F = G M1 m/r^2 direction attractive
let M1 = 650. the mass of B
then G M1 = 6.67*10^-11 *650
= 4.34* 10^-8
so I will call
k = 4.34*10^-8 and do the arithmetic later
now each force = k m/r^2
now let's do forces in the X direction
no x force from A on B
x force C on B = k(1800/80^2) = -2812 k
x force D on B = k (100/.4^2) = +625 k
total x force Fx = - 2187 k
The only y force is from A in + y direction
Fy = k (100/.5^2) = k 100/.25 = 400 k
so
Fx = -2187(4.34*10^-8)
Fy = + 400 (4.34*10^-8)
You can multiply that out
if magnitude and direction are asked then
F = sqrt(Fx^2+Fy^2)
tan angle to - x axis in quadrant 2 = Fy/|Fx|
four uniform spheres, with masses MA= 100kg, MB= 650 kg, MC= 1800 kg, and MD = 100 kg, have (x,y) coordinates of (0, 50 cm), (O,O), (-80 cm, 0), and (40,0) respectively. What is the net gravitational force F2 on spere B due to the other spheres?
1 answer
1 answer