The two wires are tied to the pole at B and D Find the momen

The two wires are tied to the pole at B and D. Find the moment about point A due to the three given forces. Use force times perpendicular distance (see video 3-13). Answer(s):

Solution

Line of action of the force F₁ = 132 N is DE and it given by

DE = AE - AD = (2150 i + 1000 k) - (1800 j) mm

===> DE = (2150 i - 1800 j + 1000 k) mm

Thus force F₁ = 132 N can be written as

F₁ = 132 / {2150² + (-1800)² + 1000²} * {2150 i - 1800 j + 1000 k} N

===> F₁ = 132 / 297 * {2150 i - 1800 j + 1000 k} N

Similarly for the force F₂ = 185 N, the line of action is BC which is given by

BC = AC - AB = {1860 i - 2300 j - 560 k} mm

Force F₂ = F₂ * BC / BC = 185 / {1860² + (-2300)² + (-560)²} * {1860 i - 2300 j - 560 k} N

===> F₂ = 185 / 3010.515 * {1860 i - 2300 j - 560 k} N

Force F₃ = - 540 i N

Resultant moment of all forces about point A is

M_A = AD x F₁ + AB x F₂ + AB x F₃

===> M_A = 1.8 j x [132 / 297 * {2150 i - 1800 j + 1000 k}] + 2.3 j x [185 / 3010.515 * {1860 i - 2300 j - 560 k}] + {2.3 j x -540 i} Nm

===> M_A = {79.812 i - 171.596 k - 79.15 i - 262.889 k + 1242 k} Nm

===> M_A = {0.663 i + 807.516 k} Nm