F232 lb F1 15 lb Y 28 65 40 in 13 in 31 in Solutionthe momen

F2=32 lb F1= 15 lb Y- 28 65 4.0 in 1.3 in 3.1 in

Solution

the moment of F1 about point A

moment = r * F

co-ordinate system

leftward - x-axis

upward - y-axis

F1 = 15 (sin 28 i + cos 28 j)

= 7.0420 i+ 13.24421j

r = position vector of point of application with respect to point about which moment is being calculated.

= (-3.1 -1.3) i + 4 j

= -4.4 i + 4 j

moment = (-4.4 i + 4 j)*(7.0420 i+ 13.24421j)

= (-4.4*13.2442-4*7.042) k

=-86.44 k (lb-inch)

similarly the moment of F2 about point A can be calculated

 F2=32 lb F1= 15 lb Y- 28 65 4.0 in 1.3 in 3.1 in Solutionthe moment of F1 about point A moment = r * F co-ordinate system leftward - x-axis upward - y-axis F1

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