Four 2.0 kg masses are located at the corners of a square 0.60 m on each side.?

Question

Find the magnitude and direction of the gravitational force on a fifth 2.0 kg mass placed at the midpoint of the bottom side of the square. N

Thanks for your help!

Answer 1

The 5th mass isn't in the center of the square but in the middle of one edge. So only two g forces cancel.
Each of the remaining forces F1, F2 acts at a distance |r| = sqrt(0.6^2+0.3^2) and is directed along each corresponding r. The resultant F = vector sum F1+F2 = (|F1|+|F2|)*0.6/r upwards
|F1| = |F2| = Gm^2/r^2

Answer 2

If the square is of negligible thickness and all the five masses are point masses and in same plane, then the fifth mass will experience equal foces due to masses on corners of that side

The pairs of forces due to masses on opposite corners of that side will cancel

The forces due to the other two corners will be equal and making angle s of 45 degree with the side.

Force due to mass on one corner = Gmm/r^2

m=2kg

r= sq rt[0.6^2+0.3^2]= sq rt 0.45 =0.6708 m

r^2=0.45

Force due to mass on one corner 'f' =6.67*10^ -11*2*2 / 0.45

Force due to mass on one corner'f' =5.93*10^-10 N

The force towards other upper is same.

We find resultant force on fifth mass by resolving the forces due to upper corners

the components along the side cancel

the upward components 'fsinO' add to give

resultant force F= 2fsinO



resultant force F=2*5.93* (0.8944.)*10^-10

resultant force F=10.698*10^-10 N upwards

Answer 3

I fail to grasp the question, I think eveyone else has the same problem.