IPhO 1969 · Theory problem 1

Three carts connected by a string over a pulley

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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 \(9.81\ \mathrm{m\,s^{-2}}\).

Figure 1

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 \(\vec{F}\).

  1. 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.

Official English text of the problem.

Answers

  • 1(a): \(T = 2.94\ \mathrm{N}\).
  • 1(b): \(F = 29.4\ \mathrm{N}\).
  • 2(a): Cart \(A\) accelerates downwards at \(5.89\ \mathrm{m\,s^{-2}}\); cart \(B\) accelerates to the right, towards the pulley, at \(5.89\ \mathrm{m\,s^{-2}}\).
  • 2(b): \(T = 1.18\ \mathrm{N}\).

Solution

The light string and ideal pulley give the same tension \(T\) throughout the string. With friction neglected, each contact force is normal to its supporting surface. We apply Newton’s second law in the ground frame.

1(a)

Since \(A\) remains at rest relative to \(C\), and \(C\) moves horizontally, \(A\) has no vertical acceleration. Its vertical forces are its weight \(m_1g\) downwards and the tension \(T\) upwards. Thus \[ T-m_1g=0, \] giving \[ \boxed{T=m_1g=0.3(9.81)=2.943\ \mathrm{N} \simeq 2.94\ \mathrm{N}.} \]

The horizontal normal force from \(C\) supplies \(A\)’s horizontal acceleration.

1(b)

Let \(a_C\) be the common horizontal acceleration of all three carts. The only horizontal force on \(B\) is the string tension, so \[ T=m_2a_C. \] Consequently, \[ a_C=\frac{m_1g}{m_2} =\frac{0.3(9.81)}{0.2} =14.715\ \mathrm{m\,s^{-2}}. \]

For the complete system of carts, string and pulley, the contact forces and string forces are internal. The only external horizontal force is \(F\). Therefore, \[ F=(m_1+m_2+m_3)a_C =\frac{m_1(m_1+m_2+m_3)}{m_2}g. \] Numerically, \[ \boxed{F=(0.3+0.2+1.5)(14.715) =29.43\ \mathrm{N}\simeq29.4\ \mathrm{N}.} \]

2(a)

With \(C\) held stationary, the pulley is fixed. If \(A\) moves downwards by a distance \(s\), the vertical string segment lengthens by \(s\). Since the string is taut and inelastic, the horizontal segment must shorten by the same amount: \(B\) moves towards the pulley by \(s\). Their accelerations therefore have equal magnitudes.

Let this magnitude be \(a\), taking downwards as positive for \(A\) and rightwards as positive for \(B\). Newton’s second law gives \[ m_1g-T=m_1a \] for \(A\), and \[ T=m_2a \] for \(B\). Adding eliminates the tension: \[ m_1g=(m_1+m_2)a. \] Hence \[ \boxed{a=\frac{m_1g}{m_1+m_2} =\frac{0.3(9.81)}{0.3+0.2} =5.886\ \mathrm{m\,s^{-2}} \simeq5.89\ \mathrm{m\,s^{-2}}.} \]

Thus \(A\) accelerates downwards and \(B\) accelerates horizontally to the right, towards the pulley.

2(b)

Using the horizontal equation for \(B\), \[ T=m_2a=\frac{m_1m_2g}{m_1+m_2}. \] Therefore, \[ \boxed{T=0.2(5.886)=1.1772\ \mathrm{N} \simeq1.18\ \mathrm{N}.} \]

The solution was checked against the official solution.