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Homework Help: Vector Addition: Find the Resultant Vector

  1. Jun 23, 2015 #1
    1. The problem statement, all variables and given/known data

    The block shown in Fig. P-223 is acted upon by its weight W = 200lb, a horizontal force Q = 600lb and the pressure P exerted by the inclined plane. The resultant R of these forces is up and parallel to the incline thereby sliding the block up it. Determine P and R. Hint: Take one axis parallel to the incline. Ans: R = 293lb


    2. Relevant equations
    R = SummationFx + SummationFy
    Fx = Fcos0
    Fy = Fsin0

    3. The attempt at a solution
    I don't know how to start the problem. In the worktext, this is the only problem that has unknown magnitude(P). To get the R I need first to P. The hint says take one axis parallel to the incline, I will be able to form an isosceles triangle with P as the leg but I still won't be able to solve for the magnitude of P.... How do I start this problem?

    This problem is from Engineering Mechanics by Singer, it is our homework #213-223, I'm already done with the rest. The previous problems has all given magnitude and angle(some you need to solve for the angle, but still are easy except this one because of unknown value of P).
    Last edited: Jun 23, 2015
  2. jcsd
  3. Jun 23, 2015 #2


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    Uhh, there's no way to find the pressure P without knowing the area of the bottom of the box. Perhaps they meant to say that P is a force.

    With that assumption, that P is a force, you can determine it's magnitude by the fact that the block is in equilibrium in the normal direction (it doesn't float away from the plane and it doesn't move inside the plane).
  4. Jun 23, 2015 #3


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    The vector P may be unknown, although you do know its direction.

    However, you do know something about the resultant R. If you follow the hint, then you know on of the components of R. Right ?
  5. Jun 23, 2015 #4

    Great!!! I got it now thanks. What I did is I made the incline my x-axis (hint). With that I now have:

    Q = 600∠-30
    W = 200∠-120

    and then:

    Qx = 519.62
    Qy = -300
    Wx = -100
    Wy = -173.21

    and since R is parallel to the incline which is now the x-axis therefore ΣRy = 0 = Qy + Wy + Py or

    Py = 473.21... using the given 15' angle, I get Px = -126.8

    R = ΣRy + ΣRx
    R = ΣRx
    R = Qx + Wx + Px

    R = 292.82

    Sweet!!! Thanks a lot :)
  6. Apr 27, 2016 #5
    Can you show the solution on how did you get the value of Py and Px?
  7. Apr 27, 2016 #6


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    You should post your own attempt first.
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