Съдържание

NBPhO 2023, Теория и експеримент — Задача 5. Force sensor

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от NBPhO 2023

Nordic-Baltic Physics Olympiad 2023 · 5 т.

Условие

— Päivo Simson.

A force sensor is constructed from a flexible beam of working length $L$ and height $h$ to which four identical electrically resistive wires of length $l \ll L$ are attached, as shown in the figure below. If a force $F$ is applied to the end of the beam, it will bend the beam and stretch the upper wires and compress the lower ones, causing changes in the electrical resistance. Using the Wheatstone bridge method, these changes can be converted into a voltmeter reading, allowing for the measurement of the applied force $F$ .

Sensor cross-section: a beam clamped at the left inside a housing and bent slightly downward by a force F (arrow) at its right end; the axes x and z are marked at the clamped end, two wires R1, R1 lie on the upper face and two wires R2, R2 on the lower face; below, a top view of the beam with the two R1 wires in the middle and the length split into L/2 and L/2.

In what follows, assume that the deflection of the beam is very small.

i) The bending moment $M(x)$ (i.e. the torque with which one fictitious half of the beam affects the other) is inversely proportional to the curvature radius $r(x)$ of the center line of the beam: $M = EI/r$ , where $E$ is Young’s modulus and $I$ is a constant depending on the geometry of the of the beam’s cross-section (both are known). Find the elongation $\Delta l$ of the upper wire when a force $F$ is applied to the sensor. Assume that the wires are placed in the middle of the beam. [2 т.]

ii) Find the resistances $R_1$ and $R_2$ when $\Delta l$ from the previous section is known and the initial resistance of the wires is $R_0$ . Assume that the volume of the wire remains constant during the deformation. [1 т.]

iii) The resistors are arranged in a Wheatstone bridge configuration shown in the figure below, where $U$ is the known battery voltage. Find the relationship between the measured voltage $V$ and the force $F$ .

Wheatstone bridge: four variable resistors R1 (top left), R2 (top right), R2 (bottom left) and R1 (bottom right) in a diamond with a voltmeter V between the left and right nodes; the battery U is connected between the top and bottom nodes with + at the top and − at the bottom.

[2 т.]

Решение

— Solution by Päivo Simson, grading schemes by ....

i) (2 т.)

To find the elongation $\Delta l$ , we need to know the curvature radius $r$ of the beam at $x = L/2$ . The torque created by the force $F$ at an arbitrary point $x$ is $F\cdot(L - x)$ . This must be balanced by the bending moment $M(x)$ of the beam. In the middle of the beam we have

M = \frac{EI}{r} = F\frac{L}{2} \implies r = \frac{2EI}{FL}.

It is easy to see that if the upper wires elongate by $\Delta l$ , the lower ones shorten by the same amount. Now we need to relate the curvature radius $r$ with the elongation $\Delta l$ .

Bent beam clamped at a hatched wall on the left with force F at the right end; axes x and z; the upper wire is labelled l+Δl and the lower wire l-Δl, the half-heights h/2 and h/2 are marked; two lines of length r from the ends of the wire segment converge downward to the centre of curvature, enclosing the angle Δφ.

The arc length of a circle is $\Delta s = r\Delta\phi$ . Knowing this, we have from the above figure

l + \Delta l = \left(r + \frac{h}{2}\right)\Delta\phi,

l - \Delta l = \left(r - \frac{h}{2}\right)\Delta\phi.

By dividing the above equations, we get an equation for $\Delta l$ that is easily solved:

\frac{l + \Delta l}{l - \Delta l} = \frac{2r + h}{2r - h} \implies \Delta l = \frac{lh}{2r}.

Combining this with the expression for $r$ we have

\Delta l = \frac{FLlh}{4EI}.

Grading:

Correct moment equation $EI/r = FL/2$ at the center of the beam (0.6 pts)
Correct geometric relations for $l \pm \Delta l$ based on $\Delta s = r\Delta\phi$ (0.6 pts)
correctly solving for $\Delta l$ (0.6 pts)
correctly expressing the final answer (0.2 pts)

ii) (1 т.)

Let $\rho$ and $S$ be the resistivity and the cross-sectional area of the wires, respectively. The initial resistance $R_0$ of all the wires is

R_0 = \frac{\rho l}{S} = \frac{\rho l^2}{lS},

where $lS$ is the volume of the wire that remains constant during the deformation. Assuming $\Delta l \ll l$ we have

R_1 = \frac{\rho(l + \Delta l)^2}{lS} \approx \frac{\rho l(1 + 2\Delta l/l)}{S} = R_0 + \Delta R,

R_2 = \frac{\rho(l - \Delta l)^2}{lS} \approx \frac{\rho l(1 - 2\Delta l/l)}{S} = R_0 - \Delta R,

where

\Delta R = \frac{2\rho\Delta l}{S} = 2R_0\frac{\Delta l}{l} = R_0\frac{FLh}{2EI}.

Grading:

$R = \rho l/S$ (0.3 pts)
Correct approximations (0.3 pts)
Correctly calculating the resistances and the change in resistance using the above relations (0.4 pts)

iii) (2 т.)

The total resistance $R$ of the circuit is

R = \frac{(R_1 + R_2)(R_1 + R_2)}{2R_1 + 2R_2} = \frac{R_1 + R_2}{2}

Since both sides of the circuit have the same total resistance $R_1 + R_2$ , the current is also the same. The voltmeter reading is therefore

\begin{aligned} V &= \frac{I}{2}R_2 - \frac{I}{2}R_1 = \frac{U}{2R}(R_1 - R_2) = \\ &= U\frac{R_1 - R_2}{R_1 + R_2} = U\frac{\Delta R}{R_0} = U\frac{FLh}{2EI}. \end{aligned}

From this, we finally have

F = \frac{2EIV}{LhU}.

Grading:

Correct total resistance of the Wheatstone circuit (0.5 pts)
Assumes that the resistance of the Voltmeter is infinite (0.5 pts)
Correctly calculating the potential difference $V$ as a function of the change in the resistance (0.5 pts)
Expressing the final answer in the form given in the solution (0.5 pts)

Оригинал в Архива: NBPhO_2023_problems.pdf · официални решения: NBPhO_2023_solutions.pdf