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Magnitude of acceleration hwk check

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data
    I wrote down in picture



    2. Relevant equations
    F=ma

    3. The attempt at a solution
    Worked out in picture
     

    Attached Files:

  2. jcsd
  3. Oct 19, 2015 #2
    North, South, East, and West, unless stated otherwise, are directions parallel to the (locally flat) surface of the Earth. In your diagram, you have the second horizontal force, whose direction is described both as being horizontal and being "62° North of East" pointing upwards, away from the ground, with an angle of 62° to the normal to the ground, and partially acting against the vertical force of weight.That direction, normal to the ground, is not called North. It is either called "the normal to the surface", perpendicular to the ground, or zenith.
    The given description, however, is meant to convey a direction that is 62° in the northward direction, starting from the direction of cardinal West. There is no vertical component described for that particular force.
    As there are no forces described acting vertically, we can assume the normal force of the ground on the block and the weight of the block add up to a net vertical force of 0. So your free body diagram should only have the two horizontal forces described.
     
  4. Oct 19, 2015 #3

    So when I take those off. My x components are still right. When I sum of all my components in the x direction I should have f1-f2cos(29)=m•a
     
  5. Oct 19, 2015 #4
    Not quite. Your diagram has the second vector's direction as 62° West of the North cardinal direction. The description in the problem, however, says the vector should be 62° North of West, which, in your diagram, means we start on the West direction (the leftwards pointing horizontal axis in your diagram) and we move 62° clockwise towards the North direction. In other words, the net East-West force should be F1 - F2cos(62°).
     
  6. Oct 19, 2015 #5
    Ohhhhhh. I see that. If it said 62 west of north. Then I would draw my angle 62 degrees from my north direction. Thanks.

    Now we have (f1-f2cos(62))/m =a. We have solved our problem.
     
  7. Oct 19, 2015 #6
    That's it! However, this is only the East-West component of the acceleration vector. There is also a component of net force in the North direction, which gives us a component of acceleration in that direction as well. Once you have both components, you can find the magnitude of the acceleration vector in the usual way.
     
  8. Oct 19, 2015 #7
    The acceleration in the y direction would be zero so that shouldn't count right.

    F2sin(62)=ma
    But a =o
     
  9. Oct 19, 2015 #8
    Why do you believe that the acceleration in the y direction should be zero ?
     
  10. Oct 19, 2015 #9
    Because the box is not accelerating in the y direction. Is only moving along the x direction as it moves across the frictionless table top.
    So since it's not moving in the y direction the acceleration is 0
     
  11. Oct 19, 2015 #10
    Hmm, this information is not given in the problem. What information did you use to conclude that the box is only moving in the x direction ? Remember the x and y axes are both horizontal if they are aligned with the cardinal directions. The vertical axis, along which the weight of the box is aligned, would be the z axis in this case.
     
  12. Oct 19, 2015 #11
    In your original diagram you chose the y-axis to be vertical, yet it's stated that the two given forces are in horizontal directions.

    Best to re-draw your diagram and realize that the forces ##m \vec{g}## and ##\vec{F}_N## don't belong there because ##m \vec{g}## points downward and ##\vec{F}_N## points upward. They lie along the z-axis, given that the xy-plane is horizontal.
     
  13. Oct 19, 2015 #12
    This is just a 2d problem so the z axis wouldn't be included. And based on the other examples we have done the a=0 with of course the block being on an object. If the block was being pulled in by a pulley then that would be a different case when a DIDNT equal zero
     
  14. Oct 19, 2015 #13
    We never do anything in 3d in my class either.
     
  15. Oct 19, 2015 #14
    Yeah the only forces that should be there are:

    F1 point in the east direction as dram
    And
    F2 point in a 62 degree angle north of west which has to be broken into its x an y components which I have done.

    But the y wouldn't matter because the box is not accelerating in the y direction which would make it zero.
     
  16. Oct 19, 2015 #15
    I agree. Which is why I said this ...

     
  17. Oct 19, 2015 #16
    There is no acceleration in the vertical direction, but the y-axis is not vertical!
     
  18. Oct 19, 2015 #17
    Okay I solved for a for my final answer
     

    Attached Files:

  19. Oct 19, 2015 #18
    Why do you show the ice rink below the puck? You are looking down on the rink from above, so all you'll see is the round shape of the puck.

    I ask this, not because I want to see a better drawing, but because you're still not getting the right answer and I'm trying to get you to see why so you can fix it.

    By the way, the northward direction is not up. It's a direction that lies in the same horizontal plane as east and west.

    I realize that nothing you do in your class is in three dimensions, but you still need to be able to orient the two-dimensional plane you are working in. That plane could be vertical or it could be horizontal. In this case it's horizontal but you are thinking of it, and drawing it, as if it's vertical.
     
    Last edited: Oct 19, 2015
  20. Oct 19, 2015 #19
    I thought we were looking at the block from the side, like if it were sitting on a table and we were trying to calculate the forces then. Then the way I drew the vectors would be right.


    Is the answer wrong like the numerical value.
     
  21. Oct 19, 2015 #20
    I would say the plane was horizontal which is why I drew the little graph showing that the y was up and the was horizontal
     
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