Answer:
Fc = [ - 4.45 * 10^-8 j ] N
Explanation:
Given:-
- The masses and the position coordinates from ( 0 , 0 ) are:
Sphere A : ma = 80 kg , ( 0 , 0 )
Sphere B : ma = 60 kg , ( 0.25 , 0 )
Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15
- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2
Find:-
what is the gravitational force on C due to A and B?
Solution:-
- The gravitational force between spheres is given by:
F = G*m1*m2 / r^2
Where, r : The distance between two bodies (sphere).
- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-
Determine the angle (α) between vectors rac and rab using cosine rule:
![cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}](https://tex.z-dn.net/?f=cos%20%28%20%5Calpha%20%29%20%3D%20%5Cfrac%7Brab%5E2%20%2B%20rac%5E2%20-%20rbc%5E2%7D%7B2%2Arab%2Arac%7D%20%5C%5C%5C%5Ccos%20%28%20%5Calpha%20%29%20%3D%20%5Cfrac%7B0.25%5E2%20%2B%200.2%5E2%20-%200.15%5E2%7D%7B2%2A0.25%2A0.2%7D%5C%5C%5C%5Ccos%20%28%20%5Calpha%20%29%20%3D%200.8%5C%5C%5C%5C%5Calpha%20%3D%2036.87%5E%7B%5Ccirc%20%5C%3A%7D)
Determine the angle (β) between vectors rbc and rab using cosine rule:
![cos ( \beta ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta ) = 0.6\\\\\beta = 53.13^{\circ \:}](https://tex.z-dn.net/?f=cos%20%28%20%5Cbeta%20%20%29%20%3D%20%5Cfrac%7Brab%5E2%20%2B%20rbc%5E2%20-%20rac%5E2%7D%7B2%2Arab%2Arbc%7D%20%5C%5C%5C%5Ccos%20%28%20%5Cbeta%20%20%29%20%3D%20%5Cfrac%7B0.25%5E2%20%2B%200.15%5E2%20-%200.2%5E2%7D%7B2%2A0.25%2A0.15%7D%5C%5C%5C%5Ccos%20%28%20%5Cbeta%20%20%29%20%3D%200.6%5C%5C%5C%5C%5Cbeta%20%20%3D%2053.13%5E%7B%5Ccirc%20%5C%3A%7D)
- Now determine the scalar gravitational forces due to sphere A and B on C:
Between sphere A and C:
Fac = G*ma*mc / rac^2
Fac = (6.674×10−11)*80*0.2 / 0.2^2
Fac = 2.67*10^-8 N
vector Fac = Fac* [ - cos (α) i + - sin (α) j ]
vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]
vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N
Between sphere B and C:
Fbc = G*mb*mc / rbc^2
Fbc = (6.674×10−11)*60*0.2 / 0.15^2
Fbc = 3.56*10^-8 N
vector Fbc = Fbc* [ cos (β) i - sin (β) j ]
vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]
vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N
- The Net gravitational force can now be determined from vector additon of Fac and Fbc:
Fc = vector Fac + vector Fbc
Fc = [ - 2.136 i - 1.602 j ]*10^-8 + [ 2.136 i - 2.848 j ]*10^-8
Fc = [ - 4.45 * 10^-8 j ] N