B1-2: Tension in a String: Find an Unknown Mass Using Equilibrium of Forces Theory



Wooden rod about \(1\text{m}\) long with eyelets as shown; about \(1.7\text{m}\) good-quality cord; 2 single pulleys; 2 x \(50\text{g}\) masses; 5 x \(100\text{g}\) masses; scale pan; unknown mass; 2 clamps & stands; 2 G-clamps; triple beam balance; metre ruler; 1 sheet graph paper; spirit level.


  1. Clamp the wooden rod firmly and horizontally, so that there is space for the scale pan and unknown mass to move a large distance vertically without touching any object. Assemble the apparatus as above, placing \(m\) = \(200\text{g}\) in the scale pan. Measure and record \(\overline{AB}\).
  2. Move \(m_1\) up and down, finally placing it in the middle of the range of possible equilibrium positions. Ensure that pulley \(P\) is directly under the mid-way mark on the rod. Measure and record values of \(m_1\), \(h\), and \(d\).
  3. Repeat 2. with \(m_1\) = \(250\text{g}\), \(300\text{g}\), \(350\text{g}\), \(400\text{g}\), \(500\text{g}\), \(600\text{g}\) each time recording \(m_1\), \(h\), and \(d\). Check that the pulley \(P\) remains under the mid-way mark on the rod.



Since point P is in equilibrium (Newton’s 1st law):

\[\text{Resultant} \ \tilde{R} = \tilde{T}_1 + \tilde{T}_2 + \tilde{W} = \tilde{0}\]

Therefore horizontally:

\[\sum{F_x} = 0 \qquad \therefore \ T_2(\cos\theta_2) - T_1(\cos\theta_1) = 0\]

However, \(\theta_1 = \theta_2\) (observation), therefore:

\[T_1 = T_2\]

And vertically:

\[\sum F_y = 0 \qquad \therefore \ T_1(\sin\theta_1) + T_2(\sin\theta_2) - W = 0\]

but \(\theta_1 = \theta_2\) and \(T_1 = T_2\), so:

\[2T_1 (\sin\theta_1) - W = 0 \label{eqn1} \tag{equation 1}\]

but \(T_1 = m_1g\), \(\sin\theta_1 = \frac{h}{d}\), and \(W = m_2g\). Hence:

\[\frac{h}{d} = \frac{m_2}{2m_1}\]


  1. Plot a graph of \(\frac{h}{d}\) against \(\frac{1}{m_1}\), and find the gradient.

  2. Use only the gradient and the formula given at the end of the theory to calculate the unknown mass \(m_2\).

  3. Measure the mass of \(m_2\): on the beam balance, and assuming this is accurate, calculate the % error in the value obtained in (2) above.

    1. Use the value of \(\frac{h}{d}\) when \(m_1 = 400\text{g}\), to calculate \(\theta_1\) at this point. Calculate \(W = m_2 g\).

    2. \(m_1\) is suddenly increased to \(500\text{g}\)...

      Assuming that at this moment \(\theta_1 = \theta_2 = \text{the value from (a) above}\), find the initial upward acceleration of \(m_2\), as it heads towards a new equilibrium position.

      (Hint: find \(T_1\) and use part of \(\ref{eqn1}\) above to find the net upward force on \(m_2.)\)