Title TOPIC 5 SINKING AND FLOATING - WazaElimu.com.pdf ✅

37 TOPIC 5: SINKING AND FLOATING CONCEPT OF UPTHRUST Consider a cork that is held just below the surface of a liquid and then released. The cork comes to the surface immediately. This shows that while inside the liquid an upward force (Upthrust) acts on the cork. This force is greater than the weight of the cork hence the cork is pushed to the surface. Hence: UPTHRUST  This is an upward force exerted by the object when is partially or totally immersed in liquid.  Upthrust is also known as buoyant force. AN APPARENT WEIGHT  This is the weight of the body when it is partially or totally immersed in liquid. REAL WEIGHT  This is the weight of a body when the body is not immersed in a fluid (when the body is in air). Upthrust = Apparent weight loss Also; Weight of liquid displaced = Upthrust Apparent loss in weight = Real weight – Apparent weight of body in liquid Therefore: Apparent loss in weight = Weight in air (WA) (real weight) – Weight in water (WL) Since; Apparent loss in weight = Upthrust Hence: Upthrust = Weight in air (WA) – Weight in liquid (WL) Up thrust = WA – WL Example 1 When a body is totally immersed in liquid its weight is recorded as 3.1N. If its weight in air is 4.9N. Calculate the up thrust acting on the body.

39 Relative densities of the substances (both liquids and solids) can be determined by applying Archimedes’ Principle. A. RELATIVE DENSITY OF A SOLID Consider the diagram below which demonstrates how to determine the relative density of a solid by Archimedes’ Principle. Procedures: 1. Measure the weight of a body in air by using a spring balance and record it as WA 2. Measure the weight of a body when totally immersed in water by using spring balance and record it as WW Results: 1. Weight in air = WA 2. Weight in water (Upthrust) = WA – WW Then: Relative density (R.D) = Weight of a substance in air Weight of the substance in water (Upthrust) Thus; R. D = Weight in air Upthrust R.D = WA WA− WW Example A stone weight 64N in air and 48N when immersed in water. Calculate the: (a) Relative density of the stone (b) Density of the stone in g/cm3 (c) Volume of the stone in cm3 Solution Data given
40 WA = 64N, WW =48N (a) R.D = WA WA− WW = 64N 64N−48N R.D = 64N 16 = 4 ∴Relative density of the stone = 4 (b) Density of the stone From; R.D =Density of the substance Density of water Density of the stone = R.D x Density of water = 4 x 1g/cm3 ∴Density of the stone = 4g/cm3 . (c) Volume of stone From; Density = Mass Volume Volume = Mass Density NOTE: To get the mass of the stone we must to consider the weight of stone when it is in air (thus 64N). Change 64N into kg the in gram (g). Thus; 1 N = 10 kg 64N =? Then; 64 N x 10 kg 1 N = 640 kg Change 640 kg into gram (g) Thus; 1 kg = 1000g 640 kg =? Then; 640 kg x 1000 g 1 kg = 640,000 g Hence; Volume = Mass Density = 640,000 g 4 g/cm3 = 160,000 cm3 ∴Volume of the stone = 160,000 cm3 B. RELATIVE DENSITY OF LIQUID (OIL) To determine the relative density of liquid (oil), consider experiment below; Liquid Procedures