#$&* course Phy 121 We looked at some interesting new applications to the concepts we've learned so far and I am eager to explore them more in depth. I am confident you will challenge to rigourous extent. 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): We used a lot of what we’ve learned here to some new ideas which took some sorting out but the concepts are fairly solid here. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Okay we start over here finding the mass of the length but we input a variable x. X * 0.015kg/cm. The gravitational force is the mass times the acceleration of gravity: (x*0.015kg/cm) * 9.8m/s^2 = x*0.147N/cm Before we subtracted our length hanging from the total length: 200cm - x The mass is therefore (200cm - X) * 0.015kg/cm. This mass experiences a gravitational force of (200cm - X) * 0.015kg/s * 9.8m/s^2 = 0.147N/cm * (200cm-x) We take this and multiply it by 0.10 to get our frictional force 0.10 * 0.147N/cm * (200cm - x) = 0.0147N/cm * (200cm - x) The forces must be equal if the chain hanging off the edge of the table is to remain stationary so we set our two expressions equal to one another. X * 0.147N/cm = 0.0147N/cm * (200cm - X) Divide by 0.0147N/cm on both sides to get: 10x = 200cm - x Add X: 11x = 200cm Divide X = 18cm. confidence rating #$&*:e rigorous algebraic concepts here and as elementary as it probably seems to you I followed the solution the entire way. It makes sense in my head but I’m not certain I could have sorted it all out on my own. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique rating: ********************************************* 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: The downward force exerted by gravity on the 5kg object is 5kg*9.8m/s^2 = 49N Terminal velocity will be reached when the forces of gravity and the air resistance are equal, so we set the two quantities equal to each other and solve 49N = 0.125v^2 We want to find velocity so we isolate v^2 by dividing 0.125 to get: 392N = v^2 Take the square root of both sides V = 19.8m/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: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!