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A 3D trigonometric problem

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Four congruent spheres, radius length 10 cm, are placed on a horizontal table so that each touches two others and their centres form a square. A fifth congruent sphere rests on them. Find the height of the top of this fifth sphere above the table.


    2. Relevant equations

    Not found.



    3. The attempt at a solution
    I have been thinking of this for 2 days, but to be honest, I have no idea have to solve this. I tried the 2D problem, which has 2 congruent circles, and 1 congruent circle on top of the them. This problem seems to be easier and I worked it out. Yet this one is different and I hope I will be given a helping hand. Thank you, everyone.
     
  2. jcsd
  3. Oct 6, 2011 #2

    vela

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    Say the centers of the spheres lie in the xy-plane. Place one with its center at the origin, one with its center on the x-axis, one with its center on the y-axis, and one with its center in first quadrant. You should be able to say what the coordinates of the centers are.

    Now think about what the x and y coordinates of the fifth sphere's center has to be. Also, how far from the origin is the center of the fifth sphere?
     
  4. Oct 6, 2011 #3
    @vela: Thank you for your help. The coordinates of the centers are (0,0), (0,20), (20,0) and (20,20). I think the coordinates of the center of the fifth sphere is (10,10) and it is [itex] 10 \sqrt{2} [/itex] from the origin.

    I checked the answer at the back of my textbook and it seems that the answer is [itex] 10 \sqrt{2} + 10 . 2 ≈ 34.14 [/itex]. I found this result is really interesting. I tried to explain it, but now I just get a "feeling" (actually, it's a picture in my head) of why it is so.
     
  5. Oct 6, 2011 #4

    HallsofIvy

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    No, the center of the third sphere must be also 20 cm from the centers of all the original spheres-one of which is at the origin. In any case, this is a three dimensional problem, not two dimensional. Writing the centers of the spheres in two components doesn't help at all. It does not include the height of the centers and height is what the questions asks.

    The centers of the four base sphere form a square, as given. The lengths of the sides are, as you say, 20. The lines connecting those four centers to the center of the fifth sphere form a "pyramid" with a square base. Those four lines also have length 20. If you draw the line from the center of the square base to the center of any one of the four base spheres, the line from the center of the square base to the center of the fifth sphere, and the line connecting the centers of those two spheres, you have a right triangle in which the length of the hypotenuse is 20, the length of one leg is half the length of the diagonal of that square, and so you can calculate the length of the other leg, the height of the center of the fifth sphere. Don't forget that the top of the fifth sphere, the question asked, is 10 cm above its center and that the center of the four original spheres, and the center of the square they form, is 10 cm. above the table.
     
  6. Oct 6, 2011 #5

    vela

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    The distance isn't correct. You found the distance to (10, 10, 0), but the center of the fifth sphere is at (10, 10, z). If you can find the distance another way and set it equal to the distance from the origin to (10, 10, z), you can solve for z.
     
  7. Oct 6, 2011 #6
    @HallsofIvy: Thank you very much. I got the answer right now!
    @vela: Actually, I did think of using three-dimensional coordinate system to solve this problem, yet I eliminate this method because I did not learn how to do it (I am in year 10 now). Anyway, thank you for your help!
     
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