course Phy 121 ylMŧ֚cassignment #022
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12:03:19 `q001. Note that this assignment contains 2 questions, which relate to a force-field experiment which is done using a computer simulation, and could for example represent the force on a spacecraft, where uphill and downhill are not relevant concepts. . An object with a mass of 4 kg is traveling in the x direction at 10 meters/second when it enters a region where it experiences a constant net force of 5 Newtons directed at 210 degrees, as measured in the counterclockwise direction from the positive x axis. How long will take before the velocity in the x direction decreases to 0? What will be the y velocity of the object at this instant?
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RESPONSE --> i have found out the acceleration and decided it will take 8 seconds before the velocity reaches 0 in the x direction. the velocity at this point would be negative according to the graph i've drawn, i can't figure out how to get the exact velocity but the y coordinate should be somewhere around14 m/s so i'll guess somewhere near 14 m/s confidence assessment: 2
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12:04:18 A constant net force of 5 Newtons on a 4 kg object will result in an acceleration of 5 Newtons/(4 kg) = 1.25 meters/second ^ 2. If the force is directed at 210 degrees then the acceleration will also be directed at 210 degrees, so that the acceleration has x component 1.25 meters/second ^ 2 * cosine (210 degrees) = -1.08 meters/second ^ 2, and a y component of 1.25 meters/second ^ 2 * sine (210 degrees) = -.63 meters/second ^ 2. We analyze the x motion first. The initial velocity in the x direction is given as 10 meters/second, we just found that the acceleration in the x direction is -1.08 meters/second ^ 2, and since we are trying to find the time required for the object to come to rest the final velocity will be zero. We easily see that the change in the next velocity is -10 meters/second. At a rate of negative -1.08 meters/second ^ 2, the time required for the -10 meters/second change in velocity is `dt = -10 meters/second / (-1.08 meters/second ^ 2) = 9.2 seconds. We next analyze the y motion. The initial velocity in the y direction is zero, since the object was initially moving solely in the x direction. The acceleration in the y direction is -.63 meters/second ^ 2. Therefore during the time interval `dt = 9.2 seconds, the y velocity changed by (-.63 meters/second ^ t) * (9.2 seconds) = -6 meters/second, approximately. Thus the y velocity changes from zero to -6 meters/second during the 9.2 seconds required for the x velocity to reach zero.
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RESPONSE --> ouch, this was a tough one. i'm saving these notes for later studying self critique assessment: 1
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12:08:16 `q002. Suppose that the mass in the preceding problem encounters a region in which the force was identical to that of the problem, but that this region extended for only 30 meters in the x direction (assume that there is the limit to the extent of the field in the y direction). What will be the magnitude and direction of the velocity of the mass as it exits this region?
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RESPONSE --> it seems to me that in this case if there is a ""bottom"" to the y direction, the mass will hit it, creating an angle bouncing back up in the positive y direction we would have to take the velocity at the time it reaches this bottom and find out the angle it creates other than that, i don't completely understand the question, this is one of those questions that makes me wish i had the in-class version of physics 121 confidence assessment: 0
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12:09:54 As we have seen in the preceding problem the object will have an acceleration of -1.08 meters/second ^ 2 in the x direction. Its initial x velocity is 10 meters/second and it will travel 30 meters in the x direction before exiting the region. Thus we have v0, a and `ds, so that you to the third or fourth equation of uniform accelerated motion will give us information. The fourth equation tells us that vf = +-`sqrt( (10 meters/second) ^ 2 + 2 * (-1.08 meters/second ^ 2) * (30 meters) ) = +-6 meters/second. Since we must exit the region in the positive x direction, we choose vf = + 6 meters/second. It follows that the average x velocity is the average of the initial 10 meters/second and the final 6 meters/second, or eight meters/second. Thus the time required to pass-through the region is 30 meters/(8 meters/second) = 3.75 seconds. During this time the y velocity is changing at -.63 meters/second ^ 2. Thus the change in the y velocity is (-.63 meters/second ^ 2) * (3.75 seconds) = -2.4 meters/second, approximately. Since the initial y velocity was zero, the y velocity upon exiting the region will be -2.4 meters/second. Thus when exiting the region the object has velocity components +6 meters/second in the x direction and -2.4 meters/second in the y direction. Its velocity therefore has magnitude `sqrt ( (6 meters/second) ^ 2 + (-2.4 meters/second) ^ 2) = 6.4 meters/second. The direction of velocity will be arctan ( (-2.4 meters/second) / (6 meters/second) ) = -22 degrees, approximately. Thus the object exits at 6.4 meters/second at an angle of 22 degrees below the positive x axis, or at angle -22 degrees + 360 degrees = 338 degrees.
