Three balls, of mass m, 5m, and 3m, are at the vertices of an equilateral triangle with side lengths equal to L. What is the magnitude of the force on mass m from the two other masses, in terms of G, m, and L
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Ответ:
F_{total}= 8.25
, θ’= 194º
Explanation:
To solve this problem we must use the law of universal gravitation and vectorly add the forces
F =![G \frac{m_{1} m_{2} }{r^{2} }](/tpl/images/1027/5081/5e32b.png)
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 =![\frac{Fx_{15y} }{F_{15} }](/tpl/images/1027/5081/53195.png)
F₁₅ₓ = F₁₅ cos 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
F_{total}= 8.25![\frac{G m^2}{L^2}](/tpl/images/1027/5081/736ed.png)
with trigonometry
tan θ =![\frac{F_{total y} }{F_{total x} }](/tpl/images/1027/5081/aa874.png)
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
Ответ:
nucleas eats?
Explanation: