You are provided with a metre rule, a weight hanger, slotted masses, M, a piece (if string, a weighing balance and a knife edge. Use the diagram above as a ...
Question 1 Report
You are provided with a metre rule, a weight hanger, slotted masses, M, a piece (if string, a weighing balance and a knife edge. Use the diagram above as a guide to perform the experiment.
(i) Using the weighing balance, determine and record the mass, \(M_o\), of the unloaded metre rule.
(ii) Determine and record the mass, m, of the weight hanger.
(ii) Suspend the metre rule horizontally on the knife edge. Adjust the knife edge to a point G on the metre rule where it balances horizontally.
(iv) Record the distance, d = AG.
(v) Suspend the weight hanger securely at a point, P, on the metre rule such that AP = 5 cm. Keep the hanger at this point throughout the experiment
(vi) Add a mass, M = 20 g to the hanger, adjust the knife edge to a point K on the metre rule such that it balances horizontally as shown in the diagram above.
(vii) Determine and record the distance z = AK.
(vii) Record M and evaluate y - (z - 5), x - (d - z] and v = \(\frac{x}{y}\)
(ix) Repeat the experiment for M = 40 g, 60 g, 80 g and 100 g. In each case, evaluate y, x and v.
(x) Tabulate the results.
(xi) Plot a graph with M on the vertical axis and v on the horizontal axis, sinning both axes from the origin (0,0).
(xii) Determine the slope, s, of the graph.
(xii) Determine the intercept, c, on the vertical axis.
(xiv) State two precautions taken to ensure accurate results.
(b) (i) Under what condition is an object said to be in a stable equilibrium
(ii) Auniform beam of weight 50 N has a body of weight 100 N hung at one end of it. If the beam is 12 m long, determine the distance of a support from a 100 N body for it to balance horizontally.
Answer Details
(a) Results and graph
Mass of unloaded metre rule, \(M_0=75.0\text{ g}\).
Mass of weight hanger, \(m=20.0\text{ g}\).
Balance point of unloaded metre rule: \(d=AG=50.0\text{ cm}\).
For each load, \(y=z-5\), \(x=d-z\), and \(v=\dfrac{x}{y}\).
| \(M\) (g) | \(z=AK\) (cm) | \(y=z-5\) (cm) | \(x=d-z\) (cm) | \(v=x/y\) |
|---|---|---|---|---|
| 20 | 34.3 | 29.3 | 15.7 | 0.536 |
| 40 | 30.0 | 25.0 | 20.0 | 0.800 |
| 60 | 26.8 | 21.8 | 23.2 | 1.064 |
| 80 | 24.3 | 19.3 | 25.7 | 1.332 |
| 100 | 22.3 | 17.3 | 27.7 | 1.601 |
The plotted graph of \(M\) against \(v\), with both axes beginning at the origin, is shown below.
Using two widely separated points on the line of best fit, \((v_1,M_1)=(0.536,20)\) and \((v_2,M_2)=(1.601,100)\):
\[s=\frac{M_2-M_1}{v_2-v_1}=\frac{100-20}{1.601-0.536}=75.1\text{ g}\approx75.0\text{ g}.\]
The vertical intercept is \(c\approx-20.0\text{ g}\).
Thus, within experimental accuracy, \(s=M_0\) and \(c=-m\).
Precautions
- The metre rule was allowed to come to rest and was balanced horizontally before each reading was taken.
- All scale readings were taken with the eye vertically above the mark to avoid parallax error.
(b)
(i) An object is in stable equilibrium if, when slightly displaced, its centre of gravity rises and a restoring moment acts to return it to its original position.
(ii) Let \(y\) be the distance of the support from the \(100\text{ N}\) body. Taking moments about the support:
\[100y=50(6-y)\]
\[100y=300-50y\]
\[150y=300\]
\[y=2.0\text{ m}.\]
Therefore, the support should be placed \(\boxed{2.0\text{ m}}\) from the \(100\text{ N}\) body.
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