Three point masses, m_1 = 150 kg, m_2 = 185 kg and m_3 = 278 kg are located in t

Question

Three point masses, m_1 = 150 kg, m_2 = 185 kg and m_3 = 278 kg are located in the x-y plane. The coordinates of the masses are the following: m_1 (-3.0 m, -6,0 m), m_2 (+4.0 m, +8.0 m) and m_3 (+5.0 m, -3.0 m). Determine the magnitude of the net gravitational force on m_3 due to the other two masses. Give your answer in the form "a.bc Times 10^(x)" N. Using the information in Problem #5a determine the angle of the net force on m_3. Give your answer as if you were moving clockwise from the + x axis. The format of your answer is "abc" degrees,

Explanation / Answer

Gravitational force between m1 and m2 is,

F1=G m1 m2/r^2
=6.67x10^-11 x150x278 /(8.54)^2 [Distance,r= sqrt[(-3-5)^2 + (-6+3)^2] = 8.54m]
=3.81x10^-8 N

Gravitational force between m2 and m3,
F2=G m2 m3/r^2
=6.67x10^-11 x185x275 /(11.045)^2 [Distance,r= sqrt[(4-5)^2 +(8+3)^2] = 11.045m]
=2.81x10^-8 N

Net gravitational force,F= (F1^2 + F2^2)
= ( (3.81x10^-8)^2 + (2.81x10^-8)^2 ) N
=4.73x10^-8 N

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