Calculating Tension for Fence Post Straightening

[fableblue]

Homework Statement

It is known that a force with a moment of 7840 lb in about D is required to straighten the fence post CD. If a=8in. ,b=52 in., and d=112 in., determine the tension that must be developed in the cable of winch puller AB to create the required moment about point D.

Homework Equations

Find angle AED = arctan (52/120) = 23.43degrees
Use M =D x F ; M=7840lb in, D=ED=112 in, F=unknown
Substitute
use algebra to find T(AB)

The Attempt at a Solution

I have attached my attempt but i think I missed something because when i check it does not check. 176.05 x sin23.43 does NOT = 7840 (given)
Now i am stuck
Would someone please give me a little nudge/ where did i go wrong? What i am not seeing?
Thanks

 

[tiny-tim]

Hi fableblue!

I can't see your attachment yet, but I notice that your answer is out by a factor of 11.2. which looks suspiciously like d … does that help?

 

[fableblue]

Not really but thanks, tiny-tim.
May be when you are able to see the attachment you see what/where my problem lies.

I am going to try it again later on with the use of the law of cosins. What is throwing me off is the pole not being straight. Can I use\angleEDC, which is at the base on the post that needs to be strightened? \angleEDC=~59\circ, I used the law of sines for that but it does not look right; \frac{112sin(23.43)}{52}
Can i use that angle?

 

[Doc Al]

When finding torque using T = rFsinθ, θ is the angle between r and F. Consider the angle ECD.

 

[fableblue]

Thank you,
I also was looking at that angle, actually i was looking at ALL the angles. But now i can see why. And i am going to assume, from your reply, that the angles are valid no matter if the post is leaning or straight. Now that i think about it the problem is looking for the intial tension in the cable.

So we get r=52sin81.25, to get the \angle of 81.25\circ i used arctan(52/8) and oppisite interior angles are =, and the used algebra and my answer is F=152.544lbs.
Does that look about right?

 

[Doc Al]

So we get r=52sin81.25, to get the \angle of 81.25\circ i used arctan(52/8) and oppisite interior angles are =, and the used algebra and my answer is F=152.544lbs.
Does that look about right?

Not exactly. The way I look at it, r is the length of the post from D to C. Solve for that distance. The angle that you need is BCD. (Finding the angle of 81.25° is a useful step along the way.)

 

[fableblue]

OK so r=\sqrt{(8^2 + 52^2)}=52.61 in the Mp=rFsin\Theta; substitute- 7840 lb in = 52.61sin81.25F; do algebra and F=147.29 lbs= tension in cable ABC.
Is this correct? How do i go about checking it?

 

[Doc Al]

OK so r=\sqrt{(8^2 + 52^2)}=52.61 in the Mp=rFsin\Theta; substitute- 7840 lb in = 52.61sin81.25F; do algebra and F=147.29 lbs= tension in cable ABC.
Is this correct? How do i go about checking it?

Your r is correct, but your angle is still wrong. Find angle BCD.

 

[fableblue]

deep breath, ok i should have looked at my drawing because then i would have seen that \angleEDC and \angle ECD were not oppisite interior angles.
What i did this time was use 180-81.25 for \angleEDC=98.75 and then i used the law of cosins to find side EC =130.78 and from there i used the law of cosines again to find \angleECD=57.83, then i used Mp=rFsin\theta \rightarrow F=7840/(52.61sin57.83) = 176.04 Is this correct? That is my intial result way back in the begining.

I see that the angle should be 89.78 then F=149.02 and 7840~(52.61sin89.78) * 149.02 i found this through some algebra but now i am trying to see why i can not get it trough trig

I really do apperciate the help you are giving me.
Thank You

 

[Doc Al]

deep breath, ok i should have looked at my drawing because then i would have seen that \angleEDC and \angle ECD were not oppisite interior angles.
What i did this time was use 180-81.25 for \angleEDC=98.75 and then i used the law of cosins to find side EC =130.78 and from there i used the law of cosines again to find \angleECD=57.83, then i used Mp=rFsin\theta \rightarrow F=7840/(52.61sin57.83) = 176.04 Is this correct? That is my intial result way back in the begining.

That looks right to me.

I see that the angle should be 89.78 then F=149.02 and 7840~(52.61sin89.78) * 149.02 i found this through some algebra but now i am trying to see why i can not get it trough trig

I don't understand what you're doing here. Where does that angle of 89.78° come from?

 

[fableblue]

I did some switching around of things to try and make it fit:shy:

So are you saying that my intial results are correct? How do i go about checking this? I substitute 176.04 in for F and they do not = each other. 7840 = ...wait a minute... SHAZAM>>>> I was checking my solution the WRONG way not using the whole equation i was leaving out sin57.83. WOW
So the way that i did it intially is that a correct way? r=112sin23.43 and TAB= F and i understood what i was doing so is the way i did W A Y back there correct?

Thanks a bunch

 

[Doc Al]

So the way that i did it intially is that a correct way? r=112sin23.43 and TAB= F and i understood what i was doing so is the way i did W A Y back there correct?

I'd say that your initial method was unorthodox (at least to me), but correct. The "obvious" way to define rFsinθ in this problem is to have r be the distance DC, since we are finding the torque on the post. But there's nothing stopping you from measuring r from D to any point along the line of the force, since the torque will be the same. You used r = the distance DE and the associated angle. That works.

 

[fableblue]

Yes, that was because i was told that a force/torque is the same anywhere in the post. It seemed to easy for me that is why I did not have faith in my result and all a long is was the way i was checking it.

Thank You