Two spheres of mass M_1 = 770 kg and M_2 = 330 kg are placed 5.20 m apart. A par

Question

Two spheres of mass M_1 = 770 kg and M_2 = 330 kg are placed 5.20 m apart. A particle of mass m = 20.0 kg is now placed midway between the two spheres, what is the net gravitational force on the particle due to the two spheres? Your response differs from the correct answer by more than 100%. At what position between the two spheres should the particle be placed so that the net gravitational force on the particle is zero? m from the sphere of mass M_1 A worker is attempting to lift a 48.0-kg palette of bricks resting on the ground by means of a rope attached to a pulley. (Assume the rope goes from the worker, directly above to the pulley and then directly down to the palette of bricks.) Before the worker pulls on the rope, what is the force exerted by the ground on the palette? (Enter your answer to at least the nearest digit.) magnitude direction the worker exerts a force of 321 N downward on his end of the rope. What is the force exerted by the ground on the palette? magnitude 263 Your response differs from the correct answer by more than 10%. Double check your calculations. N direction upward If the worker doubles the downward force, what is the force exerted by the ground on the palette? magnitude direction the magnitude is zero.

Explanation / Answer

Fnet = F1 - F2

Fnet = G*M1*m/r1^2 - G*M2/m/r2^2

r1 = r2 = 2.6 m

Fnet = 6.67*10^-11*770*20/2.6^2 - 6.67*10^-11*330*20/2.6^2

Fnet = 8.68E-8 N <<<<<<<<-----------answer

part(b)

F1 = F2

G*M1*m/r1^2 = G*M2*m/(5.2-r1^2)

M1/r1^2 = M2/(5.2-r1)^2

770/r1^2 = 330/(5.2-r1)^2

r1 = 3.14 m <<<<-------answer

=======================

(b)

along vertical

Fnet = 0

n - 480 + 321 = 0

n = 159 N   <<-----answer

upward

(c)

n - 480 + (2*321) = 0

n = -162 N

n cannot be negative

so force exerted by grond = 0 <<<-----answer

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