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

$F=960lb$
We can find the required maximum force as follows: We know that the sum of forces in the y-direction is given as $\Sigma F_y=0$ $\Sigma F_y=-Fsin(30)+T_B(\frac{4}{5})=0$ $\Sigma F_y=-Fsin(30)+(600)(\frac{4}{5})=0$..eq(1) Now the sum of forces in the x-direction is given as $\Sigma F_x=-Fcos(30)+T_B(\frac{3}{5})=0$ $\Sigma F_x=-Fcos(30)+(600)(\frac{3}{5})=0$..eq(2) Solving eq(1) and eq(2), we obtain: $F=960lb$