#$&* course Phy 121 October 24, 2011 at 4:25 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but presumably the tabletop is strong enough to support the block and so exerts exactly enough force, 49 Newtons upward, to support the block. The total of this supporting force and the gravitational force is zero. The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by any force acting on the two mass system, so we have a system of total mass 7 kg subject to a net force of 19.6 Newtons. The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8 meters/second ^ 2. STUDENT QUESTION since it is a frictionless system i thought the answer would the acceleration of gravity since that would be the force pulling the objects down. INSTRUCTOR RESPONSE A mass resting by itself on a level tabletop does now accelerate downward in response to the gravitational force exerted on it. This is because the tabletop pushes up on it. This occurs whether friction is high, low or absent. Of course if friction is very low, you have to be sure the tabletop is very nearly level, and if friction is absent you'd best be sure it is completely level. These forces are still present if you add the hanging mass to the system. The mass on the tabletop does experience the downward pull of gravity, but that force is balanced by the supporting force of the tabletop. The mass therefore does not move in the direction of the gravitational pull. It is nevertheless part of the system being accelerated by the gravitational force on the hanging mass. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA ------------------------------------------------ Self-critique rating: NA ********************************************* Question: `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5 kg * 9.8 N = 49 N * .10 = 4.9 N (friction) a = 19.6 N - 4.9 N = 14.7 N / 7 kg = 2.1 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object. The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA ------------------------------------------------ Self-critique rating: NA ********************************************* Question: `q003. Answer the same question as that of the preceding problem, but this time assume that the 5 kg object is not on a level tabletop but on an incline at an angle of 12 degrees, and with the incline descending in the direction of the pulley. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 12 deg + 270 deg = 282 deg (counterclockwise from the x axis) 5 kg * 9.8 N = 49 N 2 kg * 9.8 N = 19.6 N Now I am not sure where to go… confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Okay, now I am back and think I see more clearly now some of the holes in my thinking before: x component = 49 N * cos(282 deg) = 10.19 N = 10 N y component = 49 N * sin(282 deg) = -47.93 N = -48 N -48 N * .10 = -4.8 N (friction) 10 N - 48 N = 5.2 N 2 kg * 9.8 N = 19.6 N Fnet = 19.6 N + 5.2 N = 24.8 N a = 24.8 N / 7 kg = 3.54 m/s^2
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Given Solution: In this case you should have drawn the incline with the x axis pointing down the incline and the y axis perpendicular to the incline. Thus the x axis is directed 12 degrees below horizontal. As a result the weight vector, rather than being directed along the negative y axis, lies in the fourth quadrant of the coordinate system at an angle of 12 degrees with the negative y axis. So the weight vector makes an angle of 270 degrees + 12 degrees = 282 degrees with the positive x axis. The weight vector, which has magnitude 5 kg * 9.8 meters/second ^ 2 = 49 Newtons, therefore has x component 49 Newtons * cosine (282 degrees) = 10 Newtons approximately. Its y component is 49 Newtons * sine (282 degrees) = -48 Newtons, approximately. The incline exerts sufficient force that the net y component of the force on the block is zero. The incline therefore exerts a force of + 48 Newtons. Friction exerts a force which is .10 of this force, or .10 * 48 Newtons = 4.8 Newtons opposed to the direction of motion. Assuming that the direction of motion is down the incline, frictional force is therefore -4.8 Newtons in the x direction. The weight component in the x direction and the frictional force in this direction therefore total 10 Newtons + (-4.8 Newtons) = + 5.2 Newtons. This force tends to accelerate the system in the same direction as does the weight of the 2 kg mass. This results in a net force of 5.2 Newtons + 19.6 Newtons = 24.8 Newtons on the 7 kg system. The system therefore accelerates at rate a = (24.8 Newtons) / (7 kg) = 3.5 meters/second ^ 2, approximately. STUDENT COMMENT: i get confused every time about which angle to use and how to determine it, where did 270 deg come from, i used the right equations and was on to the right answer but used the wrong numbers in the equations, but i had the right idea self critique rating: 1 INSTRUCTOR RESPONSE: The negative y axis lies at 270 deg from the positive x axis, as measured counterclockwise. When the coordinate system is rotated 12 degrees in the manner described, the weight vector stays where it is, but the y negative axis swings 'back' 12 degrees and the angle of the weight vector becomes 282 degrees. MORE EXTENSIVE EXPLANATION The weight is in the downward vertical direction, which matches the direction of the original vertical-horizontal coordinate system. So in the original system the weight vector is at 270 degrees. However the positive x axis of the original coordinate system doesn't match the direction of motion along the incline. It's generally simpler to have the x axis parallel to the direction of motion. To accomplish this we rotate the coordinate system 12 degrees in the clockwise direction. As the