#$&* course PHY 241 8:39; Nov 24, 2010 011. Note that there are 12 questions in this set.
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Given Solution: At the surface of the Earth gravity, which is observed here near the surface to accelerate freely falling objects at 9.8 m/s^2, must exert a force of 2 kg * 9.8 m/s^2 = 19.6 Newtons on the hanging 2 kg mass. This force will tend to accelerate the system consisting of the cart, rope, hanger and suspended mass, in the 'forward' direction--the direction in which the various components of the system must accelerate if the hanging mass is to accelerate in the downward direction. This is the only force accelerating the system in this direction. All other forces, including the force of gravity pulling the cart and the remaining masses downward and the normal force exerted by the level surface to prevent gravity from accelerating the cart downward, are in a direction perpendicular to the motion of these components of the system; furthermore these forces are balanced so that they add up to 0. The mass of the system is that of the 30 kg cart plus that of the ten 2 kg masses, a total of 50 kg. The net force of 19.6 Newtons exerted on a 50 kg mass therefore results in acceleration a = Fnet / m = 19.6 Newtons / (50 kg) = 19.6 kg m/s^2 / (50 kg) = .392 m/s^2. STUDENT QUESTION: Our answers are close, but wouldn�t the 50kg actually by 48kg since the 2kg weight was taken out of the kart? INSTRUCTOR RESPONSE: The 2 kg weight is still part of the mass that's being accelerated. It's been moved from the cart to the hanger, but it's still there. All 50 kg are being accelerated by that net force. STUDENT QUESTION Am I allowed to divide Newtons by Kg? Or do I have to change the Newtons to kg*m/s^2? INSTRUCTOR RESPONSE You need to reduce everything to fundamental units. How would you know what N / kg is unless they are both expressed in compatible units? You could of course memorize the fact that N / kg gives you m/s^2, along with about 200 other shortcuts, but that would be a waste of time and wouldn't contribute much to your insight or your ability to work out units in unfamiliar situations. You can count the number of fundamental units in all of physics on one hand. If you know the definitions of the quantities, this makes it very easy to deal with questions of units. Unit calculations come down to the simple algebra of multiplying and dividing fractions, whose numerators and denominators are just products and powers of a few simple units. STUDENT QUESTION Does it matter if you define aGravity as 9.8 m/s^2 or -9.8 m/s^2? Would -.39m/s^2 still be correct since we just know it's perpendicular, and not which perpendicular direction is positive? INSTRUCTOR RESPONSE: In your solution you said 'Fnet = 2kg * -9.8m/s^2 = -19.6 N in the downward direction' You weren't completely specific about what the - sign meant and what direction is positive. If the upward direction is considered positive, then we would simply say that the gravitational force is -19.6 N. We wouldn't add 'in the downward direction' because having declared upward as positive, the negative sign already tells us that the force is downward. If we were to say that the force is 19.6 N in the downward direction, this would be a true statement, without the use of the -sign. However this system doesn't move in just the up-down direction, and we have to be careful how we define our positive direction: The signs of the displacement, velocity and acceleration depend on the direction we choose as positive. We consider the system to consist of the cart and the hanging weights. Parts of this system are moving in the vertical direction and parts are moving in the horizontal direction, so neither vertical nor horizontal can be regarded as the positive direction. Let's assume that we are oriented so that from our position the cart moves to the right as the hanging weights descend. If the cart moves to the right, the hanging weights move downward; if the cart moves to the left the hanging weight move upward. We can describe these motions as 'right-down' and 'left-up'. We have to declare our choice of positive direction. We can choose either 'right-down' or 'left-up' to be positive, whichever we find more convenient; having made this choice the opposite direction will be regarded as negative. Now gravity will pull the system in the 'right-down' direction. You might prefer to regard the gravitational force as negative, in which case you will choose 'left-up' as the positive direction. You have implicitly done so by making your acceleration negative. So choosing 'left-up' as the positive direction the force exerted by gravity on the hanging mass, which pulls the system in the 'right-down' direction, will be negative. The resulting acceleration will in this case be a = F_net / m = -19.6 N / (50 kg) = -.392 m/s^2. This means that acceleration will be opposite the direction of motion--i.e., in the 'right-down' direction. It would have been equally valid to choose 'right-down' as the positive direction. The gravitational force on the hanging mass would be +19.6 N and the acceleration +.392 N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q002. Two 1-kg masses are suspended