Skip to content

Course home

Sign up

1.5.2a Moment of a force

EasyMedium
12345678910111213
Question 12

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/Nx/cmx / \text{cm}x/cm1x/cm−1\frac{1}{x} / \text{cm}^{-1}x1​/cm−1
1.51.51.535.035.035.00.02860.02860.0286
2.02.02.026.326.326.30.03800.03800.0380
3.03.03.017.517.517.50.05710.05710.0571
4.04.04.013.113.113.1(i)
5.05.05.0(ii)0.09520.09520.0952
a.

Calculate the missing values (i) and (ii) to 3 significant figures.

[2]
b.

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}}}xmax​1​.

[1]
c.

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.

[1]
d.

Use your previous results to determine the minimum weight PPP required to balance the metre rule in this experiment.

[2]
Markscheme

1.5.2a Moment of a force Questions

  1. IGCSE
  2. /Physics
  3. /1.5.2a Moment of a force

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.

Question bank