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RESPONSE --> i do believe i made that question harder than it had to be. like i said, there are many times that i have wished i could have taken this class in the actual classroom. especially doing my experiments, i've heard yours are really fun and you go out and do cool things. my experiments are less spectacular but i am at least learning a whole lot. self critique assessment: 1
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ՏsEځh쎽韈Ϫ assignment #023 023. Forces (atwood, chains) Physics II 07-16-2007
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15:56:01 `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 --> the part lying on the table has a mass of 2.250 kg the part hanging over the edge has mass .750 kg Fnet=.75 kg x 9.8m/s^2=7.35N friction works against acceleration by .10 of the total mass of 3 kg or .3kg acceleration= F/m a=2.72 m/s^s confidence assessment: 2
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15:57:17 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 --> the friction seems to get me every time. that's one thing i need to work on; how to apply friction self critique assessment: 2
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16:08:58 `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating?
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RESPONSE --> it would have to be equal to how much F gravity exerts on the hanging part that cannot overtake how much it exerts on the lying part. accel must = 0 to get this the Fnet on the hanging part must subtract the frictional force to equal zero it would be somewhere around 275 grams which would equal 18.3 cm confidence assessment: 3
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16:09:35 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 didn't get it quite that way. i did some trial and error because i was so close to the way to figure it out. i did get it right though. self critique assessment: 3
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16:13:56 `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 --> the F exerted by gravity is 49 Newtons the acceleration must be 9.8 m/s^2 after some figuring i think the terminal velocity must be 3841.6 m/s 9.8/.125=78.4 78.4 * 49N = 3841.6 confidence assessment: 1
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16:15:08 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 --> uh-oh. i have a way of making a difficult problem extremely difficult. i need to simplify problems by breaking them down whenever possible. self critique assessment: 1
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ŘdҴy assignment #024 024. Centripetal Acceleration Physics II 07-17-2007
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16:25:37 `q001. Note that this assignment contains 4 questions. . Note that this assignment contains 4 questions. When an object moves a constant speed around a circle a force is necessary to keep changing its direction of motion. This is because any change in the direction of motion entails a change in the velocity of the object. This is because velocity is a vector quantity, and if the direction of a vector changes, then the vector and hence the velocity has changed. The acceleration of an object moving with constant speed v around a circle of radius r has magnitude v^2 / r, and the acceleration is directed toward the center of the circle. This results from a force directed toward the center of the circle. Such a force is called a centripetal (meaning toward the center) force, and the acceleration is called a centripetal acceleration. If a 12 kg mass travels at three meters/second around a circle of radius five meters, what centripetal force is required?
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RESPONSE --> i guess it would have to be the v^2/r and multiplied by the mass to get the correct amount of Force so i think it would be 1.8 * 12 = 21.6 Newtons confidence assessment: 2
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16:26:16 The centripetal acceleration of the object is v^2 / r = (3 meters/second) ^ 2/(5 meters) = 1.8 meters/second ^ 2. The centripetal force, by Newton's Second Law, must therefore be Fcent = 12 kg * 1.8 meters/second ^ 2.
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RESPONSE --> alright i got it, all except for the units which i put as newtons self critique assessment: 3
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16:30:59 `q002. How fast must a 50 g mass at the end of a string of length 70 cm be spun in a circular path in order to break the string, which has a breaking strength of 25 Newtons?
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RESPONSE --> the v^2/r would have to be greater than Fcent 25 to do it .05 kg is the mass .07 m is the length, so the radius is .035 m v^2/.035 = >25N velocity must be at least .935 m/s confidence assessment: 2
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16:34:37 The centripetal acceleration as speed v will be v^2 / r, where r = 70 cm = .7 meters. The centripetal force will therefore be m v^2 / r, where m is the 50 g = .05 kg mass. If F stands for the 25 Newton breaking force, then we have m v^2 / r = F, which we solve for v to obtain v = `sqrt(F * r / m). Substituting the given values we obtain v = `sqrt( 25 N * .7 meters / (.05 kg) ) = `sqrt( 25 kg m/s^2 * .7 m / (.05 kg) ) = `sqrt(350 m^2 / s^2) = 18.7 m/s.
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RESPONSE --> i didn't simplify my equation correctly. i think i'm on the right track with centripetal force though self critique assessment: 2
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16:39:25 `q003. What is the maximum number of times per second the mass in the preceding problem can travel around its circular path before the string breaks?
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RESPONSE --> if its v is 18.7 m/s and the length is .7m and the mass is .05kg the acceleration is V^2/r if the mass started at v0 and ended at v 18.7m/s i need to know the size of the circle that the mass is traveling don't i? suppose it is exactly .7m in that case wouldn't the max number of times per second be 26.7 before it breaks? confidence assessment: 1
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16:40:33 The maximum possible speed of the mass was found in the preceding problem to be 18.7 meters/second. The path of the mass is a circle of radius 70 cm = .7 meters. The distance traveled along this path in a single revolution is 2 `pi r = 2 `pi * .7 meters = 4.4 meters, approximately. At 18.7 meters/second, the mass will travel around the circle 18.7/4.4 = 4.25 times every second.
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RESPONSE --> dang, i forgot to apply the simple 2 'pi r to this to get the circle size. i knew to do that but i tend to overthink things and forget about a simple formula self critique assessment: 1
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16:41:30 `q004. Explain in terms of basic intuition why a force is required to keep a mass traveling any circular path.
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RESPONSE --> if a force were not there the object would keep going in , say, the horizontal direction in a ""straight"" path confidence assessment: 3
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16:42:05 We simply can't change the direction of motion of a massive object without giving it some sort of a push. Without such a force an object in motion will remain in motion along a straight line and with no change in speed. If your car coasts in a circular path, friction between the tires and the road surface pushes the car toward the center of the circle, allowing it to maintain its circular path. If you try to go too fast, friction won't be strong enough to keep you in the circular path and you will skid out of the circle. In order to maintain a circular orbit around the Earth, a satellite requires the force of gravity to keep pulling it toward the center of the circle. The satellite must travel at a speed v such that v^2 / r is equal to the acceleration provided by Earth's gravitational field.
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RESPONSE --> interesting. self critique assessment: 2
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[Tlx assignment #024 gzܞ Physics I 07-18-2007
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11:31:09 Why was it necessary to let the string go slack at the top of the circle in order to get the desired results?
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RESPONSE --> i'm not sure of this question, whether it's related to the last assignment, it's been a day or two since i did that one but i'll assume it is. if you don't let the string go slack you'll have to change the length, thereby nullifying the radius which throws the formula off and we can't get the correct answer
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11:34:09 Why do you expect that, if the string is released exactly at the top of the circle, the initial velocity of the washer will be horizontal?
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RESPONSE --> because if the weight is going in a circular direction, then its acceleration must be equal to the acceleration of gravity after an instant the washer will go down because its horizontal velocity will go down and gravity will take over
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11:35:11 ** The direction of an object moving in a circular arc is perpendicular to a radial line (i.e., a line from the center to a point on the circle). When the object is at the 'top' of its arc it is directly above the center so the radial line is vertical. Its velocity, being perpendicular to this vertical, must be entirely in the horizontal direction. **
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RESPONSE --> interesting, does it have to be at the ""top"" to be guaranteed to go in the horizontal direction then?
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11:38:21 What is the centripetal acceleration of the washer at the instant of release, assuming that it is released at the top of its arc and that it goes slack exactly at this point, and what was the source of this force?
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RESPONSE --> i thought the cent. force was equal to the v^2/r i guess the source of the force would be the sudden absence of the string. the absence of its friction, if you will.
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11:38:52 ** Under these conditions, with the string slack and not exerting any force on the object, the centripetal acceleration will be equal to the acceleration of gravity. **
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RESPONSE --> so the v^2/r would be equal to 9.8 m/s^2?
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11:41:24 Query principles of physics and general college physics 7.02. Delivery truck 18 blocks north, 10 blocks east, 16 blocks south. What is the final displacement from the origin?
