Help
with #19 pleaseHomework Answers
The unit vectors for the forces are the direction cosines. The position vector of the force F is
rOF = -1.333k ft.
The magnitude of the moment due to F is IMOFI = ex .(rOF x F) = 1.230F ft lb
The magnitude of the moment due to –F is IM-OFI = ex .(r-OF x -F) = 1.230F ft lb
The total moment about x-axis is Mx = 1.230 i^ + 1.230 i^ = 2.46 i^
From which, for a total magnitude of 32 ft lb, the force to be applied is
F = 32/2.46 = 13 lb
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Help
with #19 please
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Can somebody please help me with these 2 problems? Thanks so
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please help
A2 kN force acts on point A of rod AB as shown. Rod ABS 305 mm long. The angle a 40° Determine the magnitude in newtons and direction in degrees counterclockwise from the ex-axis of the smallest force that will create the same moment about point B as does the 2 KN magnitude 435.23 TXN counterclockwise from the director x-axis Forces F, and F, act as shown at points B and C, respectively. u F = 440 N...
please help solve for what is asked in orobkems A-H and the
free body diagrams as well. Will upvote if correct
A
B
C
D
E
F
G
H
B 4 m 6 m 3 m 12 m 2 m F=80 N Determine the projected component of the force acting along the axis AB of the pipe. 60° A The 30kg pipe is supported at A by a system of five cords. Determine the force in each cord for equilibrium....
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Can somebody please help me with these 2 problems? Thanks so
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Calculate the moment about a point CC – 4, 5, 7) ft caused by the forces F1 = – 4i + 4% – 8k lb and F, = – 3i – 7j + 7k lb acting at the points AC – 2, – 4, – 1) ft and B(2, 2,...
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Can you please help me find the distance of z with the procedure
that I have started. If you can please show all work with clear
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