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

F13 = Gm1m3/r^2 = 6.673*10^-11 * 0.3*0.3 / 0.3^2 = 6.673*10^-11 N in x direction F12 = Gm1*m2/r^2 = 6.673*10^-11 * 0.3*0.3 / 0.4^2 = 3.753*10^-11 N in y direction Net force = (6.673^2 + 3.753^2)^0.5 * 10^-11 N = 7.656 *10^-11 N Direction arctan(3.753/6.673) = 29.35 deg if weight is decreased by 10% than F13 = 6.673*10^-13 N and F12 = 3.753*10^-13 N and net will be 7.656*10^-13 N direction remains the same 29.35 deg

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