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RESPONSE --> you would have components x=10 and y=2 wouldn't the 'ds be equal to the hypotenuese of such a triangle? if so, then the 'ds from origin is 10.2 blocks
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11:42:05 The final position of the truck is 2 blocks south and 10 blocks east. This is equivalent to a displacement of +10 blocks in the x direction and -2 blocks in the y direction. The distance is therefore sqrt( (10 blocks)^2 + (-2 blocks)^2 ) = sqrt( 100 blocks^2 + 4 blocks^2) = sqrt(104 blocks^2) = sqrt(104) * sqrt(blocks^2) = 10.2 blocks. The direction makes and angle of theta = arcTan(-2 blocks / (10 blocks) ) = arcTan(-.2) = -12 degrees with the positive x axis, as measured counterclockwise from that axis. This puts the displacement at an angle of 12 degrees in the clockwise direction from that axis, or 12 degrees south of east.
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RESPONSE --> i got this one.
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11:47:01 Query principles of physics and general college physics 7.18: Diver leaves cliff traveling in the horizontal direction at 1.8 m/s, hits the water 3.0 sec later. How high is the cliff and how far from the base does he land?
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RESPONSE --> after using the givens i have, i found the final velocity vertically to be 29.4 meters, from this i determined the cliff to be 44 meters tall approx. the horizontal direction would be 1.8 m/s * 3s 5.4 meters from the base
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11:47:21 The diver's initial vertical velocity is zero, since the initial velocity is horizontal. Vertical velocity is characterized by the acceleration of gravity at 9.8 m/s^2 in the downward direction. We will choose downward as the positive direction, so the vertical motion has v0 = 0, constant acceleration 9.8 m/s^2 and time interval `dt = 3.0 seconds. The third equation of uniformly accelerated motion tells us that the vertical displacement is therefore vertical motion: `ds = v0 `dt + .5 a `dt^2 = 0 * 3.0 sec + .5 * 9.8 m/s^2 * (3.0 sec)^2 = 0 + 44 m = 44 m. The cliff is therefore 44 m high. The horizontal motion is characterized 0 net force in this direction, resulting in horizontal acceleration zero. This results in uniform horizontal velocity so in the horizontal direction v0 = vf = vAve. Since v0 = 1.8 m/s, vAve = 1.8 m/s and we have horizontal motion: `ds = vAve * `dt = 1.8 m/s * 3.0 s = 5.4 meters.
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RESPONSE --> got it, i think i'm beginning to understand projectiles pretty good.
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펱ɍДmNۘwʒ assignment #025 gzܞ Physics I 07-18-2007
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14:54:33 principles of physics and gen phy 4.26 free-body diagram of baseball at moment hit, flying toward outfield gen phy list the forces on the ball while in contact with the bat, and describe the directions of these forces
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RESPONSE --> there would be the F on it driving it in the positive horizontal direction, the same force drives it in the positive vertical direction, and gravity drives it in the negative vertical direction
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14:55:36 ** Gravity exerts a downward force equal to the weight of the ball. While in contact with the ball, and only while i contact, the bat exerts a normal force, which pushes outward along a line originating from the central axis of the bat. This force is perpendicular to the surface of the bat at the point of contact. Unless the direction of the ball is directly toward the center of the bat, which will not be the case if the ball is hit at an upward angle by a nearly level swing, there will also be a frictional force between bat and ball. This frictional force will be parallel to the surface of the bat and will act on the ball in the 'forward' direction. COMMON STUDENT ERROR: The gravitational force and the force exerted by the ball on the bat are equal and opposite. The force of the bat on the ball and the gravitational force are not equal and opposite, since this is not an equilibrium situation--the ball is definitely being accelerated by the net force, so the net force is not zero. **
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RESPONSE --> i assumed the upward force of the bat since it said it was flying towards the outfield
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|Ϙ吟꺉 assignment #025 025. More Forces Physics II 07-18-2007
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11:56:05 `q001. Note that this assignment contains 5 questions. . A pendulum consists of a 150 g mass suspended from a light string. Another light string is attached to the mass, which is then pulled back from its equilibrium position by that string until the first string makes an angle of 15 degrees with vertical. The second string remains horizontal. Let the x axis be horizontal and the y axis vertical. Assume that the mass is pulled in the positive x direction. If T stands for the tension in the pendulum string, then in terms of the variable T what are the x and y components of the tension?
