Three lead balls of mass m 1 = 16 kg, m 2 = 27 kg, and m 3 = 9.2 kg are arranged
ID: 2260068 • Letter: T
Question
Three lead balls of mass m1 = 16 kg, m2 = 27 kg, and m3 = 9.2 kg are arranged as shown in the figure below. Find the total gravitational force exerted by balls 1 and 2 on ball 3. Be sure to give the magnitude and the direction of this force.
magnitude in Newtons
Direction in degrees counter clockwise from the x-axis
Explanation / Answer
f=GMaMb/r^2
f1=G*16*9.2/( 4^2+0.5^2)= G*147.2/16.25
f2=G*27*9.2/( 1.5^2+1.5^2)= G*248.4/4.5
slope of line joining m1,m3= -0.5/3=-1/6
slope of line joining m2,m3= 1.5/1.5=1
angle between lines xtan x=(1-(-1/6))/(1+ 1*(-1/6)) =7/5x=54.46
net force=sqrt(f1^2 +f2^2 +f1f2 cos x)
=G*58.4786
=6.673*10^-11 *58.4786
=390*10^-11 N
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