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A Simple Tension Meter - Page 2

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Posted by Curbie on July 5, 2009, 8:30 pm

Looks like what I am looking for, Thanks.

Spread-sheet time.


Posted by Curbie on July 5, 2009, 7:58 pm

Thanks, I give it a try.


Posted by Tim Jackson on July 6, 2009, 7:07 am
 Curbie wrote:

Try "Triangle of forces"


Posted by daestrom on July 7, 2009, 10:24 pm
 Curbie wrote:


Well the 'math' goes along these lines (pun intended:-).

Looking at the straight line where the wire is originally and the wire
position after it's deflected, and draw a line perpendicular from the
mid-point over to the original line of the wire.  We have a right
triangle with one leg's length 1/2 inch and the other leg's length 20
inches and the hypotenuse is the wire.  With such a long narrow
triangle, we can approximate the length of the hypotenuse as also being
20 inches.  And the sin of that narrow angle is the opposite over
hypotenuse so we have (0.5)/20 = 1/40.

The tension in the wire is pulling at a very shallow angle, trying to
straighten out.  So the sideways force created by the tension in the
wire is X*sin(angle) = X/40.

Looking at the whole thing you can see that it is symmetrical and there
is a second triangle formed on the other side of the point where the
scale attaches.  The tension in *that* segment of wire is also 'X' and
it also is creating a side force of X/40.

So these two sections of wire are creating a side force of X/40 + X/40
or X/20.  And that side force is exactly countered by the pull on the scale.

So working it the other way around, the pull on the scale times 20
equals the tension in the wire.


Posted by Jim Wilkins on July 7, 2009, 11:25 pm

Nice, simple explanation. I couldn't figure out how to make vector
force resolution intuitively obvious.


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