call our mass m, the 35 does not matter, could be any amount.
we are x meters from the 190 kg
and
we are (0.4 - x) meters from the 490 kg
we have to be on the line between them so we are not pulled sideways
so
Fleft = G(190)m /x^2
Fright = G(490)m/(.4-x)^2
so for equal and opposite
190/x^2 = 490/(.4-x)^2
solve the resulting quadratic for x
A 190 kg object and a 490 kg object are separated by 0.400 m. At what position (other than an infinitely remote one) can the 35.0 kg object be placed so as to experience a net force of zero?
------------ m from the 490 kg mass
I do not understand how to attack this problem. pls help.
2 answers
when i try and solve this i get 300x^2+152x-30.4
how would i solve after this?
how would i solve after this?
1 answer