Question

The masses and coordinates of three particles are as follows: 20.0 kg, (0.50, 1.00) m; 361.0 kg, (-1.00, -1.00) m; 54.0 kg, (0.00, -0.50) m. What is the gravitational force on a 20.0 kg sphere located at the origin due to the other spheres, magnitude and direction? Give the direction as an angle in degrees counter clockwise with respect to the the + x-axis.

Answer 1

let
m1 = 20 kg at (0.50,1.00)m
m2 = 361 kg, at (-1.00, -1.00) m
m3 = 54 kg, at (0.00, -0.50 )m
m = 20.0 kg at origin

magnitude of force exerted by m1 on m, F1 = G*m1*m/r1^2

= 6.67*10^-7*20*20/(0.5^2 + 1^2)

= 2.13*10^-4 N

direction of F1: theta1 = tan^-1(1/0.5)

= 63.4 degrees above +x axis

magnitude of force exerted by m2 on m, F2 = G*m2*m/r2^2

= 6.67*10^-7*361*20/(1^2 + 1^2)

= 2.41*10^-3 N

direction of F2: theta1 = tan^-1(-1/-1)

= 45 degrees below -x axis

magnitude of force exerted by m3 on m, F3 = G*m3*m/r3^2

= 6.67*10^-7*54*20/0.5^2

= 2.88*10^-3 N

direction of F3: theta3 : along -y axis

x-component of net force, Fnetx = F1x + F2x + F3x

= 2.13*10^-4*cos(63.4) + (-2.41*10^-3)*cos(45) + 0

= -1.61*10^-3 N

y-component of net force, Fnety = F1y + F2y + F3y

= 2.13*10^-4*sin(63.4) + (-2.41*10^-3)*sin(45) + (-2.88*10^-3)

= -4.40*10^-3 N

magnitude of net force, Fnet = sqrt(Fnetx^2 + Fnety^2)

= sqrt(1.61^2 + 4.4^2)*10^-3

= 4.68*10^-3 N <<<<<<<<<<<-------------------------Answer

direction : theta = tan^-1(Fnety/Fnetx)

= tan^-1((-4.4)/(-1.61))

= 70.0 degrees below -x axis

= 250 degrees with +x axis in counter clocwise direction <<<<<<<<<<<--------------------Answer