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Homework Help: Help on some problems:

  1. Nov 29, 2004 #1
    Help on some problems:EMERGENCY!!!!!!!!

    Hi all,

    I only have 4 problems I'm stuck on. Maybe you can help me a little...please?! I know it seems like a lot of problems, but i've really tried to do these and I'm majorly stuck. If you guys'll help with these last 4, I won't ever have this many again, so just this time, you could possibly help me.


    1) Four 5.0kg spheres are located at the corners of a square of side .60 meters. Calculate the magnitude and direction of the gravitional force on one sphere due to the other three.

    2) The Sun rotates about the center of the Milky Way Glaxay at a distance of about 30000 light years from the center (1 ly=9.5 x 10^15 m) if it takes about 200 million years to make one totation, estimate the mass of our Galaxy. Assume that the mass distribution of our galaxy is conventrated most in a central uniform sphere. If all the stars had about the mass of our Sun (2 x 10^30kg) how many stars would be in our galaxy?

    3) The asteriod belt between Mars and Jupiter consists of many fragments. Which some scientists think came from a planet that once orbited the sun and then was distroyed. If the center of mass of the asteriod belt is about 3 time farther from the Sun that the Earth is , how long would it have taken this supposed planet to orbit the sun?

    4) Two blocks are in contact on a frictionless table. A horizontal force is applied to one block. If m1=2.3kg, m2=1.2kg and F=3.2 N, find the force of contact between the 2 blocks, then show that if the smae force F is applied to m2 rather than m1 the force of contact between the blocks is 2.1 N.

    Again, if you guys will help me, I will be ETERNALLY grateful! Please help!!
    Last edited: Nov 29, 2004
  2. jcsd
  3. Nov 30, 2004 #2
    1. We choose one of the spheres. You decide to mark its location as the origin. This means we have 1 sphere on the x-axis .6 m from it, another on the y-axis, and a third one right between them, .6*sqrt(2) m from it. Use the formula [tex] F=G\frac{M_{1}M_{2}}{r^{2}}[/tex] and calculate the sum of the forces.

    2. Do you know the formula for the speed of a satellite (in this case, the Sun) ? If so use it. If you don't: remember the Sun is in rotary motion, and that v=S/t.

    3. Kepler's third law should be used here.

    4. I'm sorry, I didn't quite understand this...

    Hope this helps!
  4. Nov 30, 2004 #3
    For the fourth question
    First Part (F applied on m1)
    the contact force means
    the force exerted by the block m2 on m1.
    actually this force arises due to acceleration by applying the force.

    First of all calculate the acc. of the system(divide F by total mass)

    now contact force will be the force exerted by m2 opposing your F.
    this force will be nothing but m2 * Acc.

    Second Part (F applied on m2)

    now m1 will oppose the motion.
    the contact force will be m1 * acc.

    ( The acc. does not change when you apply F on the other block)

    Try it out and let me know

    Attached Files:

  5. Nov 30, 2004 #4
    Thanks to you all!!! you've officially saved my life!!
  6. Nov 30, 2004 #5
    I'd suggest drawing a diagram here.
    | \ |
    I'll choose mass 1.
    \Sigma F_y & = -\frac{Gm_1m_3}{r_{13}^2} - \frac{Gm_1m_4}{r_{14}^2}\frac{\sqrt{2}}{2} \\
    \Sigma F_x & = \frac{Gm_1m_2}{r_{12}^2} + \frac{Gm_1m_4}{r_{14}^2}\frac{\sqrt{2}}{2} \\
    \intertext{Add the horizontal and vertical forces vectorially}
    \Sigma F^2 & = \Sigma F_x^2 + \Sigma F_y^2
    Take the square root of both sides and substitute the values from the question. Note that [itex]r_{14} = r_{13}\sqrt{2}[/itex].


    To find the mass of the galaxy, I'd use Kepler's third law in the form [itex]\frac{r^3}{T^2}= \frac{GM}{4\pi^2}[/itex] and solve for M. Divide M by 2 x 10^30kg to find an approximation for the number of stars.


    Whoa, that's a really badly worded question. However, if by centre of mass they mean the distance from the midpoint of the annular radius to the sun. Then you just apply Kepler's third for the planet and the earth. The value of \frac{GM}{4\pi^2} is the same for both planets as they both orbit around the sun, so you end up with an expression like [itex]\frac{r_1^3}{T_1^2} = \frac{r_1^3}{T_1^2}[/itex]. Since you know three of the four variables in the equation you can find the fourth.

    Just solve this with Newton's second law. I'm assuming m1 is to the left of m2 and taking right as the positive direction.
    For m1
    \Sigma F & = m_1a \\
    F - n & = m_1a \\
    \intertext{where n is the contact force}
    n & = F - m_1a
    \intertext{The only horizontal force m2 experiences is the equal and opposite reaction force to the contact force. The blocks are in contact so that they both accelerate at the same rate a,}
    \Sigma F & = m_2 a \\
    n & = m_2 a \\
    a & = \frac{n}{m_2}
    n & = F - \frac{m_1n}{m_2} \\
    n\left(1 + \frac{m_1}{m_2}\right) & = F \\
    n & = \frac{F}{1 + \frac{m_1}{m_2}}
    Now rework the problem with
    n - F & = -m_2a \\
    -n & = -m_1a \\
    a & = \frac{n}{m_1} \\
    n & = F - \frac{m_2n}{m_1} \\
    n & = \frac{F}{1+ \frac{m_2}{m_1}}
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