#$&* course PHY 121 7/25 4:30 023. Forces (atwood, chains)
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Given Solution: 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. STUDENT COMMENT: need to think of friction as a component with the force of table on chain , not chain on table even though these are the same. INSTRUCTOR RESPONSE: The frictional force arises from the mutual opposing normal forces exerted between chain and tabletop. STUDENT COMMENT: also, the normal force is not according to the total mass, just the part on the table. i would havethought that the connected part of teh chain would contribute to force. but it doesnt INSTRUCTOR RESPONSE: The normal force acts only between the tabletop and the mass of the chain supported by the tabletop. The normal force itself is balanced by the gravitational force on this segment of chain. So the combined normal and gravitational force on the chain on the table contributes nothing to the net force. However the normal force does given rise to a frictional force. This frictional force is not balanced, in the way the normal force is balanced by the gravitational force, by any other force and in this sense the frictional force contributes to the net force. (The weight of the hanging part of the chain also contributes). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I tried to figure out how to split it up. I came up with x + x + .1x = 2.1x This is splitting the total force into 3 pieces two halves and a frictional piece. The total force is 3 kg * 9.8 m/s^2 = 29.4 N 29.4 N / 2.1 = 14 N I started with a 14 N piece. m = F/a m = 14 N / 9.8 m/s^2 m = 1.428 kg or 1428 g dividing this by 15 g/cm, I got 95 cm. 95 cm is on the table, being pushed by 1.4 N of friction 105 cm is hanging off the table. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This was the approach I was looking for but never figured out. I tried to see how to isolate the friction vs. gravity component unsuccessfully. I can see why I didn't get it. It's very confusing, but I do understand the concepts. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F = m * a F = 5 kg * 9.8 m/s^2 F = 49 N 49 N = .125 v^2 v^2 = 392 m^2/s^2 v = 19.8 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT COMMENT: It would take a lot of air resistance to stop a falling 5kg object. a big fan/jet INSTRUCTOR RESPONSE: It would take 5 kg * 9.8 m/s^2 = 49 N of force to balance the gravitational force and cause the falling object to stop accelerating. However once the 49 N air resistance was achieved, the object would continue falling at whatever constant velocity was required to achieve this force. To stop it from falling would take a force in excess of 49 N. The greater the resisting force the more quickly the object would come to rest. To bring it to rest would indeed require an updraft of some sort. The necessary speed depends on the surface area of the object. For example a parachute, which is spread over a relatively large area, might well have a mass of 5 kg, and very little velocity would be required to produce an air resistance of 49 N. On the other hand a 5 kg iron cannonball has a small surface area and would have to be moving very fast to encounter 49 N of air resistance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:"