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Resultant vector help

  1. Sep 27, 2008 #1
    A person on a parachute has an upwards force of 500N and a downward force of 1000N. The wind strikes the person at 45 degrees and has a force of 300N find the resultant force to 1 decimal place.

    Hint: Resolve the force of the wind into its horizontal and vertical components, before trying to add the forces together. Remember you can’t add forces that are perpendicular, you’ll have to use Pythagoras!


    Plz help me I didn't understand the hint and plz show the working of the problem I will be really grateful to you. By the way someone told me the answer is 357.6 N but he doesn't know the working so plz help and reply soon. :)
     
  2. jcsd
  3. Sep 27, 2008 #2

    Redbelly98

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    Welcome to PF.

    Start by drawing a force diagram for the person. There are 3 forces here.

    p.s. You didn't say whether the wind force is 45° upward or downward from horizontal. Is that specified in the problem statement?
     
  4. Sep 27, 2008 #3
    Sorry forgot to specify that it is upwards actually i have the diagram but i don't know how to post it?
     
  5. Sep 27, 2008 #4

    Nabeshin

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    Just host it on photobucket or some other image hosting server and post the link. Attaching files requires them to be reviewed, which takes forever.
     
  6. Sep 27, 2008 #5
    sorry but that will take time and i want the answer qucikly sorry. Plz people help me find the answer
     
  7. Sep 27, 2008 #6
    Plz explain the answers also so that I can perform better and I have a test soon so ur info will be invauable to me thank you reply soon
     
  8. Sep 27, 2008 #7

    Redbelly98

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    Have you drawn a force diagram yet, as I suggested in post #2?

    Don't worry about posting the diagram here, just draw it for yourself.
     
  9. Sep 28, 2008 #8
    i drew it but colud u please say the steps
     
  10. Sep 28, 2008 #9
    Seems like you can just determine the delta of your upward force(1000-500) and use it as the first leg of your diagram triangle, create a second leg with a 45 degree angle to the initial leg, and give it the wind's force of 300.

    Once you draw this, you should be able to determine the value of the third leg using the laws of cosine.

    Hope that helps!
     
  11. Sep 28, 2008 #10

    Redbelly98

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    To find the net force, the steps are:

    Draw a force diagram, and label all of the forces.

    Decide what the positive-x direction is to be.

    Decide what the positive-y direction is to be.

    Figure out the x- and y-components of each force. You can either use trig, or properties of 45-degree right triangles.

    Add up the x-components of each force to get the total x-component.

    Add up the y-components of each force to get the total y-component.

    Use the Pythagorean Theorem to find the magnitude of the force.

    Use trig to find the direction of the force.
     
  12. Sep 28, 2008 #11
    didn't understand redbelly98 please could u explain in detail by doing it i will be grateful to u.
     
  13. Sep 28, 2008 #12

    Redbelly98

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    No, I can't do your homework for you ... it's forum policy.
     
  14. Sep 28, 2008 #13
    redbelly98 this is not my homework i m new to this forum and actually i m goin a bit ahead and doin more work than my teacher gave me and this question is taken from a website so please help me solve this question and i guarantee u this is not a homework question so please help me solve it if u don't believe i can even give u the site where i got the question from so please help me redbelly98
     
  15. Sep 28, 2008 #14

    Redbelly98

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    I can help but you will have to do some work as well. That's how this forum works, and that's how I work.

    You said you have drawn a force diagram. Did you label the 3 forces?

    Then the next step is to choose a direction for +x and +y. Can you do that?

    If you can find a way to post your diagram -- with forces labelled, and +x and +y directions indicated -- we could proceed to what's next.

    If you are confused by labeling the forces and +x and +y, then post what you have drawn so far.
     
  16. Sep 29, 2008 #15
  17. Sep 29, 2008 #16
  18. Sep 29, 2008 #17
    Oops I forgot to put the arrows the 300 N is going upwards North East direction and 500 N downwards
     
  19. Sep 29, 2008 #18
    plz help me finish the question soon redbelly98
     
  20. Sep 29, 2008 #19

    Redbelly98

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    There is a problem with your force diagram. The 300 N force is at a downward angle, but you said earlier it is at an upward angle. See if you can fix that.

    If the drawing was made in Word or Excel, you could put arrow heads on the lines to indicate directions more clearly.

    p.s. please stop rushing things. I get the message, you're in a hurry ... even though you have gone ahead of the class (???). I'm busy and am giving you what time I can.
     
  21. Sep 29, 2008 #20
    Sorry redbelly98 for getting u angry i was being too selfish and I forgot even u have work.

    I didn't get wht u meant by "The 300 N force is at a downward angle".Isn't the force goin upward???. So I cant amend the force diagram. So plz could u make tht point clear and then maybe i can change the force diagram. Also plz tell what is the mistake i did in the force diagram so tht i can make changes. Thanku
     
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