You are provided with a wooden block to which a hook is fixed, a set of masses, spring balance, and other necessary materials. Using the diagram above as a ...

Answer Details

(a) Static friction experiment

The block just begins to move when the horizontal pull \(F\) equals the limiting (static) friction. The limiting friction is proportional to the normal reaction, and the normal reaction equals the total weight \(M = m_0 + m\). The mass marked on the wooden block is \(m_0 = 400\,\text{g}\).

Observation / table of values (\(m_0 = 400.0\,\text{g}\)):

S/N	\(m_0\) (g)	\(m\) (g)	\(M = m_0+m\) (g)	\(R = \dfrac{M}{100}\)	\(F\) (N)
1	400.0	0.0	400.0	4.00	1.70
2	400.0	200.0	600.0	6.00	2.40
3	400.0	400.0	800.0	8.00	3.60
4	400.0	600.0	1000.0	10.00	5.20
5	400.0	800.0	1200.0	12.00	6.30

Graph of \(F\) against \(R\)

Slope of the graph

Taking two well-separated points on the line of best fit, \((R_1, F_1) = (6.0,\ 2.3)\) and \((R_2, F_2) = (12.0,\ 6.4)\):

\[ s = \frac{F_2 - F_1}{R_2 - R_1} = \frac{6.4 - 2.3}{12.0 - 6.0} = \frac{4.1}{6.0} \]\[ s = 0.68 \]

Two precautions

1. I avoided parallax error by reading the spring balance with my eye directly in line with the pointer.
2. I ensured the spring balance was pulled strictly horizontally so that the full weight acted as the normal reaction, and I noted the reading at the exact instant the block just started to move.

(b)(i) Definition

The coefficient of static friction is the ratio of the limiting (maximum) frictional force \(F\), acting just as the body is about to move, to the normal reaction \(R\) between the two surfaces in contact:

\[ \mu_s = \frac{F}{R} \]

(b)(ii) Calculation

Normal reaction:

\[ R = mg = 0.5 \times 10 = 5\,\text{N} \]

At the point of moving, the limiting friction equals the applied force, \(F = 2.5\,\text{N}\).

\[ \mu_s = \frac{F}{R} = \frac{2.5}{5} = 0.5 \]

The coefficient of static friction between the two surfaces is 0.5.