A 14.11 -kg block of metal is suspended from a scaleand immersed in water as in
ID: 1675689 • Letter: A
Question
A 14.11 -kg block of metal is suspended from a scaleand immersed in water as in Figure P9.30. The dimensions of theblock are 14.0 cm x 12.0 cm x 12.0 cm. The 14.0-cm dimension isvertical, and the top of the block is 12.0 cm below the surface ofthe water. (a) What are the forces exerted by the water on the topand bottom of the block? (Take P0 =1.013×105 N/m2.) (top force, bottomforce)(b) What is the reading of the springscale?
(c) Notice that the buoyant force equals thedifference between the forces at the top and bottom of the block.What is the buoyant force?
I was looking at the help provide on this question given by thesite and I got stuck on where the density of water came from and ifI was suppose to use it some how?
Explanation / Answer
(a) the absolute pressure at the level of the top of theblock will be Ptop = Po +water g htop =(1.0130 105 N/m2) + (103 kg /m3) (9.80 m / s2) (12.0 x 10-2m) =......... Pa at the level of the bottom of the block will be Pbottom = Po +water g hbottom =(1.0130 105 N/m2) + (103 kg /m3) (9.80 m / s2) (26.0 x 10-2m) =......... Pa so the downward force exerted on the top by the waterwill be Ftop = Ptop A = Ptop(0.12 m)2 =........ N the upward force the water exerts on the bottom of theblock is Fbottom = Pbottom A = Pbottom(0.12 m)2 =........ N (b) so the scale reads the tension in the T in the cordsupporting the block as the block is in equilibrium we can write Fy = T + Fbottom - Ftop -m g =0 T = ....... N (c) according to the archimedes principle the buoyantforce on the block equals the weight of the displaced water so we can write B = (water Vblock) g = ......... N compare the answer with answer in part (a)
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