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*6–108. The skid steer loader has a mass of 1.18 Mg, and in the position shown the center of mass is at G1. If there is a 300-kg stone in the bucket, with center of mass at G2, determine the reactions of each pair of wheels A and B on the ground and the force in the hydraulic cylinder CD and at the pin E.There is a similar linkage on each side of the loader.

1.25 m

D G2

SOLUTION

E

30

C G1

Entire system: a + ©MA = 0;

0.5 m

300 (9.81)(1.5)-1.18 (103)(9.81)(0.6)+NB (0.75) = 0 NB = 3374.6 N = 3.37 kN

+ c ©Fy = 0;

(Both wheels)

A

0.15 m

Ans. 1.5 m

3374.6 -300 (9.81) -1.18(103)(9.81)+ NA = 0 NA = 11.1 kN

B 0.75 m

Ans.

(Both wheels)

Upper member: a + ©ME = 0;

300(9.81)(2.75)- FCD sin 30° (1.25) = 0 FCD = 12 949 N = 12.9 kN F¿ CD =

+ ©F = 0; : x

FCD 12 949 = = 6.47 kN 2 2

Ans.

Ex - 12 949 cos 30° = 0 Ex = 11 214 N

+ c ©Fy = 0;

-Ey - 300(9.81) + 12 949 sin 30° = 0 Ey = 3532 N

FE = 2(11 214)2 + (3532)2 = 11 757 N Since there are two members, ¿ FE =

FE 11 757 = = 5.88 kN 2 2

Ans.

Ans: NA = 11.1 kN ( Both wheels ) F′CD = 6.47 kN F′E = 5.88 kN 600