a) net force=G34(445 -145 )/d^2 where d is half the separation distance.
b. let x be the distance from the larger object.
Then the force is zero at that point. 0=G*34(445/x^2-145/(3.9-x)^2
or 445/x^2=145/(3.9-x)^2
(3.9-x)^2=.326 x^2
take sqrt of both sides
(3.9-x)=.571 x
1.571x=3.9
x=2.48m from the larger object
check all that math.
A 145-kg object and a 445-kg object are separated by 3.90 m.
(a) Find the magnitude of the net gravitational force exerted by these objects on a 34.0-kg object placed midway between them. (N)
(b) At what position (other than an infinitely remote one) can the 34.0-kg object be placed so as to experience a net force of zero from the other two objects?
(m from the 445 kg mass toward the 145 kg mass)
1 answer
3 answers