Calculating the net force

  • Thread starter alexandria
  • Start date
  • #1
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Homework Statement


upload_2016-3-15_1-26-57.png


Homework Equations


upload_2016-3-15_1-27-24.png

upload_2016-3-15_1-27-48.png


The Attempt at a Solution



8.0 N (north) + 10.0 N (south) = 2.0 N (south)
upload_2016-3-15_1-38-38.png

i used the tip-to-tail method to add the vectors
c = ([2.0 N]^2 + [17.0 N]^2 - 2 (2.0 N) (17.0 N) cos45°)1/2
c = 15.6 N

SinA/17.0 N = Sin45°/15.6 N
A = 40°

Net force = 15.6 N (S 40° W)

can someone please verify if my answer is correct! thanks in advance :)

 

Answers and Replies

  • #2
haruspex
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SinA/17.0 N = Sin45°/15.6 N
A = 40°
I don't get 40 degrees from that, but also you have fallen into a trap.
sin(x) increases as x goes from 0 to 90 degrees, but then decreases to 0 at 180 degrees.
sin(x) and sin(180-x) produce the same number. So when you know the sine and want to find the angle, you have to decide whether it is the angle returned by the arcsin function, or 180 degrees minus that angle.
There are deterministic processes for ensuring the right answer, but in this case just look at your triangle diagram. If you draw the 2N to scale it will be much shorter.
Like I said, it's always worth checking whether the answer looks sensible.
 
  • #3
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SinA/17.0 N = Sin45°/15.6 N
how do i solve this to determine SinA?
 
  • #4
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and does my answer look 'sensible' so far? i feel like im doing something wrong
 
  • #5
haruspex
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SinA/17.0 N = Sin45°/15.6 N
how do i solve this to determine SinA?
What did you do to get 40 degrees? Please show your working.
 
  • #6
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sorry to bring back this forum again, but i re-did this question, can anyone tell me if this is correct.
39.

8.0 N (north) + 10.0 N (south) = 2.0 N (south)
upload_2016-3-31_16-34-24.png



c = ([17.0 N]2 + [2.0 N]2 - 2 (17.0 N) (2.0 N) cos45°) ½

c = 15.6 N

SinA/a = SinB/b = SinC/c

SinA/2.0 N = Sin45 degrees/15.6 N

SinA = (2.0 N) Sin45° / 15.6 N

A = sin^-1 (0.091)

A = 5.2 degrees

90 degrees – 5.2 degrees = 39.8 degrees

Fnet = 15.6 N [North 39.8 degrees West]
???
 
  • #7
haruspex
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sorry to bring back this forum again, but i re-did this question, can anyone tell me if this is correct.
39.

8.0 N (north) + 10.0 N (south) = 2.0 N (south)
View attachment 98274


c = ([17.0 N]2 + [2.0 N]2 - 2 (17.0 N) (2.0 N) cos45°) ½

c = 15.6 N

SinA/a = SinB/b = SinC/c

SinA/2.0 N = Sin45 degrees/15.6 N

SinA = (2.0 N) Sin45° / 15.6 N

A = sin^-1 (0.091)

A = 5.2 degrees

90 degrees – 5.2 degrees = 39.8 degrees

Fnet = 15.6 N [North 39.8 degrees West]
???
That all looks right except for a typo in making the post. You wrote 90- instead of 45- near the end.
 
  • #8
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so then, its 45 - 5.2 = 39.8
thanks for the help.
 

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