Съдържание

IPhO 1969, Теория — Задача 1

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IPhO 1969

3rd International Physics Olympiad 1969, Brno, Czechoslovakia

Условие

Figure 1 shows a mechanical system consisting of three carts $A$ , $B$ and $C$ of masses $m_1 = 0.3\ \mathrm{kg}$ , $m_2 = 0.2\ \mathrm{kg}$ and $m_3 = 1.5\ \mathrm{kg}$ respectively. Carts $B$ and $A$ are connected by a light taut inelastic string which passes over a light smooth pulley attaches to the cart $C$ as shown. For this problem, all resistive and frictional forces may be ignored as may the moments of inertia of the pulley and of the wheels of all three carts. Take the acceleration due to gravity $g$ to be $9.81\ \mathrm{m\ s^{-2}}$ .

Large cart C on two wheels standing on a horizontal floor; small cart B on top of C, small cart A hanging against the right side of C; a string runs from B over a pulley at the top right corner of C down to A; a horizontal force F pushes C from the left. — Figure 1:

1.

A horizontal force $\vec{F}$ is now applied to cart $C$ as shown. The size of $\vec{F}$ is such that carts $A$ and $B$ remain at rest relative to cart $C$ .

a) Find the tension in the string connecting carts $A$ and $B$ .

b) Determine the magnitude of $\vec{F}$ .

2.

Later cart $C$ is held stationary, while carts $A$ and $B$ are released from rest.

a) Determine the accelerations of carts $A$ and $B$ .

b) Calculate also the tension in the string.

Отговори

Решение

Case 1. The force $\vec{F}$ has so big magnitude that the carts $A$ and $B$ remain at the rest with respect to the cart $C$ , i.e. they are moving with the same acceleration as the cart $C$ is. Let $\vec{G_1}$ , $\vec{T_1}$ and $\vec{T_2}$ denote forces acting on particular carts as shown in the Figure 2 and let us write the equations of motion for the carts $A$ and $B$ and also for whole mechanical system. Note that certain internal forces (viz. normal reactions) are not shown.

The cart system of Figure 1 drawn in coordinate axes with origin 0, x-axis along the floor and y-axis on the left; force F pushes cart C, tension T2 acts on cart B towards the pulley, tension T1 acts upward and weight G1 downward on cart A. — Figure 2:

The cart $B$ is moving in the coordinate system $Oxy$ with an acceleration $a_x$ . The only force acting on the cart $B$ is the force $\vec{T_2}$ , thus

T_2 = m_2\, a_x\,. \qquad (1)

Since $\vec{T_1}$ and $\vec{T_2}$ denote tensions in the same cord, their magnitudes satisfy

T_1 = T_2\,.

The forces $\vec{T_1}$ and $\vec{G_1}$ act on the cart $A$ in the direction of the $y$ -axis. Since, according to condition 1, the carts $A$ and $B$ are at rest with respect to the cart $C$ , the acceleration in the direction of the $y$ -axis equals to zero, $a_y = 0$ , which yields

T_1 - m_1\, g = 0\,.

Consequently

T_2 = m_1\, g\,. \qquad (2)

So the motion of the whole mechanical system is described by the equation

F = (m_1 + m_2 + m_3)\, a_x\,, \qquad (3)

because forces between the carts $A$ and $C$ and also between the carts $B$ and $C$ are internal forces with respect to the system of all three bodies. Let us remark here that also the tension $\vec{T_2}$ is the internal force with respect to the system of all bodies, as can be easily seen from the analysis of forces acting on the pulley. From equations (1) and (2) we obtain

a_x = \frac{m_1}{m_2} g\,.

Substituting the last result to (3) we arrive at

F = (m_1 + m_2 + m_3)\, \frac{m_1}{m_2} g\,.

Numerical solution:

\begin{aligned} T_2 &= T_1 = 0.3 \cdot 9.81\ \mathrm{N} = 2.94\ \mathrm{N}\,, \\ F &= 2 \cdot \frac{3}{2} \cdot 9.81\ \mathrm{N} = 29.4\ \mathrm{N}\,. \end{aligned}

Case 2. If the cart $C$ is immovable then the cart $A$ moves with an acceleration $a_y$ and the cart $B$ with an acceleration $a_x$ . Since the cord is inextensible (i.e. it cannot lengthen), the equality

a_x = -a_y = a

holds true. Then the equations of motion for the carts $A$ , respectively $B$ , can be written in following form

\begin{aligned} T_1 &= G_1 - m_1\, a\,, && (4) \\ T_2 &= m_2\, a\,. && (5) \end{aligned}

The magnitudes of the tensions in the cord again satisfy

T_1 = T_2\,. \qquad (6)

The equalities (4), (5) and (6) immediately yield

(m_1 + m_2)\, a = m_1\, g\,.

Using the last result we can calculate

\begin{aligned} a = a_x = -a_y &= \frac{m_1}{m_1 + m_2}\, g\,, \\ T_2 = T_1 &= \frac{m_2 m_1}{m_1 + m_2}\, g\,. \end{aligned}

Numerical results:

\begin{aligned} a = a_x &= \frac{3}{5} \cdot 9.81\ \mathrm{m\ s^{-2}} = 5.89\ \mathrm{m\ s^{-2}}\,, \\ T_1 = T_2 &= 1.18\ \mathrm{N}\,. \end{aligned}

Оригинал в Архива: ipho69.pdf