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The total torque about pivot A provided by the forces shown in figure, for L=3.0m is |
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Answer» Moment of force about pivot A (80 N force) =80×3/2×sin30o = 60 N-m (anticlockwise) Moment of force about pivot A (70 N force) =70×3×sin30o=70×3×1/2 = 105 N-m (anticlockwise) Moment of force about pivot A (60 N force) =60×3/2×sin90o = 90 N-m (clockwise) Moment of force about pivot A (90 N force) =90×0×sin60o=0 Moment of force about pivot A (50 N) =50×3×sin180o=0 The total torque about pivot A T=(60+105−90) = 75 N-m |
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