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Trig..please help!

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Two forces are pushing an ice shanty along the Ice. One has a magnitude of 330 lb in a direction due east. The other force has a magnitude of 110 lb in a direction 54 degrees east of north. What are the magnitude and direction of the resulant force?

  • Trig..please help! - ,

    method 1: state them as vectors in component form

    first : 330 due east
    = (330cos 0, 330 sin 0) = (330, 0)
    second( 110cos36°, 110sin36°) = (88.99 , 64.66)

    add them to get (418.99, 64.66)

    magnitude = √(418.99^2 + 64.66) = 423.95
    direction: tanØ = 64.66/418.99
    Ø = 8.77°
    or a bearing of N 8.77° E

    method 2:
    construction the parallogram with the diagonal representing the magnitude
    I get an obtuse angled triangle with sides 110 and 330 and the contained angle of 144° ...... (90+54)

    by the cosine law:
    r^2 = 110^2 + 330^2 - 2(110)(330)cos144
    ...
    r = 423.95 , same as above

    Use the Sine Law to find the direction of the resultant.

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