Съдържание

Условие Отговори Решение

USAPhO 2004, Четвъртфинал — Задача 3

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Всички материали от USAPhO 2004

AAPT AIP United States Physics Team 2004 — Preliminary Exam, Open Response Questions · 25 т.

Указания

4 Questions, 60 minutes

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie
Q1 – 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Average Earth-Sun Distance	1 Astronomical Unit (A.U.) $= 1.5\times10^{11}\ \mathrm{m}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$

Moment of Inertia about Center of Mass – Uniform Object (will not be provided on the second screening exam)

Disk	$^1/_2\,MR^2$
Sphere	$^2/_5\,MR^2$
Rod	$^1/_{12}\,ML^2$

Solutions to Problems

Any correct solution should be awarded equivalent points. Suggested partial-credit points are presented in square brackets at the right margin. You may further break down the listed points into one point increments. If it is clear they have done an intermediate step, they should get credit for it even if they have not presented it. For example, in 4.a., if a student wrote down $\omega = (Mv\alpha L)/(2mL^2)$ , they should get 6 points credit. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.)

Условие

Side view of a flatbed truck with two wheels and a cab on the right; a board of length L leans from the flatbed against the vertical back of the cab. Its top is 0.8 L above the bed and its foot is 0.6 L from the cab wall.

A uniform board of mass $m$ and length $L$ is placed on the back of a flatbed truck as shown in the diagram above. There is no friction between the top of the board and the vertical surface of the truck. The coefficient of static friction between the bottom of the board and the horizontal surface of the truck is $\mu_s = 0.50$ . The truck always moves in the forward direction. Express your answers to the following as multiples of $g$ .

a. What is the maximum starting acceleration the truck can have if the board is not to slip or fall over? [10 т.]

b. What is the maximum stopping acceleration the truck can have if the board is not to slip or fall over? [10 т.]

c. For what value of stopping acceleration is the static frictional force equal to zero? [5 т.]

Отговори

Решение

3. a. Method I: (Inertial reference frame) The accompanying diagram shows the forces acting when the truck is accelerating to the right. The force of static friction must be to the right to provide the board’s acceleration.

There is no acceleration in the $y$ -direction, so the sum of forces in that direction must equal zero.

$\sum F_y = N - mg = 0$ . [2]

In the $x$ -direction, Newton’s Second Law yields

$\sum F_x = f - P = ma$ . [2]

There is no rotation about the center of mass (the center of the board), so the net torque about that point must equal zero. (This is not true for all points. The condition that the net torque must be zero about any axis applies only when the object is in static equilibrium, i.e., not accelerating.)

$\sum \tau = P(0.4L) + f(0.4L) - N(0.3L) = 0$ . [3]

Solving for $P$ $P = 0.75N - f$ .

Substituting into the $F_x$ equation

$f - (0.75N - f) = 2f - 0.75N = ma$ .

The acceleration is maximum when $f$ takes on its maximum value $f_{max} = \mu_sN$ [2]

$ma_{max} = 2f_{max} - 0.75N = 2\mu_sN - 0.75N = (2\mu_s - 0.75)mg$

or $a_{max} = (2(0.5) - 0.75)g = 0.25g$ [1]

a. Method II: (Accelerating reference frame attached to truck) The accompanying diagram shows the forces acting when the truck is accelerating to the right. The force of static friction must be to the right to balance the other forces. There is no acceleration in the $y$ -direction, so the sum of forces in that direction must equal zero.

$\sum F_y = N - mg = 0$ . [2]

In the $x$ -direction, Newton’s First Law yields

$\sum F_x = f - P - ma = 0$ . [2]

There is no rotation. In this frame any axis works. Selecting the bottom of the board.

$\sum \tau = P(0.8L) + ma(0.4L) - mg(0.3L) = 0$ [3]

Solving for $P$ $P = 0.375mg - 0.50ma$ .

Substituting into the $F_x$ equation

$f - (0.375mg - 0.50ma) - ma = f - 0.375mg - 0.50ma = 0$ .

The acceleration is maximum when $f$ takes on its maximum value $f_{max} = \mu_sN = \mu_smg$ [2]

$ma_{max} = 2f_{max} - 0.75mg = 2\mu_smg - 0.75mg = (2\mu_s - 0.75)mg$

or $a_{max} = (2(0.5) - 0.75)g = 0.25g$ [1]

(While we only present this method for Part (a), it can be used for Parts (b) and (c) as well.)

b. For maximum acceleration in the stopping case, $f$ must be in the same direction as the acceleration. The equations become

$\sum F_y = N - mg = 0$ [2]

$\sum F_x = f + P = ma$ [2]

$\sum \tau = P(0.4L) - f(0.4L) - N(0.3L) = 0$ [3]

Solving for $P$ $P = 0.75N + f$

Substituting into the $F_x$ equation

$f + (0.75N + f) = 2f + 0.75N = ma$ .

The acceleration is maximum when $f$ takes on its maximum value $f_{max} = \mu_sN$ [2]

$ma_{max} = 2f_{max} + 0.75N = 2\mu_sN + 0.75N = (2\mu_s + 0.75)mg$

or $a_{max} = (2(0.5) + 0.75)g = 1.75g$ [1]

c. If $f = 0$ in the Part b case, $P = ma$ [2]

and $P = 0.75N = 0.75mg$ [2]

Combining $a = 0.75g$ [1]

Free-body diagram of the board (Method I, truck accelerating to the right): P pointing left at the top, N upward and f to the right at the bottom, mg downward at the center; the top is 0.8 L above the bed and the foot is 0.6 L from the wall.

Free-body diagram of the board in the truck frame (Method II): P to the left at the top, ma to the left at the center, mg downward, N upward and f to the right at the bottom; dimensions 0.8 L and 0.6 L.

Free-body diagram of the board for the stopping case: P to the left at the top, mg downward, N upward and f to the left at the bottom; dimensions 0.8 L and 0.6 L.

Оригинал в Архива: Quarterfinal 2004.pdf · официални решения: Quarterfinal 2004 Solutions.pdf