<p>I understand up to the very last step (step 10) of the solution for problem 1

Question

<p>I understand up to the very last step (step 10) of the solution for problem 162 of chapter 4 given by cramster.</p>
<p>The one thing I don't understand with this problem is why the Tension in cable FJ is assumed to be zero.</p>
<p>Is it because the charges are equal and evenly distributed on each side of cable DH and therefore rendering cable FJ useless because it is further from A?</p>
<p>**optional question** (you don't have to answer this one)</p>
<p>Would the tension if cable FJ be the same if the member ABF would also be attached to a ball-and-socket joint at arbitrary point Z lying on the same axis as A (which would create member ABFZ)?</p>

Explanation / Answer

I think this is what your asking. Think about pulling a basket from the middle. The force F(dh) is in line with the centroid of the object, thus it is holding it up equally from the left and the right. Due to the fact that F(bg) is not acting in the Y direction there is no force opposite of F(fj) that is acting in the y direction on the opposite side, therefore a force in F(fj) would cause the member to rotate. If that explanation didn't help think of this: F(dh) can be broken into a force in the positive y and a force in the -x. therefore, it has a force in the y direction. the two 120lb loads are between the F(dh) acting in the y direction, so like a distributed load you can get the center between the points and find a resultant force.

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