TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with a uniform meter rule, a knife edge, masses and other necessary apparatus. Suspend the metre rule ...

Answer Details

Theory of the experiment. The metre rule is balanced on the knife edge at its centre of gravity G. The object W is fixed at the 15 cm mark, so the distance of W from G is D = (position of G) - 15. The mass M hangs at position Y, at a distance L from G.

By the principle of moments, at balance the anticlockwise moment of W equals the clockwise moment of M:

\[ W \times D = M \times L \]

Making M the subject:

\[ M = (W \cdot D)\,\frac{1}{L} = (WD)\,L^{-1} \]

Expected graph. A plot of M (vertical) against \(L^{-1}\) (horizontal) is a straight line through the origin whose slope is

\[ S = W \times D \]

Evaluating the slope quantity. Since \(S = WD\), the weight (in the units of M) of the object is obtained from the slope by

\[ \frac{S}{D} = W \]

So dividing the slope by the measured distance D gives the value of the load W in gram-force.

Sample table (headings).

M (g)	Y (cm)	L (cm)	\(L^{-1}\) (cm\(^{-1}\))
20	-	-	-
30	-	-	-
40	-	-	-
50	-	-	-
60	-	-	-

As M increases, L decreases so that the product ML stays constant and equal to WD.

Two precautions.

• The eye was placed vertically above the mark to avoid error due to parallax when reading positions.
• The knife edge was kept exactly at G throughout and the rule was allowed to settle to true horizontal before each reading.

(b)(i) Principle of moments. For a body 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 of the body appears to act, whatever the position of the body.