over a pulley, one on either side. A 100-gram mass is added to the 1-kg mass on one side of the pulley. How much force does gravity exert on each side of the pulley, and what is the net force acting on the entire 2.1 kg system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: One side = 1.1kg * 9.81m/s^2 = 10.791N On the other side the F = 9.81N => the F_net = 10.791N 9.81N = .981 N **It is much simpler to note that the Diff in weight affected = .1 kg .1 kg * 9.81 m/s^2 = F_net = again .981 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The 1-kg mass experiences a force of F = 1 kg * 9.8 m/s^2 = 9.8 Newtons. The other side has a total mass of 1 kg + 100 grams = 1 kg + .1 kg = 1.1 kg, so it experiences a force of F = 1.1 kg * 9.8 Newtons = 10.78 Newtons. Both of these forces are downward, so it might seem that the net force on the system would be 9.8 Newtons + 10.78 Newtons = 20.58 Newtons. However this doesn't seem quite right, because when one mass is pulled down the other is pulled up so in some sense the forces are opposing. It also doesn't make sense because if we had a 2.1 kg system with a net force of 20.58 Newtons its acceleration would be 9.8 m/s^2, since we know very well that two nearly equal masses suspended over a pulley won't both accelerate downward at the acceleration of gravity. So in this case we take note of the fact that the two forces are indeed opposing, with one tending to pull the system in one direction and the other in the opposite direction. We also see that we have to abandon the notion that the appropriate directions for motion of the system are 'up' and 'down'. We instead take the positive direction to be the direction in which the system moves when the 1.1 kg mass descends. We now see that the net force in the positive direction is 10.78 Newtons and that a force of 9.8 Newtons acts in the negative direction, so that the net force on the system is 10.78 Newtons - 9.8 Newtons =.98 Newtons. The net force of .98 Newtons on a system whose total mass is 2.1 kg results in an acceleration of .98 Newtons / (2.1 kg) = .45 m/s^2, approx.. Thus the system accelerates in the direction of the 1.1 kg mass at .45 m/s^2. Additional note on + and - directions: One force tends to accelerate the system in one direction, the other tends to accelerate it in the opposite direction. So you need to choose a positive direction and put a + or - sign on each force, consistent with your chosen positive direction. The positive direction can't be 'up' or 'down', since part of the system moves up while another part moves down. The easiest way to specify a positive direction is to specify the direction of one of the masses. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK there is more in the soln than the question! ------------------------------------------------ self-critique rating #$&*: K ********************************************* Question: `q003. If in the previous question there is friction in the pulley, as there must be in any real-world pulley, the system in the previous problem will not accelerate at the rate calculated there. Suppose that the pulley exerts a retarding frictional force on the system which is equal in magnitude to 1% of the weight of the system. In this case what will be the acceleration of the system, assuming that it is moving in the positive direction (as defined in the previous exercise)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet from above = .981 N F_fric = weight*mue = 2.1kg * 9.81m/s^2 * .01 = .206 N Fnet = F_net_grav F_fric = .981 N - .206 N = .775N A = Fnet / mass_total = .775N / 2.3 kg = .369 m/s^2 = 36.9 cm/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We first determine the force exerted a friction. The weight of the system is the force exerted by gravity on the mass of the system. The system has mass 2.1 kg, so the weight of the system must be 2.1 kg * 9.8 m/s^2 = 20.58 Newtons. 1% of this weight is .21 Newtons, rounded off to two significant figures. This force will be exerted in the direction opposite to that of the motion of the system; since the system is assume to be moving in the positive direction the force exerted by friction will be frictional force = -.21 Newtons. The net force exerted by the system will in this case be 10.78 Newtons - 9.8 Newtons - .21 Newtons = .77 Newtons, in contrast with the .98 Newton net force of the original exercise. The acceleration of the system will be .77 Newtons / (2.1 kg) = .37 m/s^2, approx.. STUDENT SOLUTION (this solution and the instructor's commentary address a common error in expression and in thinking this is worth a look the main topic is why it's not appropriate to write an expression like .98N-.21N=.77N/2.1kg=.37m/s^2 Using my numbers from the previous problem, we had .98N pulling down, so I will subtract an additional .21 from that to get the Newtons pulling down w/ friction. .98N-.21N=.77N/2.1kg=.37m/s^2. INSTRUCTOR COMMENTARY It's clear what you mean by .98N-.21N=.77N/2.1kg=.37m/s^2, and everything you said up to this point is very good and correct. However as a mathematical statement .98N-.21N=.77N/2.1kg=.37m/s^2 is incorrect. If .98N-.21N=.77N/2.1kg=.37m/s^2, then since quantities that are both equal to a third quantity