- #51

Chestermiller

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I've decided that I'm done screwing around with this problem. I'm going to solve it only using the laboratory (fixed) frame of reference, and without using the resolution of the problem into two separate problems that I've done so far. To do this, I'm going to focus on the top half of the truss, and do force and moment balances of struts AB and CD.You're right, I concluded that the torques of the ##y## components cancelled but of course they are both in the same sense.

First the kinematics. There are only two parameters that characterize the kinematics of this problem. This is the angle ##\theta## and the location of the center of mass of the overall rhombus. Using the analysis and notation in post #29, the acceleration components of the center of mass AB are:

##a_C+a_{Ax}/2## in the x direction

and

##-a_{Ax}/2## in the y direction

And for strut BC, they are

##a_C-a_{Ax}/2## in the x-direction

and

##-a_{Ax}/2## in the y direction##

where ##a_C## is the x-direction acceleration of the center of mass of the overall truss.

The figure below shows the free body diagram for the two struts comprising the upper half of the truss:

Based on this, the force- and moment balances on strut AB are:

$$\frac{F}{2}-T=m\left(a_C+\frac{a_{Ax}}{2}\right)$$

$$R_B-R_A=-m\frac{a_{Ax}}{2}$$

$$(R_B+R_A)\frac{b}{2\sqrt{2}}-\left(\frac{F}{2}+T\right)\frac{b}{2\sqrt{2}}=m\frac{b^2}{12}\frac{d^2\theta}{dt^2}$$

And for strut BC, they are:

$$T=m\left(a_C-\frac{a_{Ax}}{2}\right)$$

$$R_B+R_C=m\frac{a_{Ax}}{2}$$

$$-(R_C-R_B-T)\frac{b}{2\sqrt{2}}=-m\frac{b^2}{12}\frac{d^2\theta}{dt^2}$$

OK so far?

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