- #1
kirab
- 27
- 0
Diagram is attached.
P = forces
M = free-moving moment
C = looks like a pin connection
L = distance between A and C, and C and B
With the system in equilibrium.
Question: Write the moment equations about point A, B and C and prove that they are all the same.
so basically I end up with:
(Counter-clockwise moment is positive.)
Summation of moments about point A: P(sin(90-theta))(2L) - M - Cx(sin(90-theta))(L) + Cy(sin(theta))(L) = 0
point B: P(sin(theta))(2L) + Cx(sin(90-theta))(L) - Cy(sin(theta))(L) - M = 0
point C: P(sin(90-theta))(L) - M + P(sin(theta))(L) = 0
which means that
P(cos(theta))(2L) - Cx(cos(theta))(L) + Cy(sin(theta))(L) = P(sin(theta))(2L) + Cx(cos(theta))(L) - Cy(sin(theta))(L) = P(cos(theta))(L) + P(sin(theta))(L) = M.
I don't see how these can be equal. Any mistakes seen here in the approach to solving this problem? Maybe pin supports Cx, Cy don't create moments?
Thanks for the help!
P = forces
M = free-moving moment
C = looks like a pin connection
L = distance between A and C, and C and B
With the system in equilibrium.
Question: Write the moment equations about point A, B and C and prove that they are all the same.
so basically I end up with:
(Counter-clockwise moment is positive.)
Summation of moments about point A: P(sin(90-theta))(2L) - M - Cx(sin(90-theta))(L) + Cy(sin(theta))(L) = 0
point B: P(sin(theta))(2L) + Cx(sin(90-theta))(L) - Cy(sin(theta))(L) - M = 0
point C: P(sin(90-theta))(L) - M + P(sin(theta))(L) = 0
which means that
P(cos(theta))(2L) - Cx(cos(theta))(L) + Cy(sin(theta))(L) = P(sin(theta))(2L) + Cx(cos(theta))(L) - Cy(sin(theta))(L) = P(cos(theta))(L) + P(sin(theta))(L) = M.
I don't see how these can be equal. Any mistakes seen here in the approach to solving this problem? Maybe pin supports Cx, Cy don't create moments?
Thanks for the help!