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RESPONSE --> both components should be positive i believe the 15 degree angle should equal the arctan of y/x. that's all i know confidence assessment: 1
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11:57:29 The pendulum string makes an angle of 15 degrees with vertical. Since we have assumed that the pendulum is pulled in the positive x direction, the direction of the tension in the string will be upward and to the left at an angle of 15 degrees with vertical. The tension force will therefore be directed at 90 degrees + 15 degrees = 105 degrees as measured counterclockwise from the positive x axis. The tension will therefore have x component T cos(105 degrees) and y component T sin(105 degrees).
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RESPONSE --> i see, i didn't think there was a way would get the actual figures from what was given. i was wrong about the arctan of 15 degrees self critique assessment: 2
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12:01:25 `q002. Continuing the preceding problem, we see that we have a vertical force of T sin(105 deg) from the tension. What other vertical force(s) act on the mass? What is the magnitude and direction of each of these forces? What therefore must be the magnitude of T sin(105 deg).
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RESPONSE --> the other string acts against the mass vertically gravity acts upon it as well the magnitude of the horizontal directional force is -.2588(T) the mag of Tsin is .965 T i don't know how to get the actual figures confidence assessment: 1
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12:02:19 The only other vertical force acting on the mass will be the gravitational force, which is .150 kg * 9.8 meters/second ^ 2 = 1.47 Newtons. The direction of this force is vertically downward. Since the mass is in equilibrium, i.e., not accelerating, the net force in the y direction must be zero. Thus T sin(105 deg) - 1.47 Newtons = 0 and T sin(105 deg) = 1.47 Newtons.
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RESPONSE --> i wouldn't have gotten this because i thought there were two forces that acted on it, gravity plus the other string. see how i complicated it? self critique assessment: 1
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12:05:35 `q003. Continuing the preceding two problems, what therefore must be the tension T, and how much tension is there in the horizontal string which is holding the pendulum back?
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RESPONSE --> the T must be the square root of x^2 plus y^2 so it's 1.89 N the horizontal tension is the cos of 105 * 1.47 or -0.38 confidence assessment: 1
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12:06:26 If T sin(105 deg) = 1.47 Newtons then T = 1.47 Newtons / (sin(15 deg)) = 1.47 Newtons/.97 = 1.52 Newtons. Thus the horizontal component of the tension will be T cos(105 deg) = 1.52 Newtons * cos(105 deg) = 1.52 Newtons * (-.26) = -.39 Newtons, approximately. Since the mass is in equilibrium, the net force in the x direction must be zero. The only forces acting in the x direction are the x component of the tension, to which we just found to be -.39 Newtons, and the tension in the second string, which for the moment will call T2. Thus T2 + (-.39 N) = 0 and T2 = .39 N. That is, the tension in the second string is .39 Newtons. STUDENT COMMENT: I'm really confused now. If we started out with a .15 kg mass that is equal to 1.47 Newtons. How did we create more weight to get 1.52 Newtons? Is the horizontal string not helping support the weight or is it puling on the weight adding more force? INSTRUCTOR RESPONSE: A horizontal force has no vertical component and cannot help to support an object against a vertical force. The vertical component of the tension is what supports the weight, so the tension has to be a bit greater than the weight. The tension in the string is resisting the downward weight vector as well as the horizontal pull, so by the Pythagorean Theorem it must be greater than either.
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RESPONSE --> i got the horizontal part but missed the T, that was tough self critique assessment: 2
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12:12:45 `q004. If a 2 kg pendulum is held back at an angle of 20 degrees from vertical by a horizontal force, what is the magnitude of that horizontal force?
