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The tube supports the four parallel forces. Determine the magnitudes of forces { F }_{ C } and { F }_{ D } acting at C and D so that the equivalent resultant force of the force system acts through the midpoint O of the tube.

Step-by-step

Since the resultant force passes through point O, the resultant moment components
about x and y axes are both zero.

\sum { M}_{ x } = 0 ;    { F }_{ D }(0.4) + 600(0.4) – { F }_{ C }(0.4) – 500(0.4) = 0

{ F }_{ C }{ F }_{ D } = 100

\sum { M}_{ y } = 0 ;    500(0.2) + 600(0.2) – { F }_{ C }(0.2) – { F }_{ D }(0.2) = 0

{ F }_{ C } + { F }_{ D } = 1100

Solving Eqs. (1) and (2) yields:

{ F }_{ C } = 600 N              { F }_{ D } = 500 N

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