006 #6
    Add the two vectors (stick the tail of the second one onto the head of the second one). Use Pythagorean Theorem to find Hypotenuse and trig to find angle.
  8. Nov 7, 2006 #7
    You get 2061.55 adding the two vectors together using the pythagorean theorem, but what steps do i take next to solve this? I still cant figure it out.
  9. Nov 7, 2006 #8
    You have force. Use F=ma to find acceleration. Then use kinematics equations like what geoff posted earlier to find final position.

    Also, find the angle of the Force you found using trig. It will be the same angle as the final position you are trying to find.
  10. Nov 7, 2006 #9
    Ok so using f=ma, the acceleration should be...1.03. Which of those equations should i use? I dont have a textbook, so im not certain. Also, how does the angle help me in this problem, it isnt part of any equation is it? Can you help me set up the final equation using 1.03, or tell me which variables go to which numbers in this problem?
  11. Nov 7, 2006 #10
    You can use several of them. The one I would use is


    So you can find d. The reason I tell you to find angle is because it asks you for the location, which I assume includes an angle. The way to find it is look at the triangle you had when you added the vectors and determine the angle (easier to do it between the the Force that points West and the Total Force). Use trig to find that. Hint: [tex]tan^{-1}(\frac{500}{2000})[/tex]
    Last edited: Nov 7, 2006
  12. Nov 7, 2006 #11
    Ok so using that formula, [tex]d=v_it+\frac{1}{2}at^2[/tex]


    So would that mean the location of the motorboat is 51.5 (what unit?) away from where it was orginally after 10 seconds?
    And with the angle, i got -.004 with the tangent equation of tan.25*-1...how does that translate into an angle?
  13. Nov 7, 2006 #12
    Good, you got the distance. I gave you the forumla for the angle that it is South of West. Look for a [tex]tan^{-1}[/tex] button on your calculator and make sure it's in degrees mode, not radians.
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