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RESPONSE --> the direction of the magnitude would be up and to the left i think the angle must be 110 degrees sin 110=.939 B(.939) - 19.6 Newtons =0 y coordinate must be 20.8 cos 110 = -.342 19.6 * -.342 = --6.5 approx confidence assessment: 1
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12:14:00 At the 20 degree angle the tension in the pendulum string will have a vertical component equal and opposite to the force exerted by gravity. The tension with therefore have a horizontal component. To achieve equilibrium by exerting the horizontal force, this horizontal force must balance the horizontal component of the tension. We therefore begin by letting T stand for the tension in the pendulum string. We also assumed that the pendulum is displaced in the positive x, so that the direction of the string as measured counterclockwise from the positive x axis will be 90 degrees + 20 degrees = 110 degrees. Thus the x component of the tension will be T cos(110 deg) and the y component of the tension will be T sin(110 deg). The weight of the 2 kg pendulum is 2 kg * 9.8 meters/second ^ 2 = 19.6 Newtons, directed in the negative vertical direction. Since the pendulum are in equilibrium, the net vertical force is zero: T sin(110 deg) + (-19.6 N) = 0 This equation is easily solved for the tension: T = 19.6 N / (sin(110 deg) ) = 19.6 N / (.94) = 20.8 Newtons, approximately. The horizontal component of the tension is therefore T cos(110 deg) = 20.8 N * cos(110 deg) = 20.8 N * (-.34) = -7 N, approx.. To achieve equilibrium, the additional horizontal force needed will be + 7 Newtons.
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RESPONSE --> i got the vertical part of it right i came real close to the horizontal part, i just put in the wrong amount for T, i put the force of 19.6 rather than 20.8 self critique assessment: 2
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12:20:17 `q005. The 2 kg pendulum in the previous exercise is again pulled back to an angle of 20 degrees with vertical. This time it is held in that position by a chain of negligible mass which makes an angle of 40 degrees above horizontal. Describe your sketch of the forces acting on the mass of the pendulum. What must be the tension in the chain?
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RESPONSE --> i have a sketch of a graph, with angles of 40 degrees, 70 degrees, and 110 degrees, the 70 degrees angle is just a result of the difference of the other two angles Tsin70d-19.6=0 T.939=19.6 Ty=20.8N Tcos70d=20.8N Tx=7.11 the force is sq root of x^2 + y^2 or 22 Newtons approx. confidence assessment: 2
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12:20:42 The weight of the pendulum is partially supported by the tension in the chain. Thus the tension in the pendulum string is not the same as before. The horizontal component of the tension in the chain will be equal and opposite to the horizontal component of the tension in the pendulum string. Your picture should show the weight vector acting straight downward, the tension in the pendulum string acting upward and to the left at an angle of 20 degrees to vertical and the tension in the chain should act upward into the right at an angle of 40 degrees above horizontal. The lengths of the vectors should be adjusted so that the horizontal components of the two tensions are equal and opposite, and so that the sum of the vertical components of the two tensions is equal of opposite to the weight vector. Since both tensions are unknown we will let T1 stand for the tension in the pendulum and T2 for the tension in the chain. Then T1, as in the preceding problem, acts at an angle of 110 degrees as measured counterclockwise from the positive x axis, and T2 acts at an angle of 40 degrees. At this point whether or not we know where we are going, we should realize that we need to break everything into x and y components. It is advisable to put this information into a table something like the following: x comp y comp T1 T1 * cos(110 deg) T1 * sin(110 deg) in T2 T2 * cos(40 deg) T2 * sin(40 deg) Weight 0 -19.6 N The pendulum is held in equilibrium, so the sum of all the x components must be 0, as must the sum of all y components. We thus obtain the two equations T1 * cos(110 deg) + T2 * cos(40 deg) = 0 and T1 * sin(110 deg) + T2 * sin(40 deg) - 19.6 N = 0. The values of the sines and cosines can be substituted into the equations obtain the equations -.33 T1 + .77 T2 = 0 .95 T1 + .64 T2 - 19.6 N = 0. We solve these two simultaneous equations for T1 and T2 using one of the usual methods. Here we will solve using the method of substitution. If we solve the first equation for T1 in terms of T2 we obtain T1 = .77 T2 / .33 = 2.3 T2. Substituting 2.3 T2 for T1 in the second equation we obtain .95 * 2.3 T2 + .64 T2 - 19.6 N = 0, which we easily rearrange to obtain 2.18 T2 + .64 T2 = 19.6 Newtons, or 2.82 T2 = 19.6 N, which has solution T2 = 19.6 Newtons/2.82 = 6.9 N, approximately. Since T1 = 2.3 T2, we have T1 = 2.3 * 6.9 N = 15.9 N, approximately. Thus the pendulum string has tension approximately 15.9 Newtons and the chain the tension of approximately 6.9 Newtons.
