# Static Equilibrium and reaction forces

## Homework Statement

The J-shaped member shown in the figure is supported by a cable DE and a single journal bearing with a square shaft at A. Determine the reaction forces and at support A required to keep the system in equilibrium. The cylinder has a weight = 5.50 , and F = 1.60 is a vertical force applied to the member at C. The dimensions of the member are w = 1.50 , l = 6.00 , and h = 2.00.

Fx = 0
Fy = 0
Fz = 0

## The Attempt at a Solution

W = -5.5 k
F = -1.6 k
Coordinates:
A(0,0,0); B(0,6,0); C(-1.5,6,0); D(-3,6,0); E(0,0,2)

(AD X T_DE) + (AB X W) + (AC X F) = 0 ...eqn 1

Position vector:
DE = 3i - 6j + 2k magnitude DE = 7
AB = 6j
AC = -1.5i + 6j

Unit vector:
U_DE (3i - 6j + 2k)/7 = .43i-.86j+.29k

T_DE = T_DE(U_DE) = T_DE(.43i-.86j+.29k)

...eqn 1:
[(-3i+6j) X T_DE(.43i-.86j+.29k)] +[(6j) X (-5.5k)] + [(-1.5i+6j) X (-1.6k)] = 0

[T_DE(2.56k+.87j-2.58k+1.74i)] +[(-33i)]+ [(-2.4j-9.6i)] = 0

[T_DE(.87j+1.74i)] + [(-42.6i-2.4j)] = 0

Equating i coefficient to zero:
1.74T_DE - 42.6 = 0
T_DE = 24.48

From Equilibrium Condition:
F_z = 0
A_z + T_DE(.29) + W + F = 0
A_z + 24.48(.29) - 5.5 - 1.6 = 0
A_z + 7.1 - 5.5 - 1.6 = 0
A_z = 0

F_y = 0
A_y + T_DE(-.86) =0
A_y + 24.48(-.86) =0
A_y = 21.05

But it was wrong.

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