1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravitation, really hard! at least for me

  1. Jul 16, 2009 #1
    1. The problem statement, all variables and given/known data
    In Fig. 13-41, a particle of mass m1 = 0.23 kg is a distance d = 84 cm from one end of a uniform rod with length L = 6.4 m and mass M = 2.4 kg. What is the magnitude of the gravitational force on the particle from the rod?

    2. Relevant equations

    3. The attempt at a solution
    ok...so i figured that the desnity of the rod, M/L should be the same for dm/dr...so then dm=(M/L)dx. (i thought just to visualize easier i made dr into dx, so it looked like a coordinate system to me)

    then i figured i had to integrate F=int (dF)

    so i did F= int (Gmdm)/(x+d)^2...since dm=(M/L)dx this made the integral
    int (GmMdx)/L(x+d)^2

    further simplifying i got F= (GmM/L) integral dx(x+d)^2

    after integration i got F= GmM/L multiplied by -(x+d)^-1 from 0 to L...

    then i solved, etc. but got the wrong answer...any idea if this is totally off or if i made a silly mistake somewhere?
  2. jcsd
  3. Jul 16, 2009 #2
    It looks correct....
  4. Jul 16, 2009 #3


    User Avatar
    Homework Helper

    It looks correct to me, too. However, I would just make a comment on your integral: you should be more explicit about the limits. E.g. I did a change of variables that changed my limits to d and L+d (but anyway I got the same result as you).

    Maybe you just forgot to convert cm to m?
  5. Jul 17, 2009 #4
    i'm an idiot... the answer was about 6.05e-12, and when i plugged it into the website i left out the e-12....what a silly mistake!!! well, i have looked at it sooo many times at least that i will never forget how to do this problem...
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook