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This problem sample below is taken from theoretical problems of third IphO that's conducted in Brno, 1967, problem 1.

Problem 1
1. Figure 1 shows a mechanical system consisting of three carts A, B and C of masses m₁ =0.3 kg, m₂ =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.

1. A horizontal force is now applied to cart C as shown. The size of 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 .

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.

Solution
Case 1. The force 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 , and 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 B is moving in the coordinate system Oxy with an acceleration a_x. The only force acting on the cart B is the force , thus

T₂ = m₂ a_x (1)

Since and denote tensions in the same cord, their magnitudes satisfy

T₁ = T₂

The forces and 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, ay = 0, which yields

Consequently

T₂ = m₂ g . (2)

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

F = ( m₁ + m₂ + m₃ ) a_x , (3)

because forces between the carts A and C and also between the carts B and C are internal forces with respect to the systemof all three bodies. Let us remark here that also the tension 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



Substituting the last result to (3) we arrive at



Numerical solution:



Case 2. If the cart C is immovable then the cart Amoves with an accelera- tion ay and the cart Bwith an acceleration ax. 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

T₁ =G₁−m₁a, (4)
T₂ =m₂a. (5)

The magnitudes of the tensions in the cord again satisfy

T₁ =T₂ . (6)

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

(m₁+m₂)a=m₁ g .

Using the last result we can calculate



Numerical results:



Sources/Literatures :
-The Olympic home page www.jyu.fi/ipho