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Two forces on a point particle

  1. Oct 24, 2008 #1
    1. The problem statement, all variables and given/known data

    Besides its weight, a 4.20 kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (3.80i hat - 3.30j hat) m, where the direction of j hat is the upward vertical direction. Determine the other force.

    2. Relevant equations

    F = ma; F =[tex]\sqrt{Fx2+Fy2}[/tex]

    3. The attempt at a solution

    okay so i know that the weight is pulling down on the object at 41.16 N. So i need to find the direction and magnitude of a force countering it, so that the resultant vector will = 5.03, however i'm confused about what role 1.20 s has to do with the problem. any help?
  2. jcsd
  3. Oct 24, 2008 #2
    You can break this problem down into its component constituents... the force due to gravity acts only in the j direction, so when this particle is displaced 3.80m in the i direction, in 1.20s, for which the acceleration is constant (because force is constant), you have:

    [tex] 3.80 = \frac{1}{2}a_x(1.20)^2 [/tex]

    since the object starts at rest (this is a suvat equation). Then you have:

    [tex] F_x = ma_x [/tex]

    Under the force of gravity alone, the object will travel some distance in j direction but has in fact travelled -3.30m in that direction... find out the difference and see what extra distance you need and then set up another equation the same as the one i have in x, with

    [tex] s = 1/2a_y(1.20)^2 [/tex]

    where s is this 'extra' distance.

    The importance of the 1.20s then, is in calculating the component accelerations (and hence the forces) of the particle.
  4. Oct 24, 2008 #3
    okay i did all that, and for the Fx, i still got the wrong answer. i got 23.76, but the program i'm using said i was within 10% of the answer. for the Fy, however, it said i was off by orders of magnitude. Here is what i did, maybe you can show me where i went wrong:

    Yf = m(-g)t2 + s
    then i plugged in numbers, solved for x, and put that back into that
    s = 1/2ay(1.20)2,
    found ay and then
    plugged that into Fy = may
    and got a final answer of 376.275, but it said i was off by orders of magnitude.
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