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RESPONSE --> whew, this was a tough assignment. self critique assessment: 1
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كZg}yQ[ assignment #026 026. More Forces (buoyant) Physics II 07-18-2007
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18:38:15 `q001. Note that this assignment contains 3 questions. . Water has a density of 1 g per cm^3. If an object is immersed in water, it experiences a buoyant force which is equal to the weight of the water it displaces. Suppose that an object of mass 400 grams and volume 300 cm^3 is suspended from a string of negligible mass and volume, and is submerged in water. If the mass is suspended in equilibrium, what will be the tension in the string?
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RESPONSE --> i know that density = mass/volume so the object's density is 1.33 g/cm^3 so in order to match the density to that of the water, 100 grams need to be lifted to put the object in equilibrium i don't know how to say this in units of tension though confidence assessment: 2
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18:40:42 The 400 g mass will experience a downward gravitational force of .4 kg * 9.8 meters/second^2 = 3.92 Newtons. It will also experience in upward buoyant force equal to the weight of the 300 cm^3 of water it displaces. This volume of water, at 1 g per cm^3, will have a mass of 300 grams and therefore a weight of .3 kg * 9.8 meters/second^2 = 2.94 Newtons. The forces acting on the mass are therefore the downward 3.92 Newtons of gravity, the upward 2.94 Newtons of the buoyant force and the tension, which we will call T, in the string. Since the system is in equilibrium these forces must add up to 0. We thus have -3.92 Newtons + 2.94 Newtons + T = 0, which has solution T = .98 Newtons. In common sense terms, gravity pulls down with 3.92 Newtons of force and the buoyant force pushes of with 2.94 Newtons of force so to keep all forces balanced the string must pull up with a force equal to the .98 Newton difference.
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RESPONSE --> i see how to apply the force in a way to make it understandable. i had to see the upward and downward forces. i should have used a flow diagram, that would have helped me. self critique assessment: 3
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18:52:57 `q002. A solid cylinder has a cross-sectional area of 8 cm^2. If this cylinder is held with its axis vertical and is immersed in water to a depth of 12 cm, what will be the buoyant force on the cylinder?
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RESPONSE --> if the object goes in by 12 cm, the bouyant force will be 12cm * 8cm^2 or 96 Newtons confidence assessment: 1
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18:53:47 At a depth of 12 cm, the volume of the immersed portion will be 12 cm * 8 cm^2 = 96 cm^3. This portion will therefore displace 96 grams of water. The weight of this displace water will be .096 kg * 9.8 meters/second^2 = .94 Newtons. This will be the buoyant force on the cylinder.
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RESPONSE --> i knew i was on the right track but couldn't close the deal on this problem, i understand it now self critique assessment: 2
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18:56:20 `q003. The solid cylinder in the preceding problem has a total length of 18 cm and a mass of 80 grams. If the cylinder is immersed as before to a depth of 12 cm then released, what will be the net force acting on it at the instant of release?
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RESPONSE --> .94 Newtons is the bouyant force as we have seen. wouldn't we simply multiply that by the weight of .08 kg and then multiply that by the gravity of 9.8 m/s^2? confidence assessment: 0
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18:56:56 The buoyant force on the cylinder is still .94 Newtons, directed upward. Gravity exerts a downward force of .080 kg * 9.8 meters/second^2 = .79 Newtons, approximately. The net force on the cylinder is therefore .94 N - .79 N = .15 N, directed upward. This will give its 80 gram mass and acceleration a = F / m = .15 N / .080 kg = 1.875 m/s^2. Note that as the cylinder rises less and less of its volume is submerged, so the buoyant force will decrease while the weight remains the same. Until the buoyant force has decreased to become equal and opposite to the weight, the net force will continue to be upward and the cylinder will continue to gain velocity. After this instant the cylinder will continue to rise, but the net force will be downward so that the cylinder will begin slowing down. Eventually the cylinder will come to rest and the net downward force will cause it to start descending once more. It will continue descending until the net force is again 0, at which the time it will have a downward velocity that will carry it beyond this point until it again comes to rest and the cycle will start over again.
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RESPONSE --> this was another tough assignment, i guess they all will be from here on out self critique assessment: 1
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