At 08:08 AM 4/12/2005, Steve Maki wrote:
>Tom Rauch wrote:
>
> >>If I understand him correctly, putting some side force
> >>on the middle of my tripod will cause unequal scale
> >>readings, and I don't understand why.
> >>
> >>Steve K8LX
>
> > ....because the tower can deflect and the guys are at the
> > top.
> >
> > That will lift some leg(s) and compress others more. My
> > point is it never seems to get worse than without the guys.
>
>Yes of course, but he came up with a number WITHOUT knowing
>the deflection. That was my whole point. It seems to me
>that if you hold the top still, and the middle does not
>bow, then you have pure shear force at the bottom, and the
>three legs maintain equal compression.
Think of the problem as a single horizontal 10 ft long beam, rigidly
attached at both ends, for the mean time.
Let's just consider the forces from the weight in the middle
The beam's attached firmly to the wall at both ends
You've got a 10 pound weight at the midpoint
You've got a 5 pound shear force acting up at each end (so the overall ups
and downs match).
But, there's also a moment (torque) trying to bend the beam down in the
middle. If the beam is L feet long, and the weight (W) is at "a" feet from
End A and "b" feet from the End B (i.e. L=a+b) then the moment at end A is
MA = W * a * b^2/L^2, or 50/4 = 12.5 ft lb. Since it's symmetric, the
moment at End B is exactly the same, but in the reverse direction.
---
In your case, you've got one end fixed to the wall (call that End A) and
the other end supported by a free pivoting joint (pin joint) (call that End
B). Again, let's just consider the forces and moments from the weight in
the middle.
The reaction force at the free end is 10*5/16 (=3.125) pounds, the reaction
force at the fixed end is 10*11/16 = 6.875 pounds.
The moment at end A is W * (L+b)/2 * (a/L)*(b/L) = 10 *
(10+5)/2*(1/2)*(1/2) ft lb = 18.75 ft lb. (you can get there by summing
the two moments.. 3.125*10 = 31.25 ft lb from the end and 10*5= 50 ft lb in
the middle acting in the opposite direction, for 18.75 ft lb net)
If the base is 3 feet wide, the moment of 18.75 ft lb will correspond to a
force pulling away from the base of 6.25 pounds on one side, and pushing
with 6.25 pounds on the other side.
--
The load at the top of the tower, resisted by the guy, doesn't add any net
moment, since the up and down match (looking at the tower
horizontally). However, since the guy now needs to provide the force to
counter the 10 pounds at the top plus the 3.125 pounds from the load in the
middle, there's now a 13.125 pound down force in the tower. (The guy
tension is 18.6 lb). The downforce will divide evenly between the two legs
(Yeah, you said it's a tripod, but I'm calculating for a two legged thing,
to keep things planar)
6.5625 pounds in each leg
if you add in the forces from the moment at the fixed base,
The leg towards the guy will have a force of 6.5625-6.25 pounds (about 0.3
pounds). The leg away from the guy will have a down force of 6.5625+6.25
pounds (about 12.8 pounds).
Both legs together have a shear load of 6.875 pounds.. That, plus the
13.125 horizontal force from the guy adds up to 20 pounds, which is the
total load from your pair of 10 pound horizontal loads. That balances.
It all balances.. the net of the downforces is equal to the total imposed
by the guy (13 pounds).
Looking at the moments at the base.. 0.3125 pounds on a 1.5 ft lever arm
(back foot), -12.8125 pounds on a 1.5 ft lever arm (front foot), 10 pounds
on a 5 ft lever arm (middle load), and -3.125 pounds on 10 ft lever arm
(net of the horizontal guy load and the 10 pounds)... they all sum to zero
(of course, the back leg will tend to lift...)
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