coordinate system is rotated, the positive x axis rotates to an orientation 12 degrees below horizontal, and the negative y axis 'swings out' 12 degrees from its original vertical orientation. This leaves the vertical weight vector in the fourth quadrant, 12 degrees from the newly oriented negative y axis. As measure counterclockwise from the positive x axis, the weight vector is now at angle 282 degrees. The first figure below depicts a weight vector with its initial point at the origin of an x-y coordinate system in standard vertical-horizontal orientation. It should be clear that the vector is at angle 270 degrees, as measured counterclockwise from the positive x axis. The second figure shows the same weight vector, which is still vertical, but with the coordinate system rotated so that the positive x axis is directed down a 12-degree incline. It should be clear that negative y axis will have rotated 12 degrees from its original position so that the weight vector is now at 282 degrees. STUDENT QUESTION I have read both of the explanatons I am still not sure where the 270 degrees is coming from, how do we calculate this INSTRUCTOR RESPONSE The above discussion is fairly long, and involves an idea that is difficult for many students. The following should provide a convenient summary of the key ideas. If this point-by-point summary doesn't clarify everything for you, please submit a copy of these questions along with inserted comments, questions, etc. (insertions should be marked with &&&&). 1. If the x-y coordinate system is in its 'standard orientation', with the x axis horizontal, pointing to the right, and the y axis vertical and upward, then gravity pulls straight down, along the negative y axis. 2. The negative y axis of this system is at 270 degrees as measured counterclockwise from the positive x axis. 3. If the coordinate system is rotated 12 degrees in the clockwise direction, then the x axis now points slightly below horizontal, and the y axis is 12 degrees from vertical. 4. The gravitational force is still vertical, so the gravitational force is no longer in the direction of the negative y axis. It is now directed at an angle of 12 degrees from the negative y axis. 5. The gravitational force is no longer at 270 degrees, since the rotation of the system has rotated the y axis away from vertical. 6. The gravitational force is now in the fourth quadrant of the rotated system, directed at 282 degrees. STUDENT QUESTION i am really lost on these problems involving forces and angles. i don't understand the concepts of cosine and sin and what the mean with relationship to force. i look at the answer below and i sort of understand how to derive the angle but what i am unclear about is how you draw. it makes sense the way you have it drawn but what if you drew the incline mirrored to what is now? wouldn't that put the force vector in the third quandrant instead of the forth? so how do you know how to correctly draw it. i don't understand what the 49cosine(282) and 49 sin ( 282) actually represents. i know that we use it becase the 49N is the force resting on the incline but I dont understand what the other numbers do. I don't want to just know that I am suppose to use them . I also don't understand how you just know that """"The incline exerts sufficient force that the net y component of the force on the block is zero"""" Can you please explain? INSTRUCTOR RESPONSE You could draw the incline in the opposite direction, with the incline going up and to the right. The x axis would be directed up the incline. If you did this the force vector would be in the third quadrant. The angle would be 258 deg, and the x component of the force would therefore be negative, meaning that the x component of the gravitational force is down the incline (since the x axis points up the incline). Of course objects do tend to accelerate down inclines, so this is exactly what would be expected. A student having completed the prerequisite courses for Phy 201, 231 or 241 should have had enough trigonometry to know what the sine and cosine mean. However it has been observed over many years of teaching that this often not the case. As a result, the vector analysis in this course is built around one simple model, the circular model. This model has four rules, as presented in Introductory Problem Set 5. You should review those rules, which were part of a previous assignment. Briefly stated the rules say the following: A vector of magnitude A directed at angle theta as measured counterclockwise from the positive x axis of a right-hand coordinate system has x component A cos(theta) and y component A sin(theta). A vector with x component R_x and y component R_y has magnitude R = sqrt(R_x^2 + R_y^2) and is directed at angle arcTan(R_y / R_x), plus 180 degrees if R_x is negative. It isn't difficult to make sense of the rules, but for now we will simply apply them and try to make sense of the results. Applying the rules to the weight vector we obtain the x and y components in the given solution. Had the figure been drawn with the incline rising as we move to the right the angle of the weight vector would be 270 deg - 12 deg = 258 deg, as discussed earlier in this response. The components would then be F_x = F cos(theta) = 49 N * cos(258 deg) = -10 N (note the contrast between the given solution, in which the x axis was directed down the incline and the x component of the force was +10 N; in both cases, however, the x component of the gravitational force is directed down the incline) F_y = F sin(theta) = 49 N * sin(258 deg) = -48 N (just as before the y component is negative; in both cases the y component acts perpendicular