A 11.5 kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended fr
ID: 1294972 • Letter: A
Question
A 11.5 kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and immersed in water as shown in the figure below. The 12.0 cm dimension is vertical, and the top of the block is 5.65 cm below the surface of the water.
(b) What is the reading of the spring scale?
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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