Calculating Acceleration with Tension: A Troubleshooting Guide

[user5]

I have attached my calculations.
In the problem do they mean acceleration from K to Y?
Where did I go wrong in my calculations?
THANKS IN ADVANCE

 

[voko]

You did not go wrong anywhere, but you did not finish. You have some angle in your formula, while the expected result does not have any.

 

[user5]

I have T_x=(FCOSα)/(2SINα)=(F)/(2TANα)=[F(L²-X²)^0.5)]/2X
I have the wrong answer:(

 

[voko]

You obviously used the wrong sides of the triangle to compute the tangent function. Think about that again. If you come to the same result again, post your thoughts here, and we will discuss.

 

[user5]

Why the wrong sides?

 

[voko]

Which side is x? How is it related with the angle?

 

[user5]

tanα=x/(L²-X²)^0.5

 

[voko]

x is the perpendicular distance from the line of F. Is that the case on your diagram?

 

[user5]

Thus it is going to be tanα=(L²-X²)^0.5/X? With 2Tsinα=F it seems to give the right answer, however in the "student solution" they refer to that problem as having 2Tcosα=F, Am I right this time?
Does it matter where I decide to place α ?

 

[user5]

1. From the start x=L thus is the acceleration infinite?
2. Is a_x changing while F is constant?

 

[Cbray]

The notation in your working hurts my brain.

 

[voko]

Thus it is going to be tanα=(L²-X²)^0.5/X? With 2Tsinα=F it seems to give the right answer, however in the "student solution" they refer to that problem as having 2Tcosα=F, Am I right this time?
Does it matter where I decide to place α ?

Obviously, there are two angles that you can use, but the end result should be the same.

1. From the start x=L thus is the acceleration infinite?

Nothing ever gets infinite in the real world. If you get something infinite in a problem, that means that its mathematical model becomes invalid.

2. Is a_x changing while F is constant?

It depends on the angle, so, yes.

 

[user5]

voko, I'm grateful to you for your help!