PLEASE NOTE!!! PART A IS CORRECT!! IF YOU DO NOT GET SAME ANSWER FOR PART A THEN
ID: 1505894 • Letter: P
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PLEASE NOTE!!! PART A IS CORRECT!! IF YOU DO NOT GET SAME ANSWER FOR PART A THEN YOU ARE WRONG! STOP GIVING ME WRONG ANSWERS!!
Problem 13.26 Part A The figure shows three masses What is the magnitude of the net gravitational force on the m1 = 25kg mass? Assume m2 = 10kg and m3 = 10kg. | F= 1.72× 10-6 N Submit My Answers Give Up CORRECT Part B What is the direction of the net gravitational force on the m 25kg mass? Assume m2 10kg and m3 10kg 1403 14.03 Figure 1 ofl cw of the positive y-axis Submit My Answers Give Up m2 Incorrect; Try Again; 5 attempts remaining 20 cm Part C What is the magnitude of the net gravitational force on the m2 = 10kg mass? Assume m1-25kg and m3 = 10kg mi 10 cm F= 73055.10 11Explanation / Answer
A) Gravitational force F12=Gm1m2/R12^2
F12=6.673x10^-11 * 25 * 10 / 0.2^2
F12=4.17*10^-7 N in positive y direction
F13=Gm1m3/R13^2
F13=6.673x10^-11 * 25 * 10 / 0.1^2
F13=16.7*10^-7 N in positive x direction
Net Force F = sqrt(F12^2 + F13^2) = 17.2*10^-7 N = 1.72*10^-6 N
B) Direction of the force = tan^-1 (F12/F13) = 14.03 degrees with the x axis
C)
Gravitational force F21=Gm1m2/R12^2
F21=6.673x10^-11 * 25 * 10 / 0.2^2
F21=4.17*10^-7 N in negative y direction
F23=Gm1m3/R13^2
R13 = sqrt(0.2^2 + 0.1^2) = 0.223 m
F23=6.673x10^-11 * 10 * 10 / 0.223^2
F23=1.34*10^-7 N in the direction line connecting m2 and m3
Net Force F = sqrt(F21^2 + F23^2 + 2*F21*F23*cos())
Where is the angle between F21 and F23
= tan^-1 (0.1/0.2) = 26.565 degrees
F =sqrt(4.17*10^-7^2 + 1.34*10^-7^2 + 2*4.17*10^-7*1.34*10^-7*cos(26.565)) = 5.4*10^-7 N
D) Direction of the force
Get the x and y components of each
F21x = 0
F21y = 4.17*10^-7 N in negative y direction
F23x = 1.34*10^-7 * sin(26.565) = 0.6*10^-7 N in positive x direction
F23y = 1.34*10^-7 * cos(26.565) = 1.198*10^-7 N in negative y direction
Add them up
Fx = 0.6*10^-7
Fy = 5.368*10^-7
= arctan (0.6*10^-7 / 5.368*10^-7) = 6.37º
This is the CW angle with –y axis. Converting to CCW from the +x axis,
= 270º + 6.37º = 276.37º from the +x axis
= 90º - 6.37º = 83.63º from the -x axis
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