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    What is the magnitude of the resultant force and at what angle does it act?

    Asked Feb 15, 2010, 12:01 AM 4 Answers
    Another question regarding magnitude, an extra force is added this time.

    "For the set of forces shown in the diagram, what isd the magnitude of the resultant force and at what angle does it act?"

    Another diagram is attached.

    Cheers for any help,

    -Paaul.

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    ebaines's Avatar
    ebaines Posts: 10,896, Reputation: 1178
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    #2

    Feb 15, 2010, 07:06 AM

    Apply the same technique as with your previous question:

    http://www.askmehelpdesk.com/math-sc...ce-445787.html

    Try it - if you get stuck post back but this time show us how far you've gotten and where you're getting stuck.
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    leeroynew's Avatar
    leeroynew Posts: 41, Reputation: 3
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    #3

    Feb 15, 2010, 07:42 AM

    Thought the same method may have been used, so here's one I prepared earlier lol:

    Magnitude = 25.44
    Angle = 73.89

    If this is correct, the only place I am unsure about is the direction of the force and from which line the angle is to be meaured.
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    ebaines's Avatar
    ebaines Posts: 10,896, Reputation: 1178
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    #4

    Feb 15, 2010, 07:53 AM

    Correct - that's not so hard!

    The only "tricks" to remember are:

    (a) Always convert the angles sso that they ar measured counter-clockwise from the positive x axis. So the angle of the 40 N force is really 180-80 = 100 degrees, and the 15 N force is 180+ 45 = 225 degrees.

    (b) In taking the arc tan a positive result is either quadrant 1 or 3, and a negative result is either quadrant 2 or 4. So you need to look at the signs of Fx_total and Fy_total to figure out which quadrant. Here since both Fx_total and Fy_total are positive (right?) that means the resultant is in quadrant 1, and you are measuring from the positive x axis.
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    leeroynew's Avatar
    leeroynew Posts: 41, Reputation: 3
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    #5

    Feb 15, 2010, 08:11 AM
    Quote Originally Posted by ebaines View Post
    Correct - that's not so hard!

    The only "tricks" to remember are:

    (a) Always convert the angles sso that they ar measured counter-clockwise from the positive x axis. So the angle of the 40 N force is really 180-80 = 100 degrees, and the 15 N force is 180+ 45 = 225 degrees.

    (b) In taking the arc tan a positive result is either quadrant 1 or 3, and a negative result is either quadrant 2 or 4. So you need to look at the signs of Fx_total and Fy_total to figure out which quadrant. Here since both Fx_total and Fy_total are positive (right?) that means the resultant is in quadrant 1, and you are measuring from the positive x axis.

    Excellent. Thanks for the help. Our tutor expects us to know all this without being taught, so the help is appreciated. The explanations he does give complicate something that is really quite simple and still don't help to solve problems, for instance I've never been taught how work out which way the resultant force should go until now!
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