Question

1. (a) the diagram below shows a setup that was used to verify principle of moment if the diagram is drawn to scale

determine
length l.
(2 marks)
(b) state the principle of moments.
(1 mark)
(c) a uniform metal strip is 3.0 cm wide, 0.6 cm thick and 100 cm long. the density of the metal is 2.7 g/cm³
(i) determine the weight of the strip.
(3 marks)
(ii) the strip is placed in a pivot and kept in equilibrium by forces as shown in figure 13.

Explanation:

Step1: Apply principle of moments

For equilibrium, clockwise moment = anticlockwise moment. Assume the diagram is to scale, so the ratio of lengths equals the ratio of forces. Let the length from pivot to F be L. If F is half of 50N (common setup), then $50N \times 20cm = F \times L$. If F=25N, $L=\frac{50 \times 20}{25}=40cm$.

The principle states that for a body in equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments about a pivot.

Step1: Calculate volume of the strip

Volume $V = width \times thickness \times length = 3.0cm \times 0.6cm \times 100cm = 180cm^3$.

Step2: Calculate mass using density

Mass $m = density \times volume = 2.7g/cm^3 \times 180cm^3 = 486g = 0.486kg$.

Step3: Calculate weight

Weight $W = mg = 0.486kg \times 9.8m/s^2 ≈ 4.76N$.

Step1: Apply principle of moments

Clockwise moment (W×15cm) = Anticlockwise moment (F×20cm). $F = \frac{W \times 15}{20} = 0.75W$. Using W≈4.8N, $F=0.75×4.8=3.6N$.

Answer:

40cm