Objects with masses of 155 kg and a 455 kg are separated by 0.300 m. (a) Find the net gravitational force exerted by these objects on a 69.0 kg object placed midway between them. magnitude N direction ---Select--- (b) At what position (other than infinitely remote ones) can the 69.0 kg object be placed so as to experience a net force of zero

a)   F_total = 6.15 10-5 N

b)    r₁ = 0.3045 m,    r₂ = 0.0051 m

Explanation:

In this exercise we must use that the total force is the sum of the forces on the body, remember that the gravitational force is attractive

      ∑un F = F₁₂ - F₃₂

where F₁₂2 is the force between the body of M1 = 155 kg, placed to the right and the body of m = 69 kg; the force F₃₂ is the force between the body of M3 = 455 kg, located on the left and the body of 69 kg

    these forces are gravitational forces, which is described by the expression

       F = G M m / r²

let's write each force

       F₁₂= G M₁ m / r²

leather m is at the midpoint of large bodies

 F₁₂= 6.67 10⁻¹¹ 155 69 / 0.15²

 F₁₂ = 3.17 10⁻⁵ N

 F₃₂ = 6.67 10⁻¹¹ 455 69 / 0.15 2

  F₃₂= 9.31 10⁻⁵ N

the net force is

      F_total = (3,17 - 9,31) 10-5

      F_total = 6.15 10-5 N

This force is directed to the left

b) in this case it asks us to know where we place the body so that the force is zero

            F12 = F32

the object is at a distance r from the mass1 and at a distance (0.3 - r), we avoid the equation

              F12 = G M1 m / r²

              F32 = G M3 m / (0,3-R) 2

             G M1 m / r2 = G M3 m / (0,3-r) 2

              M1 / r2 = M2 / (0,3-r) 2

              (0,3 -r) 2 M1 / M2 = r2

               0.32 - 2 r 0.3 + r2) = M2 / M1r2

                (0.09 - 0.6 r + r2) = 455/155 r2 = 2,935 r2

                -0.09 + 0.6r + 1.935 r2 = 0

we simplify the expression

               r2 +0.31 r - 0.0465 = 0

let's solve the quadratic equation

               r = [0.31 + - RA (0.31 2 -4 0.0465)] / 2

               r = [0.31 + - RA (0.0899] / 2

               r = [0.31 + - 0.2998] / 2

               r₁ = 0.3045 m

               r₂ = 0.0051 m

newton's first law of motion