Four uniform spheres, with masses ma = 200 kg, mb = 250 kg, mc = 1700 kg, and md = 100 kg, have (x, y) coordinates of (0, 50 cm), (0, 0), (-80 cm, 0), and (40 cm, 0), respectively. what is the net gravitational force f2 on sphere b due to the other spheres?

in the anticlockwise direction from the positive x- axis.

Explanation:

Given are the masses of various spheres with their names in subscript:

Now their respective coordinate positions (in cm) are as given below:

Now force on B due to A:

Now force on B due to C:

Now force on B due to D:

We observe that the forces due to masses  C&D act opposite in direction.

So, the net force in the x-direction:

in the positive x-direction

We have only one force in y-direction due to mass A.

So,

in the positive y-direction.

Now the net force:

Now the direction of this force with respect to x-axis:

in the anticlockwise direction from the positive x- axis.

3.64 x 10^-5 N

Explanation:

mA = 200 kg

mB = 250 kg

mc = 1700 kg

mD = 100 kg

Force on B due to A

FA = 1.334 x 10^-5 N

Force on B due to C

FC = 4.43 x 10^-5 N

Force on B due to D

FD = 1.042 x 10^-5 N

FD and Fc are opposite to each other, so net of Fc and FD is F'.

F' = (4.43 - 1.042) x 10^-5 N towards left

F' = 3.39 x 10^-5 N

Now F' and FA are perpendicular to each other, So net force on B due to all other

F = 3.64 x 10^-5 N

Thus, the net force on B due to other is 3.64 x 10^-5 N.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2