Suggestions:
- FBD
- Newton/Euler: Use summation of forces in both x- and y-directions. Use Euler equation about G.
- Kinematics: Since A is moving on a straight path (x-direction), aA has only an x-component. Since we are looking for the tension force at the start of the motion, we have omega = vA = vB = 0. Also, since B moves on a circular path centered at C and since B has zero speed, the acceleration B must be perpendicular to BC (think about the path description of the acceleration of B to see this).
- Solve
Let us know if you have questions on this.
Suggestions:
6 comments:
I did a sum forces in the x, sum forces in the y, euler equation about G, then for kinematics used aG = aA + rG/A x alpha, then broke down the kinematics equation into the i and j, and ultimately ended up with 5 equations and 6 unkonwns, the unknowns being Ax, Ay, aGx, aGy, T, and alpha... and now that I think of it, since aG is perpendicular to rB/G, you can just use the magnitude aG and then use trig... I think?
you can do that, right?
You're not quite there. As described, aB is perpendicular to BC since B travels on a circular path around C and having zero speed at this instant. However, G is NOT traveling on a circular path, and you do not know the direction of aG.
What you need is to first relate aB and aA through the kinematics equation: aB = aA + alpha x rB/A - omega^2*rB/A. Use the known directions of aB and aA in this equation to relate alpha to aA. Then use aG = aA + alpha x rG/A - omega^2*rG/A to get the needed relationship between alpha and the x- and y-components of aG.
Does this help?
This was able to help me out a lot. Thank you sir.
anybody know what is 12 Mg? is it 12 * 9.81 * 10^6?
12 Mega grams or 12000 kg
yes, that did help.
gracias
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