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
[tex]\sum M__A }=0[/tex]
- 40 (180) - (80) [tex]F_E[/tex] Cos 37° + 55
-7200 + [tex]F_E[/tex] (-80 Cos 37° + 55 sin 37° ) = 0
= - 7200 - 30.79 [tex]F_E[/tex]
- 30.79 [tex]F_E[/tex] = 7200
[tex]F_E[/tex] = [tex]-\frac{7200}{30.79}[/tex]
[tex]F_E[/tex] = -233.84 lb
Thus, the force supported by the bar = -233.84 lb
Taking the equilibrium of forces in the vertical direction
[tex]\sum f_y = 0[/tex]
[tex]F_E[/tex] sin 37 + [tex]F_A[/tex] cos 20 - [tex]F_G[/tex] = 0
- 233.84 sin 37 + [tex]F_A[/tex] cos 20 - 180 = 0
[tex]F_A[/tex] cos 20 = 320.73
[tex]F_A[/tex] = [tex]\frac{320.73}{cos 20}[/tex]
[tex]F_A[/tex] = 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.
[tex]\sum f_y = 0[/tex]
[tex]F_E[/tex] cos 37 - [tex]F_{CB} + F_A[/tex] sin 20° = 0
-233.84 cos 37 - [tex]F_{CB}[/tex] + 341.31 sin 20° = 0
-186.75 - [tex]F_{CB}[/tex] + 116.74 = 0
- [tex]F_{CB}[/tex] -70.01 = 0
- [tex]F_{CB}[/tex] = 70.01
[tex]F_{CB}[/tex] = - 70.01 lb
Therefore, the force supported by the hydraulic cylinder is = -70.01 lb
