Answer:
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
