course
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16:19:11 `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 --> If the object has a mass of 15 g per 1 cm of chain then the weight of the 50cm chain hanging over the edge would be 15 * 50 = 750g or .75 kg The chain on the tables is 150cm * 15 g = 2250 g or 2.25 kg. Since the force of the chain on the table and the force of the table pushing on the chain they are equal ... The friction is 10% of this force which is 2.25 kg * 9.8 m/s/s = 22.05 kg m/s^2 22.05 N * .10 = 2.205 Newtons The part of the chain that is hanging excerts a force of .75 kg (9.8 m/s^2) = 7.35 Newtons Since F = ma; then a=Fnet/m The Fnet of the system is 7.35 - 2.205 = 5.146 N a= 5.5.146 N / 3kg = 1.715 m/s/s
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16:22:19 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
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17:12:03 `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating?
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RESPONSE --> The force of the chain hangin over the edge must equal and opposite the force of the chain on the table... thus. I probably approached this a little differently than most.. but here goes. I took m1a1 - m2a2(.10) = 0 equation and made a common value of x and x-3kg which would be the total lenght of the chain minus the part that would be hanging down, and made the following x * 9.8 m/s - (3-x)(9.8m/s)(.10) = 0 solving for this I obtain x = .278 kg .276kg / .015 kg would be the lenght of chain in cm that would equate which equals 18.5 cm or .185 m ... probably wrong but that is my stabe at it... I will look at you answer next and evaluate..
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17:15:12 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 --> I tried to work with the mass rather than the lenght.... it seems you did the same thing in trying to equal out the foces by finding a common x term. My answer is close...It seems much easier to work with the lenght x factor... I will certainly keep that in mind on future problems
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17:22:12 `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 --> F = ma = 5kg * 9.8 m/s^2 = 49 Newtons using the air resistance formula as the following 49 N = .125 v^2 v = 19.79 m/s If I understand the fomula correctly then at a velocity of 19.79 m/s the wind resistance excerts an opposite force equal to the gravitiation force of gravity.
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17:22:25 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
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