A 590N force P is applied to a bent plate as shown Determine

A 590-N force P is applied to a bent plate as shown.

Determine an equivalent force-couple system at B. (Round the final answer to one decimal place.) (You must provide an answer before moving to the next part.)

M_B =  N•m (Clockwise)

Solution

>> As there is application of Force, P at an angle 30 with the vertical,

Horizontal Componen of Force, Px = P*sin30 = 590*0.5 = 295 N

and, Vertical Component of Force, Py = P*cos30 = 590*cos30 = 510.955 N

>> Now, Moment due to this at point B, Mb = [ (0.3 i ) X ( - Py j) ] + [ (- 0.175 j ) X ( Px i ) ]

=> Mb = [ (0.3 i ) X ( - 510.995 j) ] + [ (- 0.175 j ) X ( 295 i ) ]

=> Mb = - 153.286 k + 51.625 k

=> Mb = - 101.661 k

So, at B, a Force, P = 590 N , at an angle 30 with vertical will act with a moment of 101.661 N-m, clockwise..............ANSWER.........