Answer:
F_{total}= 8.25 [tex]\frac{G m^2}{L^2}[/tex], θ’= 194º
Explanation:
To solve this problem we must use the law of universal gravitation and vectorly add the forces
F = [tex]G \frac{m_{1} m_{2} }{r^{2} }[/tex]
Let us call the mass m with the subscript 1, the mass 3m with the subscript 3 and the mass 5m with the subscript 5, the total force on particle 1 is
F_total = F₁₅ + F₁₃
The bold are vectors, in the attachment we can see a diagram of the angles and the forces, the distance between the masses is
r = L
let's find the force between m1 and m5
F₁₅ = G m₁ m₅ / r²
F₁₅ = G m 5m / L²
F₁₅ = G 5m² / L²
this force is on the line that joins the two masses, let's use trigonometry to decompose this force
cos 30 = F₁₅ₓ / F₁₅
sin 30 = [tex]\frac{Fx_{15y} }{F_{15} }[/tex]
F₁₅ₓ = F₁₅ cos 30
[tex]F_{15y}[/tex] = F₁₅ sin 30
equally with the force between mass 1 and mass 3
F₁₃ = -G 3 m² / L²
F₁₃ₓ = F₁₃ cos 30
F_{13y} = F₁₃ sin 30
to find the total force we can add each component independently, see attached
X axis
F_total x = -F₁₅ₓ + F₁₃ₓ
F_total x = -G 5m2 / L² + G 3m² / L²
F_total x = - G 2m² / L²
Y axis
F_total y = - F_{15y} - F_{13y}
F_total y = - G 5m² / L² - G 3 m² / L²
F_toal y = - G 8 m² / L²
We can give the result in two ways
1) F_total = - G m ^ 2 / L² (2 i ^ + 8 j ^)
2) in the form of module and angle.
Let's use the Pythagorean theorem
[tex]F_{total}^{2} = F_{total x}^{2} + F_{total y}^2[/tex]
[tex]F_{total}[/tex] = [tex]\frac{G m^{2} }{L^2} \sqrt{(2^2 + 8^2)}[/tex]
F_{total}= 8.25 [tex]\frac{G m^2}{L^2}[/tex]
with trigonometry
tan θ = [tex]\frac{F_{total y} }{F_{total x} }[/tex]
tan θ = 8/2
θ = tan⁻¹ 4
θ = 76º
if we measure this angle from the positive side of the x axis in a counterclockwise direction
θ’= 270 -76
θ’= 194º
