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Theoretical problems
Question 1.
A triangular prism of mass M is placed one side on a frictionless horizontal plane as shown in Fig. 1. The other two sides are inclined with respect to the plane at angles α₁ and α₂ respectively. Two blocks of masses m₂ and m₂, connected by an inextensible thread, can slide without friction on the surface of the prism. The mass of the pulley, which supports the thread, is negligible.
• Express the acceleration a of the blocks relative to the prism in terms of the acceleration a0 of the prism.
• Find the acceleration a_o of the prism in terms of quantities given and the acceleration g due to gravity.
• At what ratio m₁/m₂ the prism will be in equilibrium?

Solutions to the problems of the 5-th International Physics Olympiad, 1971, Sofia, Bulgaria
Question 1.
The blocks slide relative to the prism with accelerations a₁ and a₂, which are parallel to its sides and have the same magnitude a (see Fig. 1.1). The blocks move relative to the earth with accelerations:
(1.1) w₁ = a₁ + a₀;
(1.2) w₂ = a₂ + a₀.
Now we project w₁ and w₂ along the x- and y-axes:



(1.3) w_1x = a cos α₁ − a₀;
(1.4) w_1y = a sin α₁
(1.5) w_2x = a cos α₂ − a₀;
(1.6) w_1y = −a sin α₂

The equations of motion for the blocks and for the prism have the following vector forms (see Fig. 1.2):
(1.7) m₁w₁ = m₁g + R₁ + T₁
(1.8) m₂w₂ = m₂g + R₂ + T₂
(1.9) Ma₀ = Mg − R₁− R₂ − R − T₁ − T₂.


The forces of tension T₁ and T₂ at the ends of the thread are of the same magnitude T since the masses of the thread and that of the pulley are negligible. Note that in equation (1.9) we account for the net force –(T₁ + T₂), which the bended thread exerts on the prism through the pulley. The equations of motion result in a system of six scalar equations when projected along x and y:

(1.10) m₁ a cos α₁ − m₁a₀ = T cos α₁ − R₁ sin α₁;
(1.11) m₁ a sin α₁ = T sin α₁ + R₁ cos α₁ − m₁g;
(1.12) m₂ a cos α₂ − m₂a₀ = −T cos α₂ − R₂ sin α₂;
(1.13) m₂ a sin α₂ = T sin α₂ + R₂ sin α₂ − m₂g;
(1.14) −Ma₀ = R₁ sin α₁ − R₂ sin α₂ − T cos α₁ + T cos α₂;
(1.15) 0 = R − R₁ cos α₁ − R₂ cos α₂ − Mg.

By adding up equations (1.10), (1.12), and (1.14) all forces internal to the system cancel each other. In this way we obtain the required relation between accelerations a and a₀:

(1.16)

The straightforward elimination of the unknown forces gives the final answer for a₀:

(1.17)

It follows from equation (1.17) that the prism will be in equilibrium (a₀ = 0) if:

(1.18)

Repost from Solutions to the problems of the 5-th International Physics Olympiad, 1971, Sofia, Bulgaria

Sources/Literatures:
-The Olympic home page www.jyu.fi/ipho
-Solutions to the problems of the 5-th International Physics Olympiad, 1971, Sofia, Bulgaria
(The problems and the solutions are adapted by Victor Ivanov Sofia State University, Faculty of Physics, 5 James Bourcier Blvd., 1164 Sofia, Bulgaria)