Two forces Fu and Fv are applied to a bracket. If theresultant R of the two forc

Question

Two forces Fu and Fv are applied to a bracket. If theresultant R of the two forces has a magnitude of 725 lb and adirection as shown on the figure, determine the magnitudes oftheforces Fu and Fv. Please show formulas and steps. This is a mechanics of materials course but no one in themechanical engineering can seem to answer the questions so I amhoping it is close enough to physics. Two forces Fu and Fv are applied to a bracket. If theresultant R of the two forces has a magnitude of 725 lb and adirection as shown on the figure, determine the magnitudes oftheforces Fu and Fv. Please show formulas and steps. This is a mechanics of materials course but no one in themechanical engineering can seem to answer the questions so I amhoping it is close enough to physics.

Explanation / Answer

angle made by Fu with positive x- axis = tan-1 ( 2 / 1)                                                      = 63.43 degrees So, vector Fu = Fu cos 63.43 i + Fu sin 63.43 j                      = 0.4472 Fu i + 0.8943 Fu j angle made by Fv with positive x-axis   =tan -1 ( 1/4)                                                      = 14.03 degrees So, vector Fv = Fv cos 14.03 i + Fv sin 14.03 j                      = 0.97 Fv i + 0.2424 Fv j angle made by R with negative x -axis ' = tan -1 ( 2 /3)                                                      = 33.69 degrees angle made by Fv with positive x-axis   =tan -1 ( 1/4)                                                      = 14.03 degrees So, vector Fv = Fv cos 14.03 i + Fv sin 14.03 j                      = 0.97 Fv i + 0.2424 Fv j angle made by R with negative x -axis ' = tan -1 ( 2 /3)                                                      = 33.69 degrees angle made by R with negative x -axis ' = tan -1 ( 2 /3)                                                      = 33.69 degrees So, resulatnt vector R = R cos 33.69 * -i + R sin 33.69j                                   = -725 cos 33.69 i + 725 sin 33.69 j                                   = - 603.23 i + 402.15 j we know R = Fu + Fv - 603.23 i + 402.15 j = 0.4472 Fu i + 0.8943 Fu j +0.97 Fv i + 0.2424 Fv j from this 0.4472 Fu + 0.97 Fv =-603.23      ---( 1)             0.8943Fu+0.2424 Fv = 402.15       ----(2) (1)*0.8943 - ( 2) * 0.4472 ===> 0 + (0.97Fv *0.8943) -(0.2424Fv * 0.4472 ) =(-603.23*0.8943) -(402.15*0.4472 )        0.759 Fv = -719.31 from this Fv = -947.7 lb plug it in eq( 1) we get 0.4472 Fu -919.27=-603.23 from this Fu = 706.72 lb

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