# Forces23.pdf - M B =r F = L 2)i L 2 tan j Pi Wj = L 2 cfw_W...

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[ ] [ ] { } ( / 2) ( / 2) tan ( / 2) tan B L L P W L W P θ θ = × = × = + M r F i j i j k Example 3 . The structure shown is subjected to a force T acting at E along the line EF . Calculate the moment of T about points A and D. This example requires a lot more work. First we need to write down the force as a vector. We know the magnitude of the force is T , so we only need to work out its direction. Since the force acts along EF, the direction must be a unit vector pointing along EF . It’s not hard to see that the vector EF is 3 2 EF a a a = + i j k  We can divide by the length of EF ( 14 a ) to find a unit vector pointing in the correct direction ( 3 2 ) / 14 EF = + e i j k The force vector is ( 3 2 ) / 14 T = + F i j k Next, we need to write down the necessary position vectors Force: 2 3 a a = + r i j Point A: 2 A a = −
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