are equal to one another, .98N-.21N = .37m/s^2. However N and m/s^2 are complete different units, so the left- and right-hand sides of this equality are unlike terms. Unlike terms can't possibly be equal. And of course it's very obvious that .98N-.21N =. .37m/s^2 is simply an untrue statement. Untrue statements tend to lead to confusion, e.g., when you review your work and don't necessarily remember what you were thinking when you wrote the thing down, or when your statement is viewed by someone else who doesn't already know what to expect. If you said F_net = .98N-.21N=.77N so a = F_net / m = .77 N / (2.1kg) =.37m/s^2 then you would not have any false statements in your solution, your solution would be clear to anyone who understands Newton's Second Law, and would be much more likely useful to you when reviewing your work. (Note also that N / kg = (kg m/s^2) / kg = m/s^2 It's important to maintain the habit of reducing units to fundamental units and doing the algebra of the units.) STUDENT QUESTION &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q004. Why was it necessary in the previous version of the exercise to specify that the system was moving in the direction of the 1.1 kg mass. Doesn't the system have to move in that direction? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: no what if the system was being analyzed after the other side was moving down. Then the acceleration is still the same but it is a negative quantity. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If the system is released from rest, since acceleration is in the direction of the 1.1 kg mass its velocity will certainly always be positive. However, the system doesn't have to be released from rest. We could give a push in the negative direction before releasing it, in which case it would continue moving in the negative direction until the positive acceleration brought it to rest for an instant, after which it would begin moving faster and faster in the positive direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I said the same thing perhaps not as clear though. This is what I was trying to say! ------------------------------------------------ self-critique rating #$&*: K ********************************************* Question: `q005. If the system of the preceding series of exercises is initially moving in the negative direction, then including friction in the calculation what is its acceleration? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration is still constant but according to the direction originally given the initial v would be negative. We could switch this but it must remain constant throughout the problem. That is accel is neg, v is pos, and displacement is pos until the system comes to momentary rest. The displacement would be negative if we set rest as the new zero. We must also take into consideration that the Friction is acting /w the accel until the system comes to rest. Of course the accel is greater bc F_net = .981N+.206N=1.187N; If we divide by the mass of the system we get the new accel = .565 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Its acceleration will be due to the net force. This net force will include the 10.78 Newton force in the positive direction and the 9.8 Newton force in the negative direction. It will also include a frictional force of .21 Newtons in the direction opposed to motion. Since motion is in the negative direction, the frictional force will therefore be in the positive direction. The net force will thus be Fnet = 10.78 Newtons - 9.8 Newtons + .21 Newtons = +1.19 Newtons, in contrast to the +.98 Newtons obtained when friction was neglected and the +.77 Newtons obtained when the system was moving in the positive direction. , The acceleration of the system is therefore a = 1.19 N / (2.1 kg) = .57 m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK forgot about Ffric but fixed that totally understand what is going on! ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q006. If friction is neglected, what will be the result of adding 100 grams to a similar system which originally consists of two 10-kg masses, rather than the two 1-kg masses in the previous examples? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F_net still = .981 N But a = F_net/mt = .981 N/20.1kg = about .049 m/s^2 = 4.9 cm/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3+
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Given Solution: In this case the masses will be 10.1 kg and 10 kg. The force on the 10.1 kg mass will be 10.1 kg * 9.8 m/s^2 = 98.98 Newtons and the force on the 10 kg mass will be 10 kg * 9.8 m/s^2 = 98 Newtons. The net force will therefore be .98 Newtons, as it was in the previous example where friction was neglected. We note that this.98 Newtons is the result of the additional 100 gram mass, which is the same in both examples. The total mass of the system is 10 kg + 10.1 kg = 20.1 kg, so that the acceleration of the system is a = .98 Newtons / 20.1 kg = .048 m/s^2, approx.. Comparing this with the preceding situation, where the net force was the same (.98 N) but the total mass was 2.1 kg, we see that the same net force acting on the significantly greater mass results in significantly less