Forces - calculating the net force

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  • #26
haruspex
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c= 32.9 N (right?)
Seems about right.
this how i would solve for the angle of the net force:
SinA / 38.0 N = Sin55° /32.9 N
Which angle is your A?
 
  • #27
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what do you mean?
 
  • #28
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I have looked through your OP and I suggest you using Coordinates system axis. It will be more faster. When you use it, you can easily figure the direction, magnitude of a force. And for the answer, the net forces direction is down and to the left. The angle created by vertical axis and the force is approximately 72 degree. The magnitude is approximately 32,76 N
 
  • #29
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@The Vinh
i am aware that using Coordinates system axis is easier, however the lesson that i am doing requires that i learn their method, and it involves the cosine law etc..
they don't provide any other ways to solve and they are expecting me to solve it using their method.
 
  • #30
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what do you mean?
And by the way, he meant that A is the angle created by vertical axis or horizontal axis and the force
 
  • #31
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@The Vinh
i am aware that using Coordinates system axis is easier, however the lesson that i am doing requires that i learn their method, and it involves the cosine law etc..
they don't provide any other ways to solve and they are expecting me to solve it using their method.
Oh, ok, so if you must use it then magnitude and angle are easy to find but the real problem is the direction of net force.
 
  • #32
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Oh and you have a really " cool " teacher
 
  • #33
haruspex
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what do you mean?
You quote the sine formula, but you did not say which angle in the picture is the A in the formula.
It needs to be opposite the side you divide the sine by.
Which angle do you need to find to answer the question?
 
  • #34
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@haruspex
upload_2016-3-14_23-56-56.png

This was the previous diagram i had made, it forms a triangle instead of a parallelogram. I labelled angle A .
SinA / 38.0 N = Sin55° /32.9 N
 
  • #36
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Ok so here is my final answer


38.

c = (a^2 + b^2 – 2abcosC) ½

c = ([32.0 N]^2 + [38.0 N]^2 - 2[32.0 N] x [38.0 N]cos55°) ½

c = 32.9 N


SinA/a = SinB/b = SinC/c

SinA / 38.0 N = Sin55° /32.9 N

A = 72 degrees (approx.)

net force = 32.9 N (North 72 degrees East)

does this look right?
 
  • #37
haruspex
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(North 72 degrees East)
Draw that direction on your diagram. Does it look right?
 
  • #38
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is the correct direction (East 72 degrees North)
 
  • #39
haruspex
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is the correct direction (East 72 degrees North)
I didn't say your answer was wrong, just asked if it looked reasonable. You should always check that where possible.
 
  • #40
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oh, ok, so i based the direction of this diagram.
upload_2016-3-15_0-37-51.png

the force is pulling the object 72 degrees east of north
 
  • #42
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thank you so much for your help, i really appreciate it :)
 

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