course Phy 201
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21:07:37 `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 --> 15g/cm * 200cm = 3000g, which is 3kg for the mass of the entire chain. The total weight is therefore 3kg * 9.8m/s^2 = 29.4N. Since the chain has 150cm on the table, the weight is 2.25kg * 9.8m/s^2 = 22.05N The 50cm of chain hanging freely off the table weighs .750kg * 9.8m/s^2 = 7.35N. So for the Fnet, we have 22.05N - .10N + 7.35N = 29.3N. I would not be surprised if this is wrong though. I'm never sure if I get my signs right for the x and y directions. The acceleration will therefore be a = Fnet / m, a = 29.3N / 3kg = 9.8m/s^2
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21:14:55 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 --> I think I now finally understand which forces are supposed to be positive and negative. i know for the test 1, my biggest problem was deciding which force was supposed to be negative and now I see that the mass moving downward is positive and since the other mass is resisting it, it's negative. Before that explanation, I kept trying to decide which direction is supposed to be positive or negative, not which block prevents the other from moving. Outside of that I didn't know that I could multiply the 21N by the .10... I realize now that you told me not to multiply the Force by the angle and I got that confused with the net force in this question.
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21:17:50 `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating?
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RESPONSE --> I'm not exactly sure how to do this so I'm just guessing when I assume that 50% of the weight hanging over the table will cause it to accelerate. So with 2.25kg on the table, 1.125kg is half of that, which is 1125g / 15g/cm = 75cm of chain length...
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21:19:34 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 --> Haha well my method was at least easier though completely wrong. It's going to take a lot of work to get to where I can remember how to do all this math but I think I understand most of it.
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21:23:09 `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 --> This too is just a guess... F = 5kg * 9.8m/s^2 = 49N F = .125v^2 49N = .125v^2 v = sqrt(49N / .125) v = 19.8m/s
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21:23:29 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 --> cool
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