You are provided with a uniform meter rule, a knife edge, masses and other necessary apparatus. Suspend the metre rule horizontally on the knife edge. Read ...
Question 1 Report
You are provided with a uniform meter rule, a knife edge, masses and other necessary apparatus.
- Suspend the metre rule horizontally on the knife edge. Read and record the point of balance G of the métre rule. Keep the knife edge at this point throughout the experiment.
- Using the thread provided, suspend the object labelled W at the 15cm mark of the metre rule.
- Suspend a mass M= 20g on the other side of G. Adjust the position of the mass until the metre rule balances horizontally again.
- Read and record the position Y of the mass M on the metre rule.
- Determine and record the distance L between the mass and G. Also determine and record the distance D between W and G.
- Repeat the procedure for four other values of M = 30, 40, 50 and 60 g. In each casse ensure that W is kept Constant at the 15 cm mark and the knife edge at G.
- Evaluate L\(^{-1}\) in each case. Tabulate your readings.
- Plot a graph of M on the vertical axis against L\(^{-1}\) on the horizontal axis.
- Determine the slope S, of the graph.
- Evaluate \(\frac{4}{D}\)
State two precautions taken to obtain accurate results.
(b)i. State the principle of moments.
ii. Define centre of gravity
Answer Details
Test of Practical Knowledge - Balancing a Metre Rule
Set-up. The uniform metre rule balances horizontally on the knife edge at its centre of gravity, so the point of balance is G = 48.0 cm. The object W is suspended at the 15 cm mark throughout, and the mass M is suspended on the opposite side of G and adjusted to position Y until the rule is again horizontal.
The distance of the load from the pivot is fixed:
\[ D = G - 15 = 48.0 - 15.0 = 33.0\ \text{cm} \]The distance of the mass from the pivot in each trial is
\[ L = Y - G \]Table of readings
| S/N | M (g) | Y (cm) | L = Y - G (cm) | D (cm) | L-1 (cm-1) |
|---|---|---|---|---|---|
| 1 | 20 | 97.50 | 49.50 | 33.00 | 0.02 |
| 2 | 30 | 81.00 | 33.00 | 33.00 | 0.03 |
| 3 | 40 | 72.75 | 24.75 | 33.00 | 0.04 |
| 4 | 50 | 67.80 | 19.80 | 33.00 | 0.05 |
| 5 | 60 | 64.50 | 16.50 | 33.00 | 0.06 |
Theory
Taking moments about G, at balance the anticlockwise moment of W equals the clockwise moment of M:
\[ W \times D = M \times L \]Making M the subject:
\[ M = (W\,D)\,\frac{1}{L} = (W\,D)\,L^{-1} \]This has the form \(M = S\,L^{-1}\), a straight line through the origin of gradient \(S = W\,D\).
Graph of M against L-1
Slope of the graph
Reading two widely separated points on the line of best fit, \((L^{-1}_1, M_1) = (0.02,\ 20)\) and \((L^{-1}_2, M_2) = (0.06,\ 60)\):
\[ S = \frac{M_2 - M_1}{L^{-1}_2 - L^{-1}_1} = \frac{60 - 20}{0.06 - 0.02} = \frac{40}{0.04} = 1000\ \text{g cm} \]Evaluation of S/D
\[ \frac{S}{D} = \frac{1000}{33.0} = 30.3\ \text{g} \]Since \(S = W\,D\), the quantity \(S/D\) equals the load W, so the weight of the object is W = 30.3 g-force.
Two precautions
- The eye was placed vertically above the mark being read so as to avoid error due to parallax.
- Draught was avoided and the rule was allowed to settle to true horizontal, with the suspended masses hanging freely and not touching the bench, before each reading was taken.
(b)(i) Principle of moments
The principle of moments states that when a body is in equilibrium under the action of parallel forces, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.
(b)(ii) Centre of gravity
The centre of gravity of a body is the single point through which the whole weight (the resultant weight) of the body appears to act, whatever the position of the body.
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