Balanced and unbalanced forces

  • Thread starter pt20army
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  • #1
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The following forces act on a hockey puck (round, rubber disk) sitting on a frictionless surface: F_1 = 15.5 N at 15 degrees; F_2 = 27.9 N at 125 degrees; and F_3 = 31.7 N at 235 degrees. All the forces are in the plane of the ice. Determine the net force on the puck.


Fnetx= F1x-F2x-F3x=max, Fnety= F1y+F2y-F3y=may



F1x=15.5Ncos15=14.97
F1y=15.5Nsin15=4.01
F2x=27.9Ncos125= -16
F2y=27.9Nsin125=22.85
F3x=31.7Ncos235=-18.18
F3y=31.7Nsin235= -25.96

Fnetx=14.97+16+18.18=49.15
Fnety=4.01+22.85+25.96=52.82

Im getting this problem wrong and I am not sure where I am messing up.
 

Answers and Replies

  • #2
Doc Al
Mentor
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Don't ignore the signs of the components when you add them up.
 
  • #3
12,951
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Fnetx= F1x-F2x-F3x=max, Fnety= F1y+F2y-F3y=may

You should simply sum the forces with the values you've computed for the x and y components:

Fnetx = F1x + F2x + F3x and similarly for Fnety
 

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