Forces28.pdf - Similarly the perpendicular distance to B is...

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Similarly, the perpendicular distance to B is L/2 , and W exerts a counterclockwise moment about B. Therefore ( / 2) B LW = M k Example 2 . Member AB of a roof-truss is subjected to a vertical gravitational force W and a horizontal wind load P . Calculate the moment of the resultant force about B. The perpendicular distance from the line of action of W to B is L/2 . W exerts a counterclockwise moment about B. Therefore W exerts a moment ( / 2) B L W = M k The perpendicular distance from the line of action of P to B is ( / 2) tan L θ . P also exerts a counterclockwise moment about B. Therefore ( tan / 2) B L P θ = M k The total moment is { } ( / 2) tan B L W P θ = + M k Example 3 : It is traditional in elementary statics courses to solve lots of problems involving ladders (oh boy! Aren’t you glad you signed up for engineering?) . The picture below shows a ladder of length L and weight W resting on the top of a frictionless wall. Forces acting on the ladder are shown as well. Calculate the moments about point
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