Question

A steel plate weighing 180 lb with center of gravity at point G is supported by a roller at point A, a bar DE, and a horizontal hydraulic cylinder BC. Neglect all weights of all members except for the plate. Determine the force supported by the hydraulic cylinder, the bar, and the reaction at the roller.

Answer 1

Answer:

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

Thus, the force supported by the bar = -233.84 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Explanation:

The attached diagrams is shown in the file below:

From there; taking moments about point A

$\sum M__A }=0$

- 40 (180) - (80) $F_E$ Cos 37° + 55  

-7200 + $F_E$ (-80 Cos 37° + 55 sin 37° ) = 0

= - 7200 - 30.79  $F_E$

- 30.79  $F_E$ = 7200

$F_E$ = $-\frac{7200}{30.79}$

$F_E$  = -233.84 lb

Thus, the force supported by the bar = -233.84 lb

Taking the equilibrium of forces in the vertical direction

$\sum f_y = 0$

$F_E$ sin 37 + $F_A$ cos 20 - $F_G$ = 0

- 233.84 sin 37 + $F_A$ cos 20 - 180 = 0

$F_A$ cos 20  = 320.73

$F_A$ = $\frac{320.73}{cos 20}$

$F_A$ = 341.31 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Taking the equilibrium of forces on the horizontal direction.

$\sum f_y = 0$

$F_E$ cos 37 - $F_{CB} + F_A$ sin 20° = 0

-233.84 cos 37 - $F_{CB}$ + 341.31 sin 20° = 0  

-186.75 -  $F_{CB}$ + 116.74 = 0

-  $F_{CB}$ -70.01 = 0

-  $F_{CB}$ = 70.01

$F_{CB}$ = - 70.01 lb

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb