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Homework Help: Two forces, two angles; Net force = ?

  1. Feb 3, 2009 #1
    Greetings everyone! Thanks in advance for the help!!!

    1. The problem statement, all variables and given/known data

    Two parts to the problem. For both parts it asks for the magnitude of the net force and the direction of angle from the X axis.

    Part A:

    Two forces, F1 and F2, act on an object at rest at a 90* angle. F1 = 30N, F2 = 11N, M = 9kg


    Part B:

    /____ F2

    Theta is 60*

    3. The attempt at a solution

    Part A:

    For magnitude, I got 31.8N.
    F-net = sqrt [ f1^2 + f2^2 ]

    Angle: 24*

    Part B:

    F-net = F1 + F2
    F1 = (30N) cos 0 + (30N) sin 0 => (30N) 1 + 0 => 30N
    F2 = (11N) cos 60 + (11N) sin 60 => (11N) 1/2 + (11N) 0.8660 => 15N

    Therefore, Fnet = 45N

    Angle: 16*

    I really dont think I came close to the answers...but someone let me know, please!
  2. jcsd
  3. Feb 3, 2009 #2
    Part A:

    The way you've calculated your magnitude is correct, so I suppose your answer is as well. (although I can't for my life figure out why on earth you're given mass = 9kg.)

    How did you calculate the angle? Because I don't get the angle to be 24 degrees.. (it's a bit because the y-component is larger than the x-component - how can the net force be closer to the x than the y, if you understand what I mean?)

    Part B:

    Here you need to find the x and y components of the force F1, first. This isn't what you've done. The x-component of it will work together with F2 in the horizontal, and the y-component will be alone in the vertical. The total x and y components will be at 90 degrees to one another and you find the net force from there.
  4. Feb 3, 2009 #3
    I agree with your magnitude in Part A, but like Hannisch said, I don't think your angle is right. Remember that [tex]arctan(\frac{opposite}{adjacent})=\theta[/tex]. Also, it is imperative that you define the reference point for your angle! Is it 24o above the horizontal, below the horizontal, or is it measured from the vertical? (24 doesn't work for any of these references, but I hope you see my point).
  5. Feb 3, 2009 #4
    Wow Im sorry. Somehow I mistyped. F1 is on the X-axis(0*) and F2 is the other Tangent at the angle, not the other way around. That is how I got 24* (though I still think it is probably wrong).

    Thanks for the help guys, im trying to get this as best I can.

    side note, void, the question itself from the HW states the reference point for the angle. When it asks to solve, it solely asks for the numerical value. But I agree, reference point is always a necessity!
  6. Feb 4, 2009 #5
    Okay, so I redid everything. This is what I got.

    PART A:

    Magnitude: 31.95
    Angle: sin x = 11 / 31.95 => 20.1*

    Part B:

    Magnitude: 36.7
    Angle: sin x = 9.5/36.7 => 15*
  7. Feb 4, 2009 #6
    I say I agree with you on those values. Intuatively they make sense (net force should get greater in B, angle should get smaller since they're closer together, etc) and with my quick calculations it seems correct.
  8. Feb 4, 2009 #7
    Okay, I am done with workin on that section. I am onto a new question and I got all parts of the answer right so far, except the last part. The question is:

    Consider two vectors by A = 5 i - 3 j and B = - i - 6 j.

    (a) Calculate A + B

    4i + -9j

    (b) Calculate A - B

    6i + 3j

    (c) Calculate |A + B|


    (d) Calculate |A - B|


    (e) What is the direction of A + B? (from the +x axis)

    156* - wrong answer

    I did: sin x = 4/9.84 => x = 24 => 180 - 24 => 156*

    (f) What is the direction of A - B? (from the +x axis)

    63.5* - wrong answer

    I did: sin x = 6/6.7 => x = 63.5

    Can someone fill me in on what I did wrong?
  9. Feb 5, 2009 #8
    j is the 'vertical', right? I'd say sin(j/net) is the correct way to go.. you've done sin(i/net) which I think should be cosine.
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