to the incline, tending to compress or bend the incline inward). These results make sense. The gravitational force is quite a bit closer to perpendicular than parallel to the incline. The component of the gravitational force parallel to the incline is therefore quite a bit greater in magnitude than its component parallel to the incline. If the angle of the incline was increased a little bit the gravitational force would still be closer to perpendicular than to parallel, but not as much as before. The parallel component would increase and the perpendicular component would decrease. If the incline was at 45 degrees, the gravitational force would be just as close to parallel as to perpendicular. The magnitudes of the perpendicular and parallel components would be equal. The angle of the gravitational force would be either 315 degrees or 225 degrees, depending on which way you sketch the incline. You can check to see that this is consistent with the behavior of the sines and cosines: The magnitudes of the sine and cosine of 315 degrees are equal to one another, and are also equal to the magnitudes of the sine and cosine of 225 degrees. As the angle of the incline approaches 90 degrees, the perpendicular component shrinks toward zero while the parallel component approaches the magnitude of the gravitational force. At 90 degrees the component parallel to the incline is equal to the gravitational force, and as a result the object accelerates parallel to the incline with an acceleration equal to that of gravity. The x axis will now be pointing straight up, so that the gravitational force is at angle 180 degrees with respect to the x axis. You can check consistency again: cos(180 deg) = -1 and sin(180 deg) = 0, so that F_x would now be -49 N and F_y would be 0. Finally in response to your question about the sentence 'The incline exerts sufficient force that the net y component of the force on the block is zero.': We assume that the weight doesn't collapse the incline or cause it to bend or compress enough to significantly change its angle of elevation. Since the weight must therefore move along the incline, it doesn't move perpendicular to the incline. It therefore does not accelerated in the direction perpendicular to the incline. It follows that the net force on the weight, in the y direction, is zero. Gravity exerts a force perpendicular to the incline (this is the 'perpendicular component' discussed here), so gravity does exert a nonzero force in the y direction. But the net force in the y direction is zero. So something else is exerting a force in the y direction. That something, in this context, is the incline. Thus the incline exerts sufficient force that the net y component of the force on the block is zero. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. Answer the same question as that of the preceding problem, but this time assume that the 5 kg object is not on a level tabletop but on an incline at an angle of 12 degrees, and with the incline descending in the direction of the pulley. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 12 deg + 270 deg = 282 deg (counterclockwise from the x axis) 5 kg * 9.8 N = 49 N 2 kg * 9.8 N = 19.6 N Now I am not sure where to go… confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Okay, now I am back and think I see more clearly now some of the holes in my thinking before: x component = 49 N * cos(282 deg) = 10.19 N = 10 N y component = 49 N * sin(282 deg) = -47.93 N = -48 N -48 N * .10 = -4.8 N (friction) 10 N - 48 N = 5.2 N 2 kg * 9.8 N = 19.6 N Fnet = 19.6 N + 5.2 N = 24.8 N a = 24.8 N / 7 kg = 3.54 m/s^2
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Given Solution: In this case you should have drawn the incline with the x axis pointing down the incline and the y axis perpendicular to the incline. Thus the x axis is directed 12 degrees below horizontal. As a result the weight vector, rather than being directed along the negative y axis, lies in the fourth quadrant of the coordinate system at an angle of 12 degrees with the negative y axis. So the weight vector makes an angle of 270 degrees + 12 degrees = 282 degrees with the positive x axis. The weight vector, which has magnitude 5 kg * 9.8 meters/second ^ 2 = 49 Newtons, therefore has x component 49 Newtons * cosine (282 degrees) = 10 Newtons approximately. Its y component is 49 Newtons * sine (282 degrees) = -48 Newtons, approximately. The incline exerts sufficient force that the net y component of the force on the block is zero. The incline therefore exerts a force of + 48 Newtons. Friction exerts a force which is .10 of this force, or .10 * 48 Newtons = 4.8 Newtons opposed to the direction of motion. Assuming that the direction of motion is down the incline, frictional force is therefore -4.8 Newtons in the x direction. The weight component in the x direction and the frictional force in this direction therefore total 10 Newtons + (-4.8 Newtons) = + 5.2 Newtons. This force tends to accelerate the system in the same direction as does the weight of the 2 kg mass. This results in a net force of 5.2 Newtons + 19.6 Newtons = 24.8 Newtons on the 7 kg system. The system therefore accelerates at rate a = (24.8 Newtons) / (7 kg) = 3.5 meters/second ^ 2, approximately. STUDENT COMMENT: i get confused every time about which angle to use and how to determine it, where did 270 deg come from, i used the right equations and was on to the right answer but used the wrong numbers in the equations, but i had the right idea self critique rating: 1 INSTRUCTOR RESPONSE: The negative y axis lies at 270 deg from the positive x axis, as measured counterclockwise. When the coordinate system is rotated 12 degrees in the manner described, the weight vector stays where it is, but the y negative axis swings 'back' 12 degrees and the angle of the weight vector becomes 282 degrees. MORE EXTENSIVE EXPLANATION The weight is in the downward vertical direction, which matches the direction of the original vertical-horizontal coordinate system. So in the original system the weight vector is at 270 degrees. However the positive x axis of the original coordinate system doesn't match the direction of motion along the incline. It's generally simpler to have the x axis parallel to the direction of motion. To accomplish this we rotate the coordinate system 12 degrees in the clockwise direction. As the coordinate system is rotated, the positive x axis rotates to an orientation 12 degrees below horizontal, and the negative y axis 'swings out' 12 degrees from its original vertical orientation. This leaves the vertical weight vector in the fourth quadrant, 12 degrees from the newly oriented negative y axis. As measure counterclockwise from the positive x axis, the weight vector is now at angle 282 degrees. The first figure below depicts a weight vector with its initial point at the origin of an x-y coordinate system in standard vertical-horizontal orientation. It should be clear that the vector is at angle 270 degrees, as measured counterclockwise from the positive x axis. The second figure shows the same weight vector, which is still vertical, but with the coordinate system rotated so that the positive x axis is directed down a 12-degree incline. It should be clear that negative y axis will have rotated 12 degrees from its original position so that the weight vector is now at 282 degrees. STUDENT QUESTION I have read both of the explanatons I am still not sure where the 270 degrees is coming from, how do we calculate this INSTRUCTOR RESPONSE The above discussion is fairly long, and involves an idea that is difficult for many students. The following should provide a convenient summary of the key ideas. If this point-by-point summary doesn't clarify everything for you, please submit a copy of these questions along with inserted comments, questions, etc. (insertions should be marked with &&&&). 1. If the x-y coordinate system is in its 'standard orientation', with the x axis horizontal, pointing to the right, and the y axis vertical and upward, then gravity pulls straight down, along the negative y axis. 2. The negative y axis of this system is at 270 degrees as measured counterclockwise from the positive x axis. 3. If the coordinate system is rotated 12 degrees in the clockwise direction, then the x axis now points slightly below horizontal, and the y axis is 12 degrees from vertical. 4. The gravitational force is still vertical, so the gravitational force is no longer in the direction of the negative y axis. It is now directed at an angle of 12 degrees from the negative y axis. 5. The gravitational force is no longer at 270 degrees, since the rotation of the system has rotated the y axis away from vertical. 6. The gravitational force is now in the fourth quadrant of the rotated system, directed at 282 degrees. STUDENT QUESTION i am really lost on these problems involving forces and angles. i don't understand the concepts of cosine and sin and what the mean with relationship to force. i look at the answer below and i sort of understand how to derive the angle but what i am unclear about is how you draw. it makes sense the way you have it drawn but what if you drew the incline mirrored to what is now? wouldn't that put the force vector in the third quandrant instead of the forth? so how do you know how to correctly draw it. i don't understand what the 49cosine(282) and 49 sin ( 282) actually represents. i know that we use it becase the 49N is the force resting on the incline but I dont understand what the other numbers do. I don't want to just know that I am suppose to use them . I also don't understand how you just know that """"The incline exerts sufficient force that the net y component of the force on the block is zero"""" Can you please explain? INSTRUCTOR RESPONSE You could draw the incline in the opposite direction, with the incline going up and to the right. The x axis would be directed up the incline. If you did this the force vector would be in the third quadrant. The angle would be 258 deg, and the x component of the force would therefore be negative, meaning that the x component of the gravitational force is down the incline (since the x axis points up the incline). Of course objects do tend to accelerate down inclines, so this is exactly what would be expected. A student having completed the prerequisite courses for Phy 201, 231 or 241 should have had enough trigonometry to know what the sine and cosine mean. However it has been observed over many years of teaching that this often not the case. As a result, the vector analysis in this course is built around one simple model, the circular model. This model has four rules, as presented in Introductory Problem Set 5. You should review those rules, which were part of a previous assignment. Briefly stated the rules say the following: A vector of magnitude A directed at angle theta as measured counterclockwise from the positive x axis of a right-hand coordinate system has x component A cos(theta) and y component A sin(theta). A vector with x component R_x and y component R_y has magnitude R = sqrt(R_x^2 + R_y^2) and is directed at angle arcTan(R_y / R_x), plus 180 degrees if R_x is negative. It isn't difficult to make sense of the rules, but for now we will simply apply them and try to make sense of the results. Applying the rules to the weight