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BPO 2020, Теория — Задача 3. Friction Coefficients

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Balkan Physics Olympiad – BPO 2, online contest, 24 October 2020, Craiova, ROMANIA · 24 октомври 2020 г.

Условие

A body of negligible size slides down on an inclined plane. The distance between the initial position of the body (where it was at rest) and the base of the inclined plane is $s_1$ . The angle between the inclined plane and a horizontal plane is $\alpha$ . The friction coefficient between the body and the inclined plane is $k_1$ , and the one between the body and the horizontal plane is $k_2$ , respectively. At the end of its motion the body stops at distance $s_2$ from the base of the inclined plane (from the beginning of the horizontal plane, see the figure). There is no energy loss at the transition of the body from the inclined to the horizontal plane.

$\alpha/{}^{\circ}$	$s_2$ / cm
15	8.0
20	19.5
25	30.3
30	40.6
35	51.0
40	61.4
45	70.5
50	79.9
55	88.3
60	95.5

An inclined plane (friction coefficient k1) at angle α to the horizontal dashed line meets a horizontal plane (friction coefficient k2); arrow s1 along the incline from the start position to the base; a small body rests on the horizontal plane at distance s2 from the base; a vertical arrow g⃗ points down.

a) Derive a formula expressing the dependence of distance $s_2$ on the given parameters ( $\alpha, s_1, k_1$ , and $k_2$ ). [4 т.]

b) At a fixed distance $s_1 = 0.500\ \mathrm{m}$ it was experimentally investigated the dependence of the distance $s_2$ on the angle $\alpha$ . The measurements are tabulated in the table. Using suitable new variables present the experimental dependence in such a way that it can be easily processed graphically to determine the friction coefficients. Give the expressions of the new variables as functions of the given parameters. [3 т.]

c) Using the empty columns in the table, calculate the values of the new variables for each measurement. Draw the dependence on the graph paper provided. [9 т.]

d) Using the obtained graph calculate the values of the friction coefficients $k_1$ and $k_2$ . [6 т.]

e) From the obtained data calculate the value of the maximal angle $\alpha_{max}$ at which the body will still stand at rest on the inclined plane. [3 т.]

Отговори

Решение

a) The friction force $F_{fr1}$ , acting on the body when it slides on the inclined plane, is $F_{fr1} = k_1 R_1$ , where $R_1$ is the force of the reaction. As $R_1 = mg\cos\alpha$ , then $F_{fr1} = k_1 mg\cos\alpha$ . [1 point] In the same way the friction force $F_{fr2}$ , acting on the body when it slides on the horizontal plane, is $F_{fr2} = k_2 R_2$ , where $R_2$ is the force of the reaction. As $R_2 = mg$ , then $F_{fr2} = k_2 mg$ . [1 point] Using the law of the change of the mechanical energy, $\Delta E_{mech} = A_{friction}$ , i.e. $-mgs_1\sin\alpha = -k_1 mg\cos\alpha\, s_1 - k_2 mgs_2$ . [1 point] After a simplification $s_2 = s_1(\sin\alpha - k_1\cos\alpha)/k_2$ . (1) [1 point]

b) The formula (1) can be transformed in the form $\frac{s_2}{s_1\cos\alpha} = -\frac{k_1}{k_2} + \frac{1}{k_2}\tan\alpha$ . (2) So, if the new variables are $x = \tan\alpha$ , [1.5 points] and $y = \frac{s_2}{s_1\cos\alpha}$ , [1.5 points] it can be seen that the equation (2) is a linear dependence $y = a + bx$ , where $a = -\frac{k_1}{k_2}$ and $b = \frac{1}{k_2}$ . This linear dependence can be easily processed graphically.

c) The calculated values of the new variables for each measurement are filled on the table. [3 points] The data are presented in the figure below. [6 points] (2 x 0.5 p for writing the represented quantities on the axes, 4 points for correct representation of the points on the graph and 1 p for drawing the line)

d) From the obtained graph the parameters of the straight line are $a \approx -0.50$ [1 point] and $b \approx 2.50$ . [2 points] After that the friction coefficients can be calculated, $k_2 = \frac{1}{b} = 0.40$ [1.5 points] and $k_1 = -a.k_2 = 0.20$ . [1.5 points]

$\alpha$ , °	$s_2$ , cm	$\tan\alpha$	$\frac{s_2}{s_1\cos\alpha}$
15	8.0	0.268	0.166
20	19.5	0.364	0.415
25	30.3	0.466	0.669
30	40.6	0.577	0.938
35	51.0	0.700	1.245
40	61.4	0.839	1.603
45	70.5	1.000	1.994
50	79.9	1.192	2.486
55	88.3	1.428	3.079
60	95.5	1.732	3.820

Graph of s2/(s1.cos(α)) against tan(α) from 0.0 to 2.0 with ten square data points on a straight line; the line crosses the vertical axis at a = - 0.5 (arrow) and a dashed slope triangle is labelled b = 2.0/0.8 = 2.5.

e) From the formula (1) it can be seen that the rest of the body on the inclined plane is possible only if $\tan\alpha \le k_1$ . [1 point] So, the value of the maximal angle is $\alpha_{max} = \tan^{-1}(0.2) \approx 11.3^{\circ}$ [2 points]

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf