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sgvaibhav
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Moment about an axis - Statics - 3 Dimensional Equilibrium
The member is supported by a pin at A and a cable BC. If the load at D is 300, determine the x, y, z components of reaction at these supports.
i.e Find the reactions at A and the tension in the cable BC
Since it is in equilbrium.
[tex]\Sigma F = 0[/tex]
[tex]\Sigma M = 0[/tex]
The most confusing part for me, is moment about an axis. (If possible, please tell me, how to find moment about an axis - the direct method and through the vector matrix).
Correct me, if i am wrong.
rBC = 0.3i - 0.6j + 0.2k
|rBC| = 0.7
uBC = 1/0.7 (0.3 i - 0.6j + 0.2k)
TBC = |TBC|/0.7 (0.3i - 0.6j + 0.2k)
Now
[tex]\Sigma Fz = 0[/tex]
[tex] = -300 + Az + TBC(0.2 / 0.7)[/tex]
Az is unknown, so to determine Az - we find moment about the x-axis of B.
This will give, Az = 0
[tex]\Sigma Fz = 0[/tex]
[tex]= -300 + TBC(0.2/0.7) = 0[/tex]
TBC = 1050 N.
[tex]\Sigma Fx = 0[/tex]
Ax + 1050(0.3/0.7) = 0
Ax = -450 N
[tex]\Sigma Fy = 0[/tex]
Ay - 1050(0.6/0.7) = 0
Ay = 900N
Thats what i could do max (i think I am right), now only moment about Y-axis w.r.t a and moment about Z-axis is left, w.r.t a again. (May and Maz).
Please tell me how to get this part, with 2-3 methods to get moment about an axis (Direct and through matrix).
Thanks a lot =)
Homework Statement
The member is supported by a pin at A and a cable BC. If the load at D is 300, determine the x, y, z components of reaction at these supports.
i.e Find the reactions at A and the tension in the cable BC
Homework Equations
Since it is in equilbrium.
[tex]\Sigma F = 0[/tex]
[tex]\Sigma M = 0[/tex]
The Attempt at a Solution
The most confusing part for me, is moment about an axis. (If possible, please tell me, how to find moment about an axis - the direct method and through the vector matrix).
Correct me, if i am wrong.
rBC = 0.3i - 0.6j + 0.2k
|rBC| = 0.7
uBC = 1/0.7 (0.3 i - 0.6j + 0.2k)
TBC = |TBC|/0.7 (0.3i - 0.6j + 0.2k)
Now
[tex]\Sigma Fz = 0[/tex]
[tex] = -300 + Az + TBC(0.2 / 0.7)[/tex]
Az is unknown, so to determine Az - we find moment about the x-axis of B.
This will give, Az = 0
[tex]\Sigma Fz = 0[/tex]
[tex]= -300 + TBC(0.2/0.7) = 0[/tex]
TBC = 1050 N.
[tex]\Sigma Fx = 0[/tex]
Ax + 1050(0.3/0.7) = 0
Ax = -450 N
[tex]\Sigma Fy = 0[/tex]
Ay - 1050(0.6/0.7) = 0
Ay = 900N
Thats what i could do max (i think I am right), now only moment about Y-axis w.r.t a and moment about Z-axis is left, w.r.t a again. (May and Maz).
Please tell me how to get this part, with 2-3 methods to get moment about an axis (Direct and through matrix).
Thanks a lot =)
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