vector we obtain the x and y components in the given solution. Had the figure been drawn with the incline rising as we move to the right the angle of the weight vector would be 270 deg - 12 deg = 258 deg, as discussed earlier in this response. The components would then be F_x = F cos(theta) = 49 N * cos(258 deg) = -10 N (note the contrast between the given solution, in which the x axis was directed down the incline and the x component of the force was +10 N; in both cases, however, the x component of the gravitational force is directed down the incline) F_y = F sin(theta) = 49 N * sin(258 deg) = -48 N (just as before the y component is negative; in both cases the y component acts perpendicular to the incline, tending to compress or bend the incline inward). These results make sense. The gravitational force is quite a bit closer to perpendicular than parallel to the incline. The component of the gravitational force parallel to the incline is therefore quite a bit greater in magnitude than its component parallel to the incline. If the angle of the incline was increased a little bit the gravitational force would still be closer to perpendicular than to parallel, but not as much as before. The parallel component would increase and the perpendicular component would decrease. If the incline was at 45 degrees, the gravitational force would be just as close to parallel as to perpendicular. The magnitudes of the perpendicular and parallel components would be equal. The angle of the gravitational force would be either 315 degrees or 225 degrees, depending on which way you sketch the incline. You can check to see that this is consistent with the behavior of the sines and cosines: The magnitudes of the sine and cosine of 315 degrees are equal to one another, and are also equal to the magnitudes of the sine and cosine of 225 degrees. As the angle of the incline approaches 90 degrees, the perpendicular component shrinks toward zero while the parallel component approaches the magnitude of the gravitational force. At 90 degrees the component parallel to the incline is equal to the gravitational force, and as a result the object accelerates parallel to the incline with an acceleration equal to that of gravity. The x axis will now be pointing straight up, so that the gravitational force is at angle 180 degrees with respect to the x axis. You can check consistency again: cos(180 deg) = -1 and sin(180 deg) = 0, so that F_x would now be -49 N and F_y would be 0. Finally in response to your question about the sentence 'The incline exerts sufficient force that the net y component of the force on the block is zero.': We assume that the weight doesn't collapse the incline or cause it to bend or compress enough to significantly change its angle of elevation. Since the weight must therefore move along the incline, it doesn't move perpendicular to the incline. It therefore does not accelerated in the direction perpendicular to the incline. It follows that the net force on the weight, in the y direction, is zero. Gravity exerts a force perpendicular to the incline (this is the 'perpendicular component' discussed here), so gravity does exert a nonzero force in the y direction. But the net force in the y direction is zero. So something else is exerting a force in the y direction. That something, in this context, is the incline. Thus the incline exerts sufficient force that the net y component of the force on the block is zero. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `q003. Answer the same question as that of the preceding problem, but this time assume that the 5 kg object is not on a level tabletop but on an incline at an angle of 12 degrees, and with the incline descending in the direction of the pulley. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 12 deg + 270 deg = 282 deg (counterclockwise from the x axis) 5 kg * 9.8 N = 49 N 2 kg * 9.8 N = 19.6 N Now I am not sure where to go… confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Okay, now I am back and think I see more clearly now some of the holes in my thinking before: x component = 49 N * cos(282 deg) = 10.19 N = 10 N y component = 49 N * sin(282 deg) = -47.93 N = -48 N -48 N * .10 = -4.8 N (friction) 10 N - 48 N = 5.2 N 2 kg * 9.8 N = 19.6 N Fnet = 19.6 N + 5.2 N = 24.8 N a = 24.8 N / 7 kg = 3.54 m/s^2
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Given Solution: In this case you should have drawn the incline with the x axis pointing down the incline and the y axis perpendicular to the incline. Thus the x axis is directed 12 degrees below horizontal. As a result the weight vector, rather than being directed along the negative y axis, lies in the fourth quadrant of the coordinate system at an angle of 12 degrees with the negative y axis. So the weight vector makes an angle of 270 degrees + 12 degrees = 282 degrees with the positive x axis. The weight vector, which has magnitude 5 kg * 9.8 meters/second ^ 2 = 49 Newtons, therefore has x component 49 Newtons * cosine (282 degrees) = 10 Newtons approximately. Its y component is 49 Newtons * sine (282 degrees) = -48 Newtons, approximately. The incline exerts sufficient force that the net y component of the force on the block is zero. The incline therefore exerts a force of + 48 Newtons. Friction exerts a force which is .10 of this force, or .10 * 48 Newtons = 4.8 Newtons opposed to the direction of motion. Assuming that the direction of motion is down the incline, frictional force is therefore -4.8 Newtons in the x direction. The weight component in the x direction and the frictional force in this direction therefore total 10 Newtons + (-4.8 Newtons) = + 5.2 Newtons. This force tends