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Finding Resultant Force using Component Method

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data

    This is the force table lab that we are doing.

    [itex]F _1 = 300g @ 30^o[/itex]
    [itex]F _2 = 450g @ 110^o[/itex]
    [itex]F _3 = 400g @ 230^o[/itex]
    [itex]F _4 = 270g @ 298^o[/itex]

    Finding resultant force using Component Method

    1. Express each force F1, F2, and F3 in unit vector notation. Take the origin to be at the center of the force table (at pivot point) with the +x axis along [itex]0^o[/itex] and +y-axis along [itex]90^o[/itex].

    2. Use the component method to obtain the resultant force vector Fcomp in unit vector notation. Calculate the magnitude and direction.

    2. Relevant equations
    sinθ=opp/hyp
    cosθ=adj/hyp
    F=ma

    3. The attempt at a solution

    His directions are kind of confusing to me.
    For the step 1 I found:

    I converted the mass into kg and then found the force.

    [itex]F _1 = 2.94N [/itex]
    [itex]F _2 = -4.41N [/itex]
    [itex]F _3 = -3.92N [/itex]

    Then I express each force into unit vector notation.

    [itex]F _1 = 2.94cos30i +2.94sin30j [/itex]
    [itex]F _2 = -4.41cos70i +4.41sin70j [/itex]
    [itex]F _3 = -3.92cos50i -3.92sin50j [/itex]

    Did I do this part right? Please check it for me.

    I don't know where to start for step 2. Do I find F4?
     
  2. jcsd
  3. Oct 19, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    You should leave the signs off of the force magnitudes. Force magnitudes are always positive (absolute) values. It's the information from the angles where you form the components that will turn them into vectors with direction.
    Okay, I see you've sorted out how to use the trig functions with the given angles. Well done. At this point you should reduce the vectors to purely numerical component values. That is, evaluate each component, getting rid of the trig.
    The instructions don't mention F4 at this point, so I'd just combine F1,F2, and F3 into a resultant. Do you no how to add vectors in unit vector form?
     
  4. Oct 19, 2015 #3
    Is the formula:
    R=A+B+C which would be R=(F1x+F1y)i+(F2x+F2y)j+(F3x+F3y)k?
     
  5. Oct 19, 2015 #4

    gneill

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    Staff: Mentor

    No, you only add like with like. That is, sum all the x-components to yield the net x component, sum all the y-components to yield the net y-component,...
     
  6. Oct 19, 2015 #5
    Okay,
    R=(-1.48)i+(2.61)j
    Magnitude:
    R=2.99
    but what is the direction? Is that right?

    I just want to clarify the R that we found is considered the F4 but found mathematically right? Because in the lab we were given F1, F2, ,F3 and we had to find F4 using the weights.
     
  7. Oct 19, 2015 #6

    gneill

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    Staff: Mentor

    Looks okay. The direction is encoded in the components. Just sketch the resultant using the components and you'll see the direction. You can use these rectangular components to work out the magnitude and the angle if you want the polar version of the vector. Oh, always include units on results! That should be R=[(-1.48)i+(2.61)j] N, and |R| = 2.99 N.
    I don't know, I wasn't in your lab :smile: But it seems a likely scenario that some fourth force was employed to balance the force table setup. Did you have to play with a fourth force to try to find an equilibrium? If so, in order to balance the resultant of the forces F1 through F3 you'd have to apply a force of the same magnitude as the resultant but in the opposite direction (i.e. oppose it equally).
     
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