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Newton's 3rd Law and Acceleration

  1. Jun 9, 2009 #1
    To give a 19-kg child a ride, two teenagers pull on a 3.6 kg- sled with ropes. Both teenagers pull with a force of 55 N at an angle of 35 degrees relative to the forward direction, which is the direction of motion. In addition, the snow exerts a retarding force on the sled that points opposite to the direction of motion, and has a magnitude of 57 N .

    Find the acceleration of the sled and child.

    What I am doing is adding the weight of the child and the sled and using that as my mass and then finding the acceleration of the two kids pulling the kid by using a=f/m and then I did that same thing for the kid..... I don't know what to do from there....any suggestions? What am I forgetting?
  2. jcsd
  3. Jun 9, 2009 #2


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    You want to use F=ma indeed. Don't forget that the formula in reality is [itex]\sum F=ma[/itex] which we usually don't write down. There is only an acceleration if there is a net force acting on an object. What is the net force on the sled+child in the x-direction?
  4. Jun 9, 2009 #3


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    What you should do is resolve the 55N forces into x and y components, then find the resultant force(F) in the x direction. Then use F=ma where m is the mass of the child+sled.
  5. Jun 9, 2009 #4
    isn't the net force 55+57?
  6. Jun 9, 2009 #5
    wait a minute, are you telling me to use the angles to find an x and y value?
  7. Jun 9, 2009 #6


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    No the net force is not 55N+57N. Read the problem statement precisely. It says that the retarding force opposite to the direction of motion, F is a vector after all.

    Yes technically you have to calculate both the x and y components of the force. But intuitively you know that the pullers won't lift the sled vertically, so you can just calculate the x-component only. You can of course always check if mg>F_y.

    Tip: If a force problem involves angles then you can be pretty certain you actually have to use those angles.
  8. Jun 9, 2009 #7
    OKay so to find Fx I just did Fx=57sin(35) and that equals 32.69 and that is what I divided by the total mass 22.6 and came to the answer (may not be the right answer) a=1.45....is that basically it or did I just get lucky?
  9. Jun 9, 2009 #8


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    You calculated the y-component instead of the x-component. You also only calculated one force and not the net force.

    net force= mass times acceleration
  10. Jun 9, 2009 #9
    Cyosis is right, you want to solve for the x component of the force which is the cosine of the angles, also you have two teenagers pulling in that direct thus two such forces. Add those resulting forces to the opposite of the force created by the snow and that is your net force.

    and as Cyosis said

    net Force = total mass * acceleration. That should give you a much more reasonable answer.
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