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(4) 8i+5j-2k22.Moment of a force of magnitude 20 N acting along positive x direction at point (3m, 0, 0) about the point (0.2,0(in Nm) is(1) 20(3) 40(2) 60(4) 3023. A flywheel of moment of inertia 2 kn-m2 is rntatodl nf |
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Answer» By the definition of torque,г = r x FGiven points are(3m,0,0) and(0,2,0)r = (0 - 3m) i + (2-0) j + (0-0) kr = -3m i + 2 jNow,г = (-3m i + 2 j) x (20 i)г = (0-0) i - (0-0) j + (0-40) kг = - 40 kΙ гΙ = 40 N |
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