Moments - How to Calculate Tension in a Cable to Supply a Given Moment?

[zrome]

Homework Statement

"In raising the pole from the position show, the tension T in the cable must supply a moment about O of 72 Knm. Determine T."

Homework Equations

M = Fd

The Attempt at a Solution

d = 12sin(30) = 6m
T = 72/d
==> 12m

I do not know what I am doing wrong here. I'm assuming there is a problem with "12sin(30)" but I checked atleast 3 times and I still get the same eq. Please assist in any manner possible. Thank You in advance.

 

[PhanthomJay]

The moment of a force about a point O is the force times the perpendicular distance from the line of action of that force to the point O (M=F(d_perp)). Alternatively, the moment of a force about point O is the force times the distance from the point of the application of the force to point O, times the sine of the included angle between the two (M = rF sin theta = 30Tsin theta in this example). This becomes a geometrry/trig problem, where you must determine theta using the appropraite trig functions (as in using the law of sines and cosines, for example).

 

[zrome]

The moment of a force about a point O is the force times the perpendicular distance from the line of action of that force to the point O (M=F(d_perp)). Alternatively, the moment of a force about point O is the force times the distance from the point of the application of the force to point O, times the sine of the included angle between the two (M = rF sin theta = 30Tsin theta in this example). This becomes a geometrry/trig problem, where you must determine theta using the appropraite trig functions (as in using the law of sines and cosines, for example).

Now I'm not too sure if I understand you correctly here, but what I make out of your response is that the angle between, I believe it to be 30 degrees; you take the sin of the angle and then multiply that by the force, in this case T.

Doing that I arrive at the conclusion that 30Tsin(30) = 72 and solving for T results a 4.8 knm

I see the answer is wrong and am still unable to make out how to correctly solve this problem. Unfortunately I have a quiz on this material tomorrow evening at 6 and I need your help in trying to understand how to do this problem, for which the author of the book decided not to give the EDIT: workto.

The final answer is 8.65 knm. Could you kindly walk me step by step through the problem so that I can clearly understand how it is done? (I agree my trig is horrible)

 

[tiny-tim]

Hi zrome!

… the angle between, I believe it to be 30 degrees; you take the sin of the angle and then multiply that by the force, in this case T.

Where do you get 30º from?

Hint: draw the vertical line from the top of the cable to the ground … so you have a right-angled triangle, and you want to know the sine of one of the angles.

 

[nvn]

That's a good hint by tiny-tim, to get the height of the vertical line. And, you can compute the horizontal projection of the pole on the ground, right? After that... Hint: Use definition of tan to compute an angle.

 

[tiny-tim]

… Use definition of tan to compute an angle.

Hi nvn!

Actually, we don't need to know the angle …

we only want its sine, whch we can get from the sides of the right-angled triangle (and Pythagoras)

 

[zrome]

Hi nvn!

Actually, we don't need to know the angle …

we only want its sine, whch we can get from the sides of the right-angled triangle (and Pythagoras)

Ya computed the vertical line and was able to see the big triangle I was missing. Thank you for all the help.