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Static Equilibrium... Please help
The Jshaped 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
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)
Take moment about A:
(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
AD = 3i + 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+.87j2.58k+1.74i)] +[(33i)]+ [(2.4j9.6i)] = 0
[T_DE(.87j+1.74i)] + [(42.6i2.4j)] = 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.
Homework Statement
The Jshaped 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.
Homework Equations
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)
Take moment about A:
(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
AD = 3i + 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+.87j2.58k+1.74i)] +[(33i)]+ [(2.4j9.6i)] = 0
[T_DE(.87j+1.74i)] + [(42.6i2.4j)] = 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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