to accelerate the system in the same direction as does the weight of the 2 kg mass. This results in a net force of 5.2 Newtons + 19.6 Newtons = 24.8 Newtons on the 7 kg system. The system therefore accelerates at rate a = (24.8 Newtons) / (7 kg) = 3.5 meters/second ^ 2, approximately. STUDENT COMMENT: i get confused every time about which angle to use and how to determine it, where did 270 deg come from, i used the right equations and was on to the right answer but used the wrong numbers in the equations, but i had the right idea self critique rating: 1 INSTRUCTOR RESPONSE: The negative y axis lies at 270 deg from the positive x axis, as measured counterclockwise. When the coordinate system is rotated 12 degrees in the manner described, the weight vector stays where it is, but the y negative axis swings 'back' 12 degrees and the angle of the weight vector becomes 282 degrees. MORE EXTENSIVE EXPLANATION The weight is in the downward vertical direction, which matches the direction of the original vertical-horizontal coordinate system. So in the original system the weight vector is at 270 degrees. However the positive x axis of the original coordinate system doesn't match the direction of motion along the incline. It's generally simpler to have the x axis parallel to the direction of motion. To accomplish this we rotate the coordinate system 12 degrees in the clockwise direction. As the coordinate system is rotated, the positive x axis rotates to an orientation 12 degrees below horizontal, and the negative y axis 'swings out' 12 degrees from its original vertical orientation. This leaves the vertical weight vector in the fourth quadrant, 12 degrees from the newly oriented negative y axis. As measure counterclockwise from the positive x axis, the weight vector is now at angle 282 degrees. The first figure below depicts a weight vector with its initial point at the origin of an x-y coordinate system in standard vertical-horizontal orientation. It should be clear that the vector is at angle 270 degrees, as measured counterclockwise from the positive x axis. The second figure shows the same weight vector, which is still vertical, but with the coordinate system rotated so that the positive x axis is directed down a 12-degree incline. It should be clear that negative y axis will have rotated 12 degrees from its original position so that the weight vector is now at 282 degrees. STUDENT QUESTION I have read both of the explanatons I am still not sure where the 270 degrees is coming from, how do we calculate this INSTRUCTOR RESPONSE The above discussion is fairly long, and involves an idea that is difficult for many students. The following should provide a convenient summary of the key ideas. If this point-by-point summary doesn't clarify everything for you, please submit a copy of these questions along with inserted comments, questions, etc. (insertions should be marked with &&&&). 1. If the x-y coordinate system is in its 'standard orientation', with the x axis horizontal, pointing to the right, and the y axis vertical and upward, then gravity pulls straight down, along the negative y axis. 2. The negative y axis of this system is at 270 degrees as measured counterclockwise from the positive x axis. 3. If the coordinate system is rotated 12 degrees in the clockwise direction, then the x axis now points slightly below horizontal, and the y axis is 12 degrees from vertical. 4. The gravitational force is still vertical, so the gravitational force is no longer in the direction of the negative y axis. It is now directed at an angle of 12 degrees from the negative y axis. 5. The gravitational force is no longer at 270 degrees, since the rotation of the system has rotated the y axis away from vertical. 6. The gravitational force is now in the fourth quadrant of the rotated system, directed at 282 degrees. STUDENT QUESTION i am really lost on these problems involving forces and angles. i don't understand the concepts of cosine and sin and what the mean with relationship to force. i look at the answer below and i sort of understand how to derive the angle but what i am unclear about is how you draw. it makes sense the way you have it drawn but what if you drew the incline mirrored to what is now? wouldn't that put the force vector in the third quandrant instead of the forth? so how do you know how to correctly draw it. i don't understand what the 49cosine(282) and 49 sin ( 282) actually represents. i know that we use it becase the 49N is the force resting on the incline but I dont understand what the other numbers do. I don't want to just know that I am suppose to use them . I also don't understand how you just know that """"The incline exerts sufficient force that the net y component of the force on the block is zero"""" Can you please explain? INSTRUCTOR RESPONSE You could draw the incline in the opposite direction, with the incline going up and to the right. The x axis would be directed up the incline. If you did this the force vector would be in the third quadrant. The angle would be 258 deg, and the x component of the force would therefore be negative, meaning that the x component of the gravitational force is down the incline (since the x axis points up the incline). Of course objects do tend to accelerate down inclines, so this is exactly what would be expected. A student having completed the prerequisite courses for Phy 201, 231 or 241 should have had enough trigonometry to know what the sine and cosine mean. However it has been observed over many years of teaching that this often not the case. As a result, the vector analysis in this course is built around one simple model, the circular model. This