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

the skeletons of most mammals reach a certain size and then stop growing. however, many animals, including some mammals, keep growing throughout their lives. kangaroos, for example, just keep growing and growing until they die. also, the longer an animal lives the greater the chance it might catch some deadly disease.

well clustered together.

points close on a scatter diagram