Verizon 12:18 AM session. masteringphysics.com C Practice Problem 13.7 In this e
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Verizon 12:18 AM session. masteringphysics.com C Practice Problem 13.7 In this example we will see how to determine the buoyant force given the mass and density of a submerged object. A 15.0 kg solid-gold statue is being raised from a sunken treasure ship, as shown in (a) Find the tension in the hoisting cable when the statue is completely immersed. (b) Find the tension when the statue is completely out of the water SET UP When immersed in water, the statue is acted upon by an upward buoyant force with magnitude FB. From Archimedes's principle, this force is equal in magnitude to the weight of water displaced. shows a free-body diagram for the statue, including its weight, the tension T, and the upward buoyant force FB SOLVE Part (a): To find Fa, we first find the volume of the statue, using its mass and the density of gold: 15.0 kg Poll 19.3x10 kg/m3 = 7.77 x 10-4 m The weight of an equall amount of water is (1.00 × 103 kg/)(7.77 x 10.4 )(9.80 m/s2) 7.61 N = = This weight is equal to the magnitude FB of the buoyant force. In equilibrium, the tension in the cable equals the weight of the statue minus the magnitude of the buoyant force. Taking the +y direction as upward, we find that T+7.61 N-147 N = 0 T = 139 N We note that this tension is less than the weight of the statue. Part (b): To find the tension when the statue is in air, we ignore the very small buoyant force of air (as we have done in earlier chapters). The tension when the block is in air is equal to its weight: m,tatueg = (15.0 kg)(9.80 m/s*) = 147 N REFLECT The tensions calculated here are for a statue at rest or being hoisted up at constant speed. Here is a shortcut leading to the same result for the buoyant force: The density of water is less than that of gold by a factor of 1.00x10k/m 19.3x10 kg/m =0.0518 - That is, the weight of a given volume of water is 0.0518 times that of an equal volume of gold, and the weight of the displaced water is 0.0518(147 N) 7.62 NExplanation / Answer
Volume of brass statue
V=m/p ==0.53/8000=6.625*10-5 m3
Buoyant force is
FB=pwaterVg=1000*6.625*10-5*9.8=0.64925 N
Tension in the string is
T=mg-FB =(0.53*9.8)-0.64925
T=4.54 N
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