IPhO 1971 · Theory problem 1

Two blocks on a triangular prism on a frictionless plane

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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 \(\alpha_1\) and \(\alpha_2\) respectively. Two blocks of masses \(m_1\) and \(m_2\), 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 \(a_0\) of the prism.
  • Find the acceleration \(a_0\) of the prism in terms of quantities given and the acceleration \(g\) due to gravity.
  • At what ratio \(m_1/m_2\) the prism will be in equilibrium?
Figure 1

Fig. 1

Official English text of the problem.

Answers

The source has three unnumbered bullets and gives no point values; they are labelled (1)–(3) below.

Take \(a_0\) as positive to the right, and \(a\) as positive when \(m_1\) accelerates down its face and \(m_2\) accelerates up its face, relative to the prism.

  1. \[ \boxed{ a=\frac{g(m_1\sin\alpha_1-m_2\sin\alpha_2) +a_0(m_1\cos\alpha_1+m_2\cos\alpha_2)} {m_1+m_2}} \]

  2. \[ \boxed{ a_0= \frac{g(m_1\cos\alpha_1+m_2\cos\alpha_2) (m_1\sin\alpha_1-m_2\sin\alpha_2)} {(m_1+m_2)(M+m_1+m_2) -(m_1\cos\alpha_1+m_2\cos\alpha_2)^2}} \]

  3. \[ \boxed{\frac{m_1}{m_2}=\frac{\sin\alpha_2}{\sin\alpha_1}} \]

Solution

We use the usual ideal-thread approximation: the thread is massless and taut, and the pulley turns without friction and has negligible inertia. Consequently, the tension \(T\) is the same on both sides. The prism translates horizontally, and the blocks remain in contact with their faces.

Because the thread is inextensible and the pulley is fixed to the prism, the blocks have equal signed relative accelerations when measured down the left face and up the right face.

(1) Relative acceleration in terms of \(a_0\)

Work in the inertial frame of the horizontal plane. Each block’s absolute acceleration equals the prism’s acceleration plus its acceleration relative to the prism.

For \(m_1\), take the positive direction down its incline. The projection of the rightward prism acceleration onto this direction is \(-a_0\cos\alpha_1\). Newton’s second law along the incline therefore gives \[ m_1(a-a_0\cos\alpha_1)=m_1g\sin\alpha_1-T. \]

For \(m_2\), take the positive direction up its incline. The corresponding equation is \[ m_2(a-a_0\cos\alpha_2)=T-m_2g\sin\alpha_2. \]

The normal forces have no components along the inclines. Adding the equations eliminates \(T\): \[ (m_1+m_2)a =g(m_1\sin\alpha_1-m_2\sin\alpha_2) +a_0(m_1\cos\alpha_1+m_2\cos\alpha_2). \] Dividing by \(m_1+m_2\) gives answer (1).

(2) Acceleration of the prism

There is no external horizontal force on the complete system: gravity and the supporting plane’s normal force are vertical. Hence its horizontal momentum is conserved, so its centre of mass has zero horizontal acceleration.

For positive \(a\), both blocks accelerate leftward relative to the prism. Their absolute horizontal accelerations are \[ a_{1x}=a_0-a\cos\alpha_1,\qquad a_{2x}=a_0-a\cos\alpha_2. \] Thus \[ Ma_0+m_1(a_0-a\cos\alpha_1) +m_2(a_0-a\cos\alpha_2)=0, \] or \[ (M+m_1+m_2)a_0 =(m_1\cos\alpha_1+m_2\cos\alpha_2)a. \]

For compactness, define \[ S=m_1+m_2,\qquad C=m_1\cos\alpha_1+m_2\cos\alpha_2,\qquad D=m_1\sin\alpha_1-m_2\sin\alpha_2. \] The two equations become \[ Sa=gD+Ca_0,\qquad (M+S)a_0=Ca. \] Eliminating \(a\), \[ \bigl[S(M+S)-C^2\bigr]a_0=gCD, \] which gives answer (2).

For the inclined faces shown, \(0<C<S\), so \[ S(M+S)-C^2=MS+(S^2-C^2)>0. \] Therefore the prism accelerates rightward if \(m_1\sin\alpha_1>m_2\sin\alpha_2\), and leftward if the inequality is reversed.

(3) Condition for equilibrium of the prism

Equilibrium requires \(a_0=0\). Since \(C>0\), horizontal momentum conservation then also requires \(a=0\). The equations along the inclines reduce to \[ T=m_1g\sin\alpha_1=m_2g\sin\alpha_2. \] Consequently, \[ \boxed{\frac{m_1}{m_2}=\frac{\sin\alpha_2}{\sin\alpha_1}}. \] At this ratio, a prism initially at rest remains at rest.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.