Question

480XN 0 mm 120 mm For the system shown, P has a magnitude of 500 N 1. Calculate the moment of P about O 2. Calculate the moment of P about B

Answer 1

12500N x , 480 D 200mm - 1400 120mm

We can resolve the force P as horizontal and vertical forces. Also we can create triangles BB'O and AA'O as shown in the figure.

500 Cos30 500 Sin 30 48

Consider triangle BB'O,

Angle at O is 48˚

B’O = BO cos(48˚)

        = 120 cos(48˚)

        = 80.3 mm

BB’ = BO sin(48˚)

        = 120 sin(48˚)

        = 89.2 mm

Consider triangle AA’O,

Angle at O is 40˚

A’O = OA cos(40˚)

        = 200 cos(40˚)

        = 153.21 mm

AA’ = OA sin(40˚)

        = 200 sin(40˚)

        = 128.56 mm

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1)   Calculation of moment of P with respect to O

Moment of P with respect to O = ( 500 cos(30˚) × A’O ) - (500 sin(30˚) × AA’)

                                                   = ( 500 cos(30˚) × 153.21 ) - (500 sin(30˚) × 128.56)

                                                   = 34201.9 Nmm ( anti-clockwise)

                                                             or

                                                   = 34.202 Nm ( anti-clockwise)

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2)   Calculation of moment of P with respect to B

Moment of P with respect to B = ( 500 cos(30˚) × (A’O + B’O)) - (500 sin(30˚) × (AA’ – BB’)

                                                   = ( 500 cos(30˚) × (153.21 + 80.3) ) - (500 sin(30˚) × (128.56 – 89.2))

                                                   = 91272.8 Nmm ( anti-clockwise)

                                                       or   91.3 Nm

                                                   =   91.3 Nm ( anti-clockwise)

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