An equilateral triangle 97.0 m on a side has a m 1 = 45.00 kg mass at one corner

Question

An equilateral triangle 97.0 m on a side has a m1 = 45.00 kg mass at one corner, a m2 = 60.00 kg mass at another corner, and a m3 = 125.00 kg mass at the third corner. Find the magnitude and direction of the net force acting on the 45.00 kg mass.
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An equilateral triangle 97.0 m on a side has a m1 = 45.00 kg mass at one corner, a m2 = 60.00 kg mass at another corner, and a m3 = 125.00 kg mass at the third corner. Find the magnitude and direction of the net force acting on the 45.00 kg mass. N degree counterclockwise from the positive x-axis

Explanation / Answer

F21 = Gm1m2/r^2 = 6.67*10^-11*45*60/97^2 = 1.91*10^-11 N

F31 = Gm1m3/r^2 = 6.67*10^-11*45*125/97^2 = 3.98*10^-11 N

angle between the forces = 60 degrees

Let m3 and m2 lie on postive x-axis

F21 = 120 degrees, F31 = 60degrees with x-axis

F21 = -1.91cos60i + 1.91sin60j *10^-11

F31 = 3.98cos60i + 3.98sin60j *10^-11

net = 2.07i + 5.89j *10^-11

Magnitude = 10^-11*sqrt(2.07^2 + 5.89^2) = 6.243*10^-11 N

Angle = tan^-1( 5.89/2.07) = 70.633 degrees

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