Calculating the Resultant

  • Thread starter godkills
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  • #1
godkills
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Force(N) Direction (deg) x Component y Component

F3 110 30 95.3 55

F4 100 125 122.6 81.9

F5 85 200 100.1 150.9

Resultant ? ? ? ?


Can somebody help me check on my data because I think I did the whole thing wrong and also when I got the resultant of x component it was 318 and 287.8

then I calculated the force which was like 428.

When I was solving for x component and y for F3 i got a few negatives and somewhere in the book said i had to add 180 degrees.
I am kind of confused had a bad prof for lab.
 

Answers and Replies

  • #2
PhanthomJay
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You should draw a rough sketch first on a typical x-y 4 quadrant graph. . For the F3 force of 110 N, your components are correct . For the F4 force of 100 N at 125 degrees ccw from positive x axis, your sketch should show for example that the x component of that force is -100cos55. (it lies in the 2nd quadrant, and points up and to the left..it's x componnet is the magnitude of the x leg of the right triangle formed using F4 as the hypotenuse...and the x component points left, so throw a negative sign on it. Continue..
 
  • #3
godkills
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Oh so when i add up all the x components it would be 95.3 + (-) 57.4 + (-) 79.87 = -42.27?

and how about calculating the resultant for force and direction you use pythagoream therom and the tan-1(opp/adj)?
 
  • #4
PhanthomJay
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Oh so when i add up all the x components it would be 95.3 + (-) 57.4 + (-) 79.87 = -42.27?
yes, good.
and how about calculating the resultant for force and direction you use pythagoream therom and the tan-1(opp/adj)?
Yes, first find the sum of the y components, add 'em up, and then calculate the resultant force and direction. Be sure to draw a sketch that will help deternine what is opposite, what is adjacent, and what the angle is.
 
  • #5
godkills
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It would seem weird to have a negative angle wouldn't it?

final components -42.27 and 107.8
 
  • #6
PhanthomJay
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What angle are you getting for a result? Are you drawing a sketch of the x (pointing left) and and y (pointing up) components of the resultant? In what quadrant does the resultant lie?
 
  • #7
godkills
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Well the triangle would like on quadrant 2 where x is negative and y is positive.
 
  • #8
PhanthomJay
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Well the triangle would like on quadrant 2 where x is negative and y is positive.
Yes, and what is the angle measured ccw from the positive x axis?
 
  • #9
godkills
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tan-1(-42.27/107.8) = -21.4
 
  • #10
PhanthomJay
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Calculators don't know too much about quadrants :wink: . That's why it is important to draw a sketch. Using tan theta = opp/adj, that value of 21.4 degrees is measured in which direction from which axis?
 
  • #11
godkills
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from top to bottom
 
  • #12
PhanthomJay
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from top to bottom
I am unsure what you mean by 'from top to bottom'.
 
  • #13
godkills
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from the y axis to the x axis the angle above the x axis
 
  • #14
PhanthomJay
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If the x comp is -42 and the y comp is + 108, and you are using tan theta = x/y, then then it's 21.2 degrees ccw from the pos y axis, right?
 
  • #15
godkills
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yeah thats what i was trying to say
 
  • #16
PhanthomJay
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OK, that's correct. Which is how many degrees counterclockwise from the positive x axis (conventionally, the way angles are measured).?
 
  • #17
godkills
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90+21.2 = 111.2
 
  • #18
PhanthomJay
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Yes, correct...actually I miscopied the angle , it should be 111.4 N, but considering significant figures and such, 111 degrees ccw from pos x axis is OK. It is also a good idea when drawing a sketch of the problem to graphically, to scale, add the vectors by placing the tail of the second vector on the arrow of the first, then the tail of the 3rd vector on the arrow of the second, then draw a line from the tail of the first vector to the arrow of the third vector..that line is the resultant, and will give you a rough idea of the magnitude and direction of the resultant, as a sanity check.
 

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