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 ...

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

1. The eye was placed vertically above the mark being read so as to avoid error due to parallax.
2. 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.