1. Limited time only! Sign up for a free 30min personal 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!

Newton's Laws direction and acceleration

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Two forces are acting on a 0.250-kg hockey puck as it slides along the ice. The first force has a magnitude of 0.405 N and points 25.0° north of east. The second force has a magnitude of 0.565 N and points 75.0° north of east. If these are the only two forces acting on the puck, what will be the magnitude and direction of the puck\'s acceleration? Enter the direction as an angle measured in degrees counterclockwise from due east.

    2. Relevant equations



    3. The attempt at a solution

    for the first force I got 0.405cos(25) = 0.367 (right) 0.405sin(25)=0.171 (up)
    then for the second force 0.565cos(75)=0.146 (right) 0.565sin(75)=0.546 (up)
    So next I would add all the forces 0.367+0.146=0.513N and 0.171+0.546=0.717N
    then I don't know where to go from there
     
  2. jcsd
  3. Oct 17, 2013 #2

    jedishrfu

    Staff: Mentor

    Since you have the total force as x and y components you can figure out the angle with respect to the x axis and you can use the pythagorean theorem to figure out the forces magnitude.

    from there use f=ma to determine the acceleration...
     
  4. Oct 17, 2013 #3
    Okay so I did 0.513^2 + 0.717^2 = c^2
    After solving I got c=0.882
    Then I used the arctan(0.717/0.513)=54.4 degrees
    How would I find acceleration though would I just use the the x components of the force for f=ma?
     
  5. Oct 17, 2013 #4

    jedishrfu

    Staff: Mentor

    Newtons law give you a clue if you consider F and a as vectors then what can you conclude?

    As a hint if I push a box north in what direction is the box accelerating?

    (Think parallel)
     
  6. Oct 17, 2013 #5

    Radarithm

    User Avatar
    Gold Member

    Fnet = ma
    F and a are vectors; since F is equal to ma, what can you conclude about their vector relationship?
     
  7. Oct 17, 2013 #6
    So their vector relationship would cancel out?
     
  8. Oct 17, 2013 #7

    Radarithm

    User Avatar
    Gold Member

    Not quite. Two lines that would never intersect on a 2D coordinate system. What do you call those?

    hint: the word starts with a "P" and ends with an "L"
     
  9. Oct 17, 2013 #8
    So they are parallel vectors
     
  10. Oct 17, 2013 #9

    Radarithm

    User Avatar
    Gold Member

    Exactly. The direction of the force is parallel to the direction of the acceleration.
     
  11. Oct 17, 2013 #10
    Okay, I'm still a little lost how this relates to the question. If direction of the force is parallel to the direction of the acceleration wouldn't they cancel out though? one would be in a negative direction and one in the positive
     
  12. Oct 17, 2013 #11

    Radarithm

    User Avatar
    Gold Member

    Can you explain what would be in the negative direction and what would be in the positive direction? The direction of the force is the direction of the acceleration, so they wouldn't cancel out unless something of equal and opposite magnitude of the force was in the other direction.
     
  13. Oct 17, 2013 #12
    Okay well thank you anyway, I'm getting more confused as we keep talking about it.
     
  14. Oct 17, 2013 #13

    Radarithm

    User Avatar
    Gold Member

    I apologize for confusing you. Good luck with the problem.
     
  15. Oct 17, 2013 #14
    I think I might have made sense with it so i just use my original forces to find the acceleration. So .405N/0.25kg=1.62m/s^2 and for the second case 0.565N/0.245kg=2.26 m/s^s
     
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: Newton's Laws direction and acceleration
Loading...