Newton's Laws direction and acceleration

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Homework Statement



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.

Homework Equations





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
 

Answers and Replies

  • #2
12,828
6,707
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...
 
  • #3
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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?
 
  • #4
12,828
6,707
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)
 
  • #5
Radarithm
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Fnet = ma
F and a are vectors; since F is equal to ma, what can you conclude about their vector relationship?
 
  • #6
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So their vector relationship would cancel out?
 
  • #7
Radarithm
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So their vector relationship would cancel out?

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"
 
  • #8
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So they are parallel vectors
 
  • #9
Radarithm
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So they are parallel vectors

Exactly. The direction of the force is parallel to the direction of the acceleration.
 
  • #10
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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
 
  • #11
Radarithm
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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

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.
 
  • #12
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Okay well thank you anyway, I'm getting more confused as we keep talking about it.
 
  • #13
Radarithm
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I apologize for confusing you. Good luck with the problem.
 
  • #14
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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
 

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