model has four rules, as presented in Introductory Problem Set 5. You should review those rules, which were part of a previous assignment. Briefly stated the rules say the following: A vector of magnitude A directed at angle theta as measured counterclockwise from the positive x axis of a right-hand coordinate system has x component A cos(theta) and y component A sin(theta). A vector with x component R_x and y component R_y has magnitude R = sqrt(R_x^2 + R_y^2) and is directed at angle arcTan(R_y / R_x), plus 180 degrees if R_x is negative. It isn't difficult to make sense of the rules, but for now we will simply apply them and try to make sense of the results. Applying the rules to the weight vector we obtain the x and y components in the given solution. Had the figure been drawn with the incline rising as we move to the right the angle of the weight vector would be 270 deg - 12 deg = 258 deg, as discussed earlier in this response. The components would then be F_x = F cos(theta) = 49 N * cos(258 deg) = -10 N (note the contrast between the given solution, in which the x axis was directed down the incline and the x component of the force was +10 N; in both cases, however, the x component of the gravitational force is directed down the incline) F_y = F sin(theta) = 49 N * sin(258 deg) = -48 N (just as before the y component is negative; in both cases the y component acts perpendicular to the incline, tending to compress or bend the incline inward). These results make sense. The gravitational force is quite a bit closer to perpendicular than parallel to the incline. The component of the gravitational force parallel to the incline is therefore quite a bit greater in magnitude than its component parallel to the incline. If the angle of the incline was increased a little bit the gravitational force would still be closer to perpendicular than to parallel, but not as much as before. The parallel component would increase and the perpendicular component would decrease. If the incline was at 45 degrees, the gravitational force would be just as close to parallel as to perpendicular. The magnitudes of the perpendicular and parallel components would be equal. The angle of the gravitational force would be either 315 degrees or 225 degrees, depending on which way you sketch the incline. You can check to see that this is consistent with the behavior of the sines and cosines: The magnitudes of the sine and cosine of 315 degrees are equal to one another, and are also equal to the magnitudes of the sine and cosine of 225 degrees. As the angle of the incline approaches 90 degrees, the perpendicular component shrinks toward zero while the parallel component approaches the magnitude of the gravitational force. At 90 degrees the component parallel to the incline is equal to the gravitational force, and as a result the object accelerates parallel to the incline with an acceleration equal to that of gravity. The x axis will now be pointing straight up, so that the gravitational force is at angle 180 degrees with respect to the x axis. You can check consistency again: cos(180 deg) = -1 and sin(180 deg) = 0, so that F_x would now be -49 N and F_y would be 0. Finally in response to your question about the sentence 'The incline exerts sufficient force that the net y component of the force on the block is zero.': We assume that the weight doesn't collapse the incline or cause it to bend or compress enough to significantly change its angle of elevation. Since the weight must therefore move along the incline, it doesn't move perpendicular to the incline. It therefore does not accelerated in the direction perpendicular to the incline. It follows that the net force on the weight, in the y direction, is zero. Gravity exerts a force perpendicular to the incline (this is the 'perpendicular component' discussed here), so gravity does exert a nonzero force in the y direction. But the net force in the y direction is zero. So something else is exerting a force in the y direction. That something, in this context, is the incline. Thus the incline exerts sufficient force that the net y component of the force on the block is zero. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!
#$&* course Phy 121 October 24, 2011 at 5:52 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** If we add the x components of the original two vectors we get the x component of the resultant. If we add the y components of the original two vectors we get the y component of the resultant. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA ------------------------------------------------ Self-critique rating: NA ********************************************* Question: `qExplain how we get the components of a vector from its angle and magnitude. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To get the x component, multiply the magnitude by the cos(angle) To get the y component, multiply the magnitude by the sin(angle) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis). To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). ** NOTE REGARDING THE CIRCULAR DEFINITION OF TRIGONOMETRIC FUNCTIONS VS. THE RIGHT-TRIANGLE DEFINITIONS: Students with a background in trigonometry often use the right-triangle definitions of sine and cosine (sine and cosine defined in terms of opposite and adjacent sides and hypotenuse), as opposed to the circular definition (using a coordinate system, with angles measured counterclockwise from the positive x axis--the definition used in this course). The two definitions are pretty much equivalent and completely consistent. The circular definition is a bit more general for two reasons: • The circular definition