Weight distribution on both ends of a ramp?

[dcb]

I now understand your argument about the scale. It seemed to me like the scale is designed to only measure Y forces but I got inconsistent results in an experiment. Intuitively it seems like the X forces are 0 at 0 degrees reach a maximum at 45 and go back to 0 at 90 (at least for H=0).

I grabbed a 2x8 sitting around in the garage and a device measures angles by using gravity to point a needle straight up. I put double sided tape on the scale so the 2x8 wouldn't slip. The 2x8 had some knots in it so the weight distribution was likely not perfectly uniform.

So as to not bias the experiment I did the scale measurements first. I measured the bottom left edge of the 2x8 at 20 degrees and 45 degrees and got 785 grams and 1160 grams. I used my finger at the bottom right edge to push it up. I pretty much kept the force on my finger perpendicular to the H edge.

I then measured the length to be 452 mm and the height to be 184 mm and the board weighed 1368 grams.

When I did the math I got 787 grams at 20 degrees which is pretty close. At 45 degrees I got 962 grams so the scale is measuring some X force at least at the worst angle.

When I get time I'll build form a nail out of some tubular metal at the right height for 45 degrees (plus the scale height), put it under the right edge and redo the experiment.

 

[jbriggs444]

I pretty much kept the force on my finger perpendicular to the H edge.

Yes, this is important.

I assumed a vertical support force on both ends -- that being the relevant case for a buoy supporting one end of a dock. You are assuming a perpendicular support force at the free end with a hinge joint (e.g. tape) at the scale end.

In your scenario there is nothing indeterminate. We can solve for the support forces and their angles at both ends as a function of the beam angle. There are no angles for which a solution is unavailable.

We begin by labelling the forces and endpoints. We have a beam with one end free and the other end hinged to the floor. Let us arbitrarily say that it is the left end that is hinged. Call this point "L". The right end is free. Call it "R".

We can imagine that "L" is at the origin of our coordinate system. The beam is making some angle $\theta$ that probably lies in the first quadrant. (The analysis also works for the 2nd, 3rd or 4th quadrants).

The beam has mass $m$ and length $l$.

We want to solve for unknown forces $F_\text{Lx}$ and $F_\text{Ly}$ for the horizontal and vertical components of the support force from hinge on beam at point L and for the unknown force $F_\text{Rt}$ for tangential component of the force of your hand on the beam at point R.

We are assuming that the radial component of the force of your hand on the beam is zero.

We can begin with a torque balance, summing torques about the hinge at point L. The result should be a zero if you are exerting just enough force to hold the beam in place having lifted it.

There are only two forces on the beam that have lines of action that do not pass through the chosen axis. One is gravity ($\vec{mg}$). The other is the tangential force from your hand, $F_\text{Rt}$. The moment arm for gravity has length $\frac{l}{2}$. But that moment arm may not be perpendicular to the force. So we have to multiply by $\cos \theta$. The moment arm for the tangential force for your hand is $l$. That moment arm is perpendicular already. Gravity acts clockwise. Your hands act counterclockwise.

Put it together and we have:$$0 = \frac{1}{2} mgl \cos \theta - F_\text{Rt}l$$We want to solve this for $F_\text{Rt}$. That easy -- we just divide both sides by $l$ and add $F_\text{Rt}$ to both sides:$$F_\text{Rt} = \frac{1}{2}mg \cos \theta$$But this is not the only number we care about.

Let us now do a force balance in the $y$ direction, summing the vertical components of all forces acting on the beam. There are three such forces, $F_\text{Ly}$ from the hinge, $mg$ from gravity and the $y$ component of $F_\text{Rt}$ from your hand. That $y$ component will be given by $F_\text{Rt} \cos \theta$. Put it togther and substitute in our solution for $F_\text{Rt}$ above and we get:$$0 = F_\text{Ly} - mg + \frac{1}{2} mg \cos^2 \theta$$We can move the terms around and arrive at:$$F_\text{Ly} = mg (1 - \frac{1}{2} \cos^2 \theta)$$Then we can apply a trig identity: $2 cos^2 \theta - 1 =\cos 2 \theta$ (or, equivalently $\cos^2 \theta\ = \frac{\cos 2 \theta + 1}{2}$) to arrive at:$$F_\text{Ly} = mg \frac{3 - cos 2 \theta}{4}$$That was a bit more algebraic manipulation than I can usually complete correctly, so a quick sanity check is in order.

For $\theta$ = 0, the result is $\frac{mg}{2}$ as expected.
For $\theta$ = 90, the result is $mg$ as expected
For $\theta$ = -90, the result is $mg$ as expected.

Let us finish with a force balance in the $x$ direction, summing the horizontal components of all forces acting on the beam. There are two such forces, $F_\text{Lx}$ from the hinge and the $x$ component of $F_\text{Rt}$ from your hand. That $x$ component will be given by $-F_\text{Rt} \sin \theta$. Put it together and we have:$$F_\text{Lx} = \frac{1}{2}mg \sin \theta \cos \theta$$We can simplify this using the double angle formula for the sine function: $\sin 2 \theta = 2 \sin \theta \cos \theta$ (or $\frac{1}{2} \sin \theta \cos \theta = \frac{\sin 2 \theta}{4}$). This will yield:
$$F_\text{Lx} = \frac {mg \sin 2 \theta} {4}$$For $\theta$ = 0, the result is 0 as expected.
For $\theta$ = 45, the result is maximized at $\frac{mg}{4}$
For $\theta$ = 90, the result is back to 0 as expected.

Your result that the x component of the tangential force is maximized at 45 degrees is confirmed.

Intuitively it seems like the X forces are 0 at 0 degrees reach a maximum at 45 and go back to 0 at 90 (at least for H=0).

Edit: Fixed the sign convention on the $x$ components to right = positive.