Three identical masses of 560 kg each are placed on the x axis. One mass is at x

Question

Three identical masses of 560 kg each are placed on thex axis. One mass is at x1 = -12.0 cm, oneis at the origin, and one is at x2 = 38.0 cm. a) What is the magnitude of the netgravitational force on the mass at the origin due to the other twomasses? Take the gravitational constant to beG = 6.67×1011 N*m^2/kg^2.
b) What is the direction of the netgravitational force on the mass at the origin due to the other twomasses? Three identical masses of 560 kg each are placed on thex axis. One mass is at x1 = -12.0 cm, oneis at the origin, and one is at x2 = 38.0 cm. a) What is the magnitude of the netgravitational force on the mass at the origin due to the other twomasses? Take the gravitational constant to beG = 6.67×1011 N*m^2/kg^2.
b) What is the direction of the netgravitational force on the mass at the origin due to the other twomasses? a) What is the magnitude of the netgravitational force on the mass at the origin due to the other twomasses? Take the gravitational constant to beG = 6.67×1011 N*m^2/kg^2.
b) What is the direction of the netgravitational force on the mass at the origin due to the other twomasses? b) What is the direction of the netgravitational force on the mass at the origin due to the other twomasses?

Explanation / Answer

a)the force between the mass at x1 = -12.0 cm and the mass atthe origin is F1 = G = 6.67 * 10^-11 Nm^2/kg^2 m = 560 kg r = 12.0 cm = 12.0 * 10^-2 m the force between the mass at the origin and the mass at x2 =38.0 cm = 38.0 * 10^-2 m is F2 = (G * m * m/r1^2) r1 = 38.0 cm = 38.0 * 10^-2 m the magnitude of the net graviational force on the mass at theorigin due to the other two masses is Fnet= (Fx^2 + Fy^2)^(1/2) ------------(1) Fx = F1x + F2x = F1 * cos1 + F2 * cos2 ad Fy = F1y + F2y = F1 * sin1 + F2 * sin2 here,1 = 2 = 0o or Fx = F1 * cos(0o) + F2 * cos(0o) = F1+ F2 ad Fy = F1 * sin(0o) + F2 *sin(0o) = 0 + 0 = 0 substituting the above values in equation (1) weget Fnet= ((F1 + F2)^2)^(1/2) = F1 + F2 b)the direction of the net gravitational force on the mass atthe origin due to the other two masses tan = (Fy/Fx) = (0/F1 + F2) = 0 or = tan-1(0) = 0o therefore,the direction of the net gravitational force isalong the x-axis (radial direction).

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