## Engineering Mechanics: Statics & Dynamics (14th Edition)

$C_x=4.97KN$ $D_x=2.23KN$ $D_y=7.2KN$
According to the principle of impulse and momentum $\Sigma M_A=\frac{dm}{dt}(d_{DB}v_B-d_{DA}v_A)$ We plug in the known values to obtain: $-C_x\times 2=600(0-(1.5-0.12)\times 12)$ $\implies C_x=4968N=4.97KN$ We know that the resultant force in the $x$-direction is given as $\Sigma F_x=\frac{dm}{dt}(v_B)_x-(v_A)_x$ $\implies D_x+C_x=600(12-0)$ This simplifies to: $D_x=2232N=2.23KN$ Now the resultant force in the $y$-direction is given as $\Sigma F_y=\frac{dm}{dt}[(v_B)_y-(v_A)_y]$ We plug in the known values to obtain: $D_y=600(0-(-12))$ This simplifies to: $D_y=7200N=7.2KN$