A student investigates the balancing of a metre rule pivoted at the 50.0 cm50.0\text{ cm}50.0 cm mark.
An object QQQ is placed at the 85.0 cm85.0\text{ cm}85.0 cm mark on the rule. A load of variable weight PPP is placed on the other side of the pivot at a distance xxx from the pivot to keep the metre rule balanced and horizontal.
The table below shows the results of the measurements:
| P/NP / \text{N}P/N | x/cmx / \text{cm}x/cm | 1x/cm−1\frac{1}{x} / \text{cm}^{-1}x1/cm−1 |
|---|---|---|
| 1.51.51.5 | 35.035.035.0 | 0.02860.02860.0286 |
| 2.02.02.0 | 26.326.326.3 | 0.03800.03800.0380 |
| 3.03.03.0 | 17.517.517.5 | 0.05710.05710.0571 |
| 4.04.04.0 | 13.113.113.1 | (i) |
| 5.05.05.0 | (ii) | 0.09520.09520.0952 |
Calculate the missing values (i) and (ii) to 3 significant figures.
The maximum possible distance of the load from the pivot in this setup is xmax=50.0 cmx_{\text{max}} = 50.0\text{ cm}xmax=50.0 cm. Calculate the value of 1xmax\frac{1}{x_{\text{max}}}xmax1.
The relationship between PPP and xxx is given by:
P=k(1x) P = k \left(\frac{1}{x}\right) P=k(x1)where kkk is a constant. Use the data for P=3.0 NP = 3.0\text{ N}P=3.0 N to calculate the value of kkk in N⋅cm\text{N}\cdot\text{cm}N⋅cm.
Use your previous results to determine the minimum weight PPP required to balance the metre rule in this experiment.
43 exam-style questions on CIE IGCSE Physics 1.5.2a Moment of a force. Each one has a worked solution and a mark scheme showing where the marks go.