3rd-ipho_mfo69 - 3 rd International Physics Olympiad 1969...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3 rd International Physics Olympiad 1969, Brno, Czechoslovakia Problem 1. Figure 1 shows a mechanical system consisting of three carts A , B and C of masses m 1 = 0 . 3 kg, m 2 = 0 . 2 kg and m 3 = 1 . 5 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 m s- 2 . · ¶‡ · ¶‡ i ¡ e e e e- C B A ~ F Figure 1: 1. A horizontal force ~ F is now applied to cart C as shown. The size of ~ 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 ~ 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. 1 Solution: Case 1. The force ~ 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 ~ G 1 , ~ T 1 and ~ 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.- 6 x y · ¶‡ · ¶‡ i ¡ e e e e- 6 ?- C B A ~ F ~ T 2 ~ T 1 ~ G 1 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 ~ T 2 , thus T 2 = m 2 a x . (1) Since ~ T 1 and ~ T 2 denote tensions in the same cord, their magnitudes satisfy T 1 = T 2 . The forces ~ T 1 and ~ 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 . (2) So the motion of the whole mechanical system is described by the equation F = ( m 1 + m 2 + m 3 ) a x , (3) 2 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 ~ 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 = m 1 m 2 g ....
View Full Document

This note was uploaded on 11/08/2011 for the course PHYS 0000 taught by Professor Na during the Spring '11 term at Rensselaer Polytechnic Institute.

Page1 / 9

3rd-ipho_mfo69 - 3 rd International Physics Olympiad 1969...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online