A 15.5 kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended fr
ID: 1514174 • Letter: A
Question
A 15.5 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 5.80 cm below the surface of the water. What are the magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water? (Use P_0 = 1.0130 Times 10^5 N/m^2.) What is the reading of the spring scale? Show that the buoyant force equals the difference between the forces at the top and bottom of the block. (Do this on paper. Your instructor may ask you to turn in this work.)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 = 12cm x 10cm x 10cm = 1200 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) 15.5 kg - 1.2 kg = 14.3 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*5.8 + 10.13)*10^2 = 6697 N
Fbottom = (9.8*(5.8+12) + 10.13)*10^2 = 18457 N
c) Fbottom - Ftop = 18457 - 6697 = 11760 N
In the preamble we showed the volume was 1200 cm^3 and that water weighs 1g/cc for a weight of 1200 g. We convert that to Newtons by multiplying by gravity - 9.8 m/sec^2 or
1200 * 9.8 = 11760 which is equal to Fbottom - Ftop = 11760 N
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.