Question

Objects with masses of 165 kg and a 465 kg are separated by 0.390 m. (a) Find the net gravitational force exerted by these objects on a 41.0 kg object placed midway between them. magnitude? direction? (b) At what position (other than infinitely remote ones) can the 41.0 kg object be placed so as to experience a net force of zero?

Answer 1

magntiude of gravitational force between two masses m1 and m2 seprated by a distance of d m is given by

F=G*m1*m2/d^2

where G=universal gravitational constant=6.673*10^(-11)

the force is always attractive in nature and will be along the direction of the object exerting the force.

part A:

force due to 165 kg:

distance between 165 kg object and 41 kg object=0.39/2=0.195 m

then force magnitude=6.673*10^(-11)*165*41/(0.195^2)=1.1871*10^(-5) N

force will directed away from the 41 kg mass and towards the 165 kg mass.

force due to 465 kg:

distance between 465 kg mass and 41 kg object=0.39/2=0.195 m

then force magnitude=6.673*10^(-11)*465*41/0.195^2=3.3457*10^(-5) N

direction will be away from 41 kg object and towards the 465 kg object.

hence both the forces are opposing each other.

as force due to 465 kg has more magnitude, net force will be along the direction towards 465 kg.

net force magnitude=3.3457*10^(-5) -1.1871*10^(-5)=2.1586*10^(-5) N

part b:

as we saw that, in between the two objects, the forces are opposing each other.

hence let , at a distance of d meters from the 165 kg object, net force is 0.

==>G*165*41/d^2=G*465*41/(0.39-d)^2

==>165*(0.39-d)^2=465*d^2

==>(0.39-d)/d=sqrt(465/165)=1.6787

==>0.39-d=1.6787*d

==>0.39=2.6787*d

==>d=0.1456 m

hence at a distance of 0.1456 m from 165 kg object, net force will be zero.