An 11.6 kg block of metal is suspended from a scale and immersed in water, as in

Question

An 11.6 kg block of metal is suspended from a scale and immersed in water, as in the figure below. The dimensions of the block are 12.0 cm multiplied by 10.3 cm multiplied by 10.3 cm. The 12.0 cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the water. (a) What are the forces exerted by the water on the top and bottom of the block? Take P0 = 1.0130 105 N/m2. Ftop 6273.1: Your answer is incorrect. N Fbottom 18749.2 : Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N (b) What is the reading of the spring scale? 10.327 kg is the wrong answer....N

Explanation / Answer

Archimedes principal says the buoyant force on an object is exactly equal to the volume of the water displaced. So, the buoyant force is V = 10.3cm x 10.3cm x 12cm = 1273 cm^3. Since 1 cc of water weighs 1 g, the weight of the block, less the buoyancy is the weight that will register on the scale, or b) 11.1 kg - 1.273 kg = 9.827 kg a) they give us air pressure in N/m^2, we want to convert to cm^2 or 1.013 x 10^5 N/m^2 /(100 cm/M)^2 = 10.13 N/cm^2, to that air pressure, we want to add the pressure of the water. Water weighs 1 g per cubic centimenter, so it puts a pressure of 1g per square centimeter for every centimeter of height. Convert g to N by multiplying by Gravity = 9.8 m/s^2 so the pressure of water is 9.8 N/cm^2 for every cm below the surface. Then, Force = Pressure * Area Ftop = (9.8*5 + 10.13)*10.3^2 = 6273.1N Fbottom = (9.8*(5+12) + 10.13)*10.3^2 = 18749.3 N c) Fbottom - Ftop = 18749.3 - 6273.1 = 12476.3 N In the preamble we showed the volume was 1273 cm^3 and that water weighs 1g/cc for a weight of 1273 g. We convert that to Newtons by multiplying by gravity - 9.8 m/sec^2 or 1273 * 9.8 = 12475.5 which is within rounding error of equalling 12476.3N

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