Question

A 225-kg object and a 525-kg object are separated by 3.30 m.

(a) Find the magnitude of the net gravitational force exerted by these objects on a 38.0-kg object placed midway between them.

(b)At what position (other than an infinitely remote one) can the 38.0-kg object be placed so as to experience a net force of zero from the other two objects?
______________ m from the 525 kg mass toward the 225 kg mass

Answer 1

Part A)

F = Gmm/r² - Gmm/r²

F = (6.67 X 10^-11)(525)(38)/(1.65)² - (6.67 X 10^-11)(225)(38)/(1.65)²

F = 2.79 X 10^-7 N (.000000279 N)

Part B)

In this case, Gmm/r² will need to equal each other and G and the 38 kg mass will cancel leaving

m/r² = m/r²

225/x² = 525/(3.3 - x)²

Cross multiply

225(3.3 - x)² = 525x²

2450.25 - 1485x + 225x² = 525x²

In standard form, that is 300x² + 1485x - 2450.25 = 0

Find the roots using the quadratic equation

The roots are...1.30 and -6.26

The negative root can be eliminated since that distance is outside the masses

Thus the distance is 1.30 m from the 225 kg mass

Thus 3.3 = 1.3 = 2.0 m from the 525 kg mass

Answer 2

(225 kg) ------1.15 m------ (38 kg) ------1.15 m------ (525 kg)

(a) For the Left Half:

(225 kg) ------1.15 m------ (38 kg)

F = G M m / d^2
= 4.312E-7 N
For the Right Half:

(38 kg) --6----1.15 m------ (525 kg)

F = G N m / d^2
=1.00617E-6 N

The difference, or net force, is 5.74973E-7 N.

b).

(225 kg) ------3.3-x m------ (38 kg) ------x m------ (525 kg)

F-f =0

F - force exerted by 225 on 38kg

and f - force exerted by 525 on 38kg

G *38*525 / x^2 = G * 38*225 / (3.3-x)^2

(3.3-x)^2 / x^2 = 225 /525

(10.89 +x^2 -6.6x ) *525 = 225*x^2

5717.25 + 525x^2 -3465x = 225x^2

300x^2 -3465x + 5717.25 = 0

x1=1.99437m from 525kg towards 225kg

Answer 3

Grav force = Gm1m2/r2

So net force = F1-F2

=4.9* 10-7 -2.1 * 10-7 = 2.8 * 10-7 N

B) F1= F2

525/R2 = 225/ (3.3-R) 2

R= 2m approx from 525 kg