can be applied to positive or negative angles, and to angles greater that 180 degrees, whereas triangles are limited to positive angles less than 180 degrees. • The circular definition can yield positive or negative components, whereas the sides of triangles are all positive. In most applications it is your choice which definition you use. Some applications are easier if you use the right-triangle definition, others are easier of you use the circular definition, and some simply require the circular definition. In developing this course I chose to express all trigonometric solutions in terms of one of the definitions, in order to avoid confusion for students with a weak background in trigonometry. If only one of the definitions is to be used, it must be the more general circular definition with its four simple rules • (x coordinate = magnitude * cos(angle), • y coordinate = magnitude * sin(angle), • magnitude = sqrt ( (x coordinate)^2 + (y coordinate)^2 ), • angle = arcTan ( y coord / x coord), plus 180 deg or pi rad if x coord is negative). The circular definition is sufficient for Principles of Physics or General College Physics.. However General College Physics students are to have completed a year of precalculus or equivalent, which includes trigonometry, and it is expected that these students can reconcile the circular and right-triangle definitions and approaches, and understand the right-angle trigonometry in their text. University Physics students are of course expected to already be familiar with trigonometry and the use of vectors (though in reality some refreshing is usually required, and is provided in the first chapter of the University Physics text). However students at the level of University Physics should encounter no serious obstacle with the trigonometry. In a nutshell, here is a summary of how the right-triangle definitions are related to the circular definitions: On a circle of radius r centered at the origin, any first- or second-quadrant angle gives us a triangle in the upper half-plane having that base angle. • The hypotenuse of this triangle is r, • the adjacent side is the x coordinate r cos(theta), and • the opposite side is the y coordinate r sin(theta). Thus • adjacent side / hypotenuse = r cos(theta) / r = cos(theta), • opposite side / hypotenuse = r sin(theta) / r = sin(theta), and • opposite side / adjacent side = r sin(theta) / ( r cos(theta) ) = sin(theta) / (cos(theta)) = tan(theta). The definitions of the cosecant, secant and cotangent functions are then made in the usual manner. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA ------------------------------------------------ Self-critique rating: NA ********************************************* Question: `qprin phy, gen phy 7.02. Const frict force of 25 N on a 65 kg skiier for 20 sec; what is change in vel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 25 N * 20 s = 500 N ‘dv = 500 N / 65 kg = 7.7 m/s^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf the direction of the velocity is taken to be positive, then the directio of the frictional force is negative. A constant frictional force of -25 N for 20 sec delivers an impulse of -25 N * 20 sec = -500 N sec in the direction opposite the velocity. By the impulse-momentum theorem we have impulse = change in momentum, so the change in momentum must be -500 N sec. The change in momentum is m * `dv, so we have m `dv = impulse and `dv = impulse / m = -500 N s / (65 kg) = -7.7 N s / kg = -7.7 kg m/s^2 * s / kg = -7.7 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA ------------------------------------------------ Self-critique rating: NA ********************************************* Question: `qgen phy #7.12 23 g bullet 230 m/s 2-kg block emerges at 170 m/s speed of block YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a**STUDENT SOLUTION: Momentum conservation gives us m1 v1 + m2 v2 = m1 v1' + m2 v2' so if we let v2' = v, we have (.023kg)(230m/s)+(2kg)(0m/s) = (.023kg)(170m/s)+(2kg)(v). Solving for v: (5.29kg m/s)+0 = (3.91 kg m/s)+(2kg)(v) .78kg m/s = 2kg * v v = 1.38 kg m/s / (2 kg) = .69 m/s. INSTRUCTOR COMMENT: It's probably easier to solve for the variable v2 ': Starting with m1 v1 + m2 v2 = m1 v1 ' + m2 v2 ' we add -m1 v1 ' to both sides to get m1 v1 + m2 v2 - m1 v1 ' = m2 v2 ', then multiply both sides by 1 / m2 to get v2 ` = (m1 v1 + m2 v2 - m1 v1 ' ) / m2. Substituting for m1, v1, m2, v2 we will get the result you obtained.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q**** Univ. 8.75 (11th edition 8.70) (8.68 10th edition). 8 g bullet into .992 kg block, compresses spring 15 cm. .75 N force compresses .25 cm. Vel of block just after impact, vel of bullet? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The spring ideally obeys Hook's Law F = k x. It follows that k = .75 N / .25 cm = 3 N / cm. At compression 15 cm the potential energy of the system is PE = .5 k x^2 = .5 * 3 N/cm * (15 cm)^2 = 337.5 N cm, or 3.375 N m = 3.375 Joules, which we round to three significant figures to get 3.38 J. The KE of the 1 kg mass (block + bullet) just after impact is in the ideal case all converted to this PE so the velocity of the block was v such that .5 m v^2 = PE, or v = sqrt(2 PE / m) = sqrt(2 * 3.38 J / ( 1 kg) ) = 2.6 m/s, approx. The momentum of the 1 kg mass was therefore 2.6 m/s * .992 kg = 2.6 kg m/s, approx., just after collision with the bullet. Just before collision the momentum of the block was zero so by conservation of momentum the momentum of the bullet was 2.6 kg m/s. So we have mBullet * vBullet = 2.6 kg m/s and vBullet = 2.6 kg m/s / (.008 kg) = 330 m/s, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: