course Phy 121 ?????????????kw?P?assignment #026026. More Forces (buoyant)
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14:01:27 `q001. Note that this assignment contains 3 questions. . Water has a density of 1 g per cm^3. If an object is immersed in water, it experiences a buoyant force which is equal to the weight of the water it displaces. Suppose that an object of mass 400 grams and volume 300 cm^3 is suspended from a string of negligible mass and volume, and is submerged in water. If the mass is suspended in equilibrium, what will be the tension in the string?
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RESPONSE --> The density of this object is 1.33 g/cm^3. Therefore I think the tension in the string is equal to this: 1.33 g/cm^3. confidence assessment: 3
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14:07:34 The 400 g mass will experience a downward gravitational force of .4 kg * 9.8 meters/second^2 = 3.92 Newtons. It will also experience in upward buoyant force equal to the weight of the 300 cm^3 of water it displaces. This volume of water, at 1 g per cm^3, will have a mass of 300 grams and therefore a weight of .3 kg * 9.8 meters/second^2 = 2.94 Newtons. The forces acting on the mass are therefore the downward 3.92 Newtons of gravity, the upward 2.94 Newtons of the buoyant force and the tension, which we will call T, in the string. Since the system is in equilibrium these forces must add up to 0. We thus have -3.92 Newtons + 2.94 Newtons + T = 0, which has solution T = .98 Newtons. In common sense terms, gravity pulls down with 3.92 Newtons of force and the buoyant force pushes of with 2.94 Newtons of force so to keep all forces balanced the string must pull up with a force equal to the .98 Newton difference.
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RESPONSE --> That last paragraph makes a lot of sense. Gravity is pushing down, water is pushing up so the tension has to balance that out. self critique assessment: 2
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14:15:39 At a depth of 12 cm, the volume of the immersed portion will be 12 cm * 8 cm^2 = 96 cm^3. This portion will therefore displace 96 grams of water. The weight of this displace water will be .096 kg * 9.8 meters/second^2 = .94 Newtons. This will be the buoyant force on the cylinder.
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RESPONSE --> I accidently hit the Next button. But I am getting that the water will kind of balance out the gravitational force on the object. So in this question you can find the volume which equals the grams and then you can multiply this to find the weight which will be balanced out by the water. self critique assessment: 2
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14:24:17 `q003. The solid cylinder in the preceding problem has a total length of 18 cm and a mass of 80 grams. If the cylinder is immersed as before to a depth of 12 cm then released, what will be the net force acting on it at the instant of release?
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RESPONSE --> If it has a mass of 80 grams, then I think it has a volume of 80 cm^3. But I am not completely sure. If it was then I think it would have a weight of 784 and a buoyant force of 784 as well. I don't think this is right however. confidence assessment: 1
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14:28:30 The buoyant force on the cylinder is still .94 Newtons, directed upward. Gravity exerts a downward force of .080 kg * 9.8 meters/second^2 = .79 Newtons, approximately. The net force on the cylinder is therefore .94 N - .79 N = .15 N, directed upward. This will give its 80 gram mass and acceleration a = F / m = .15 N / .080 kg = 1.875 m/s^2. Note that as the cylinder rises less and less of its volume is submerged, so the buoyant force will decrease while the weight remains the same. Until the buoyant force has decreased to become equal and opposite to the weight, the net force will continue to be upward and the cylinder will continue to gain velocity. After this instant the cylinder will continue to rise, but the net force will be downward so that the cylinder will begin slowing down. Eventually the cylinder will come to rest and the net downward force will cause it to start descending once more. It will continue descending until the net force is again 0, at which the time it will have a downward velocity that will carry it beyond this point until it again comes to rest and the cycle will start over again.
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RESPONSE --> I didn't chnage the grams from the density to kilograms. Also, I didn't know we were looking for an acceleration here, but the buoyant force. But now I see that we take the buyonat force and subtract the gravitational force. Once we get this it's .15 in the upward direction. That's the force and we have the mass so we can get the acceleration. self critique assessment: 2
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