course Phy 201 ãg— Õ‰ôY§€†Çš½åüm‘ÿºÂ™assignment #023 023. Forces (atwood, chains) Physics II 10-27-2008
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02:27:18 `q001. Note that this assignment contains 3 questions. . A chain 200 cm long has a density of 15 g/cm. Part of the chain lies on a tabletop, with which it has a coefficient of friction equal to .10. The other part of the chain hangs over the edge of the tabletop. If 50 cm of chain hang over the edge of the tabletop, what will be the acceleration of the chain?
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RESPONSE --> Since the chain as 50cm hanging ove rthe edge and for every cm it weighs 15 gram then the total mass will be 50cm*15g/cm=750grams. We convert this to .75kg. Te acceleration of gravity is 9.8m/s^2 and multiply that by the mass we get the force, .75kg*9.8m/s^2=7.35newtons.We now find the Ffrict by taking the 150cm of chain that is on the table top. 150cm*.015kg/cm=2.25kg. 2.25kg*9.8m/s^2=22.05newton. We have a coefficient of friction of.10 so we get a Frict of 22.05newtons*.10=2.21newtons. We now have a Fnet of 7.35newtons-2.21newtons=5.14newtons. The total mass of the system is 2.25kg+.75kg=3kg. We take the Fnet of 5.14newtons/3kg mass we get a=1.71m/s^2 confidence assessment: 1
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02:31:24 The part of the chain hanging over the edge of the table will experience an unbalanced force from gravity and will therefore tend to accelerate chain in the direction of the hanging portion. The remainder of the chain will also experience the gravitational force, but this force will be countered by the upward force exerted on the chain by the table. The force between the chain and the table will give rise to a frictional force which will resist motion toward the hanging portion of the chain. If 50 cm of chain hang over the edge of the tabletop, then we have 50 cm * (15 g/cm) = 750 grams = .75 kg of chain hanging over the edge. Gravity will exert a force of 9.8 meters/second ^ 2 * .75 kg = 7.3 Newtons of force on this mass, and this force will tend to accelerate the chain. The remaining 150 cm of chain lie on the tabletop. This portion of the chain has a mass which is easily shown to be 2.25 kg, so gravity exerts a force of approximately 21 Newtons on this portion of the chain. The tabletop pushes backup with a 21 Newton force, and this force between tabletop and chain results in a frictional force of .10 * 21 Newtons = 2.1 Newtons. We thus have the 7.3 Newton gravitational force on the hanging portion of the chain, resisted by the 2.1 Newton force of friction to give is a net force of 5.2 Newtons. Since the chain has a total mass of 3 kg, this net force results in an acceleration of 5.2 N / (3 kg) = 1.7 meters/second ^ 2, approximately.
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RESPONSE --> ok self critique assessment: 3
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02:56:55 `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating?
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RESPONSE --> First we need to find how much force per cm of chain. We take .015kg/cm and multiply that by 9.8m/s^2 we get .147newton/cm. This will apply to the chain hanging and to the chain on the table top. Since the Ffrict is what keeps the chain on the table then the force on the chain on the table will be .10*.147newtons/cm=.0147newtons/cm. With this being said then .147newtons/cm* x=.0147*200cm-x where x=length of chain. .147newtons/cm*x=.0147newtons/cm*200cm-x, If we divide both sides by .0147newtons/cm we get 10*x=200cm-x, we then add x to both sides to get 11*x=200cm, we then divide 11 by both sides and get x=18.2cm confidence assessment: 1
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02:58:54 The maximum length that can hang over is the length for which the frictional force opposing motion is precisely equal to the gravitational force on the hanging portion of the chain. If x stands for the length in cm of the portion of chain hanging over the edge of the table, then the mass of the length is x * .015 kg / cm and it experiences a gravitational force of (x * .015 kg / cm) * 9.8 m/s^2 = x * .147 N / cm. The portion of chain remaining on the tabletop is 200 cm - x. The mass of this portion is (200 cm - x) * .015 kg / cm and gravity exerts a force of (200 cm - x) * .015 kg / cm * 9.8 meters/second ^ 2 = .147 N / cm * (200 cm - x) on this portion. This will result in a frictional force of .10 * .147 N / cm * (200 cm - x) = .0147 N / cm * (200 cm - x). Since the maximum length that can hang over is the length for which the frictional force opposing motion is precisely equal to the gravitational force on the hanging portion of the chain, we set the to forces equal and solve for x. Our equation is .0147 N / cm * (200 cm - x) = .147 N/cm * x. Dividing both sides by .0147 N/cm we obtain 200 cm - x = 10 * x. Adding x to both sides we obtain 200 cm = 11 x so that x = 200 cm / 11 = 18 cm, approx..
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RESPONSE --> ok self critique assessment: 3
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03:02:09 `q003. The air resistance encountered by a certain falling object of mass 5 kg is given in Newtons by the formula F = .125 v^2, where the force F is in Newtons when the velocity v is in meters/second. As the object falls its velocity increases, and keeps increasing as it approaches its terminal velocity at which the net force on the falling object is zero, which by Newton's Second Law results in zero acceleration and hence in constant velocity. What is the terminal velocity of this object?
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RESPONSE --> Since we have the formula F=.125v^2 then we can find the force by taking th 5kg mass multiplying by the acceleration of gravity 9.8m/s^2=49newtons. We now have the equation 49newtons=.125v^2. To find the velocity we divide both sides by .125, 392newtons=v^2, we take the sqrt of both sides which is v=19.8m/s confidence assessment: 3
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03:03:34 Only two forces act on this object, the downward force exerted on it by gravity and the upward force exerted by air resistance. The downward force exerted by gravity remains constant at 5 kg * 9.8 meters/second ^ 2 = 49 Newtons. When this force is equal to the .125 v^2 Newton force of friction the object will be at terminal velocity. Setting .125 v^2 Newtons = 49 Newtons, we divide both sides by .125 Newtons to obtain v^2 = 49 Newtons/(.125 Newtons) = 392. Taking square roots we obtain v = `sqrt (392) = 19.8, which represents 19.8 meters/second.
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RESPONSE --> ok self critique assessment: 3
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