Question

To develop the ability to break a frame or machine down into subsystems and to determine the forces developed at internal pin connections Part D A tractor shovel Frames and machines are systems of pin-connected, multiforce members. Frames are designed to support loads, whereas machines are designed to transmiltl o alter the effects of loads. For a frame or machine to be in equiibrium, each member of the trame or machine system must be in equilibrium. Free-body diagrams of the overall system as well as individual members, groups of members, and The tractor shovel shown (Figure 4) carries a 505 kg load that has its center of mass at H. The shovel's dimensions are: a-55.0 mm, b 220 mm,c 330 mm, d 110 mm, and e 385 mm Find the reaction force of the pin at E on the member ACED. Assume that the positive direction of the x and y axes is to the right and upward, respectively Enter the Cartesian components of the reaction force at E separated by a comma. Express your answers in newtons to three significant figures Figure 4o14 View Available Hints) 0 02 The sign of each component enoodes the direction of the force No credit lost. Try again. Submit 2h Return to Assignment Provide Feedback

Answer 1

to calculate the angle

$\theta$ = tan^-1 ( c/b )

= tan^-1 ( 330/220 )

$\theta$ = 56.31^o

take momentum about point " D "

$\Sigma$ M_D = 0

- W x 2 b + F_GI sin$\theta$ x b + F_GI cos$\theta$ ( c + d ) = 0

- ( 530 x 9.8 x 2 x 220 ) + ( F_GI x sin56.31 x 220 ) + ( F_GI x cos56.31 x ( 330 + 110 ) ) = 0

F_GI = 5356.1 N

to find the length

EI = 2 a / sin30

= 2 x 55 / sin30

EI = 220 mm

at point " E " the momentum is

$\Sigma$ M_E = 0

F_KJ x EI/2 - F_GI cos56.31 x ( EI x cos30 ) - F_GI x sin56.31 x 220 x sin30 ) = 0

( F_KJ x 220/2 ) - ( 5356.1 x cos56.31 x 220 x cos30 ) - ( 5356.1 x sin56.31 x 220 x sin30 ) = 0

F_KJ = 9602.51 N

$\Sigma$ F_x = 0

E_x - F_KJ cos30 + F_GI x cos56.31 = 0

E_x - ( 9602.51 x cos30 ) + ( 5356.1 x cos56.31 ) = 0

E_x = 5345 N

at y - axis

$\Sigma$ F_y = 0

- E_y - F_KJ sin30 + F_GI sin56.31 = 0

- E_y - ( 9602.51 x sin30 ) + ( 5356.1 x sin56.31 ) = 0

E_y = - 344.7 N