Three balls, with masses fo 4m, 2m, and m, aer equally spaced along a line. The

Question

Three balls, with masses fo 4m, 2m, and m, aer equally spaced along a line. The spacing between neighboring balls is r. We can arrange the balls in three different ways, as shown in the figure. In each case, the balls are in an isolated region of space very far from anything else. Rank teh three cases, from largest to smallest, based on teh magnitude of teh net gravitational force experienced by the ball with teh mass of 4m. (use only *>* or "="' symbols. Do not include any parentheses around the letters or symbols.) If m is 6.00 kg and r is 80.0 cm, calculate the magnitude and deirection of the net gravitational force acting on the ball with the mass of 4m in case Define the positive direction as pointing to the right, so your answer should have a positive sign if the net force is directed right, and a negative sign if the net force is directed left.

Explanation / Answer

gravitational force is given as G*m1*m2/d^2

where m1 and m2 are two masses and d is distance between them

G=universal gravitational constant=6.674*10^(-11)

the force is always attractive in nature.

set up 1:

force on 4m due to both 2m and m mass will be to the right and hence will be positive.

then total force=(G*4*m*2*m/r^2)+(G*4*m*m/(2*r)^2)

=(8*G*m^2/r^2)+(G*m^2/r^2)=9*G*m^2/r^2

set up 2:

force on 4m due to 2m is is along left direction and hence negative and due to m is along right direction and hence positive

net force=(-G*2*m*4*m/r^2)+(G*m*4*m/r^2)=-4*G*m^2/r^2

set up 3:

force on 4m due to both 2m and m mass will be to the right and hence will be positive.

then total force=(G*4*m*m/r^2)+(G*4*m*2*m/(2*r)^2)

=(4*G*m^2/r^2)+(2*G*m^2/r^2)=6*G*m^2/r^2

hence magnitude wise, set up 1>set up 3>set up 2

in terms of magnitude of net gravitational force on mass 4m

part b:

in set up 2, force=-4*G*m^2/r^2

=-4*6.674*10^(-11)*6^2/0.8^2=-1.50165*10^(-8) N

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