Three 0.300-kg billiard balls are placed on a table at the corners of a right tr

Question

Three 0.300-kg billiard balls are placed on a table at the corners of a right triangle as shown in the figure. The sides of the triangle are of lengthsa= 0.400 m,b= 0.300 m, andc= 0.500 m. Calculate the gravitational force vector on the cue ball (designatedm1) resulting from the other two balls as well as the magnitude and direction of this force.

What if we replaced all the three balls with smaller ones that are10%lighter? Find the magnitude and direction (angle between -180? and 180?) of the force exerted bym1andm3onm2.

Explanation / Answer

F31 = Gm*m/b^2 = (G * .3^2)/.3^2 = G F21 = Gm*m/a^2 = (G * .3^2)/.4^2 = 9G/16 Force Vector = (6.67*10^-11)i + (3.75*10^-11)j Direction = tan (theta) = (3.75*10^-11)/(6.67*10^-11) = 29.35 degrees SO direction is 29.35 degrees from the horizontal. You can proceed the same way. If you are stuck somewhere please let me know. I can tell you some method you have to actually divide the force component F23 into horizontal and vertical axis by using F23 cos (theta') + F23 sin(theta') and then proceed. You get in the coordinate axis and then it becomes easy to solve. If you face problem let me know.

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