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Force vectors

  • Thread starter caityjo
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  • #1
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a force of 60N due east and a force of 40N due north act concurrently on an object. What is te magnitude and direction of their resultant? Find: Magnitude (graphically and using the pythagorean theorem) and the direction (graphically and triginomically)

HELP! My brain is not working...I can find the magnitude, but I can't remember how to find the direction!
 

Answers and Replies

  • #2
OlderDan
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caityjo said:
a force of 60N due east and a force of 40N due north act concurrently on an object. What is te magnitude and direction of their resultant? Find: Magnitude (graphically and using the pythagorean theorem) and the direction (graphically and triginomically)

HELP! My brain is not working...I can find the magnitude, but I can't remember how to find the direction!
Do you remember the definitions of sine cosine and tangent?
 
  • #3
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OlderDan said:
Do you remember the definitions of sine cosine and tangent?
Kinda?? I am just learning this stuff...and my teacher sucks, so I am kinda lost as to how I do this.
 
  • #4
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OlderDan said:
Do you remember the definitions of sine cosine and tangent?
But like, I know how to use them for the most part....I can find the magnitude ok, at least I think so...(72N right?) but the direction I cant get
 
  • #5
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To find the angle, use a protractor.

Hey, you can solve this graphically

Use a scale and protractor. Start at an origin, draw the forces tip to tail, draw the resultant, measure your answer. It's easy.
 
  • #6
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civil_dude said:
Hey, you can solve this graphically

Use a scale and protractor. Start at an origin, draw the forces tip to tail, draw the resultant, measure your answer. It's easy.
I know how to do that part, I got the magnitude of R both graphically and using the pythagorean theorem...but I can't remember how to find the direction using trip...I know I have to break it down and stuff...but I dont remember how
 
  • #7
OlderDan
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caityjo said:
I know how to do that part, I got the magnitude of R both graphically and using the pythagorean theorem...but I can't remember how to find the direction using trip...I know I have to break it down and stuff...but I dont remember how
The components of the resultant are just the vectors you started with in this problem. Each of them is the leg of a right triangle, and the resultant is the hypotenuse. Here are the trig relations

sin(theta) = opposite side/hypotenuse
cos(theta) = adjacent side/hypotenuse
tan(theta) = opposite side/adjacent side

Some people memorize the acronym SOH-CAH-TOA to help remember these

On your calculator, you can use the inverse trig functions to find the angles if you know the ratios (you do in this problem).
 
  • #8
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OlderDan said:
The components of the resultant are just the vectors you started with in this problem. Each of them is the leg of a right triangle, and the resultant is the hypotenuse. Here are the trig relations

sin(theta) = opposite side/hypotenuse
cos(theta) = adjacent side/hypotenuse
tan(theta) = opposite side/adjacent side

Some people memorize the acronym SOH-CAH-TOA to help remember these

On your calculator, you can use the inverse trig functions to find the angles if you know the ratios (you do in this problem).
ok...so lemme see if I get this... the opposite is 60N east right? and the Adjacent is 40N north. So....

sin(60/72)=.740
cos(40/72)=.850
tan(60/40)=1.5

right?? and what part do I have to break down the angles into the x and y components??

I have been working on this ONE problem for like 2 hours before I found this site...I really dont understand alot of what I am trying to do here...so your help is greatly appreciated!
 
  • #9
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caityjo said:
ok...so lemme see if I get this... the opposite is 60N east right? and the Adjacent is 40N north. So....

sin(60/72)=.740
cos(40/72)=.850
tan(60/40)=1.5

right?? and what part do I have to break down the angles into the x and y components??

I have been working on this ONE problem for like 2 hours before I found this site...I really dont understand alot of what I am trying to do here...so your help is greatly appreciated!
I think I had my calculator in the wrong mode...

sin(60/72)=.014
cos(40/72)=.999
tan(60/40)=.026

I think that's right??? but what do I do now??
 
  • #10
OlderDan
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caityjo said:
ok...so lemme see if I get this... the opposite is 60N east right? and the Adjacent is 40N north. So....

sin(60/72)=.740
cos(40/72)=.850
tan(60/40)=1.5

right?? and what part do I have to break down the angles into the x and y components??

I have been working on this ONE problem for like 2 hours before I found this site...I really dont understand alot of what I am trying to do here...so your help is greatly appreciated!
Your calculations are just the ratios, not the trig values related to those ratios.

Which side you call opposite and which you call adjacent depends on which angle you are talking about. The choice you made for opposite and adjacent is for the angle between the resultant and the positve y-axis. It is conventional to work with the angle between the resultant and the positive x-axis when possible. In that case, you would have to switch your opposite and adjacent. There is nothing wrong with the way you did it, as long as you know what angle you are talking about.

Use your calculator to find the angle that has a tangent of 1.5. For most calculators you do this by pressing the 2nd key and then tan and them put in the ratio. You should get 56.3 degrees. If you had chosen your opposite and adjacent the other way, the ratio woudl ahve been 2/3 and the angle would be 33.7 degrees. Notice how these two angles add up to 90 degrees. The direction of your resultant is 33.7 degrees North of East or 56.3 degrees East of North.

You really need to see some graphics instead of just words for this stuff. Take a look at this site

http://hyperphysics.phy-astr.gsu.edu/HBASE/vect.html
 
  • #11
OlderDan
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caityjo said:
I think I had my calculator in the wrong mode...

sin(60/72)=.014
cos(40/72)=.999
tan(60/40)=.026

I think that's right??? but what do I do now??
Those ratios are not angles. You could calculate

sin^(-1) (60/72) = 56.4 degrees etc.

sin^(-1) means to inverse sine. The direct operation sine gives you the ratio associated with an angle. The inverse operation gives you the angle when you know the ratio.
 

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