acceleration. Note on the direction of the frictional force: It's not quite accurate to say that the frictional force is always in the direction opposite motion. I'm not really telling you the whole story here--trying to keep things simple. Friction can indeed speed things up, depending on your frame of reference. The more accurate statement is that forces exerted by kinetic friction act in the direction opposite the relative motion of the two surfaces. (Forces exerted by static friction act in the direction opposite the sum of all other forces). For example a concrete block, free to slide around in the bed of a pickup truck which is accelerating forward, is accelerated by the frictional force between it and the truckbed. So the frictional force is in its direction of motion. If the block doesn't slide, it is static friction that accelerates it and there is no relative motion between the surfaces of the block and the truckbed. If the block does slide, the frictional force is still pushing it forward relative to the road, and relative to the road it accelerates in its direction of motion, but the frictional force isn't sufficient to accelerate it at the same rate as the truck; it therefore slides backward relative to the truckbed. Relative to the truckbed the block slides backward while the frictional force pushes it forward--the frictional force is in the direction opposite the relative motion. If the block is sliding, it is moving toward the back of the truck while friction is pushing it toward the front. So in this case the frictional force acts in the direction opposite the relative motion of the two surfaces. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q007. If friction is not neglected, what will be the result for the system with the two 10-kg masses with .1 kg added to one side? Note that by following what has gone before you could, with no error and through no fault of your own, possibly get an absurd result here, which will be repeated in the explanation then resolved at the end of the explanation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F_net = F_net_grav F_fric = about 1 N => the system is not moving bc the friction is creating a creating a greater force than the change in weight from one side to the other. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If friction is still equal to 1% of the total weight of the system, which in this case is 20.1 kg * 9.8 m/s^2 = 197 Newtons, then the frictional force will be .01 * 197 Newtons = 1.97 Newtons. This frictional force will oppose the motion of the system. For the moment assume the motion of the system to be in the positive direction. This will result in a frictional force of -1.97 Newtons. The net force on the system is therefore 98.98 Newtons - 98 Newtons - 1.97 Newtons = -.99 Newtons. This net force is in the negative direction, opposite to the direction of the net gravitational force. If the system is moving this is perfectly all right--the frictional force being greater in magnitude than the net gravitational force, the system can slow down. Suppose the system is released from rest. Then we might expect that as a result of the greater weight on the positive side it will begin accelerating in the positive direction. However, if it moves at all the frictional force would result in a -.99 Newton net force, which would accelerate it in the negative direction and very quickly cause motion in that direction. Of course friction can't do this--its force is always exerted in a direction opposite to that of motion--so friction merely exerts just enough force to keep the object from moving at all. Friction acts as though it is quite willing to exert any force up to 1.97 Newtons to oppose motion, and up to this limit the frictional force can be used to keep motion from beginning. In fact, the force that friction can exert to keep motion from beginning is usually greater than the force it exerts to oppose motion once it is started. STUDENT QUESTION I understand how we got the 197 newtons, but I do not understand why we need to find the 1 percent, Is friction always 1 percent of of the total weight of the object, or were we just to assume from the previous problem?????? I understand why it is negative and how we get 1.97 newtons INSTRUCTOR RESPONSE Friction is a percent of the force 'pressing' two objects together; the percent depends on the nature of the two surfaces (e.g., ice sliding over smooth plastic will be a small percent, while rubber sliding over asphalt is much higher). That percent is usually called the 'coefficient of friction', and is generally expressed in decimal form (e.g., a 1% coefficient of friction would be .01). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK I understand or at least all I know is not diff from the above statement. ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q008. A cart on an incline is subject to the force of gravity. Depending on the incline, some of the force of gravity is balanced by the incline. On a horizontal surface, the force of gravity is completely balanced by the upward force exerted by the incline. If the incline has a nonzero slope, the gravitational force (i.e., the weight of the object) can be thought of as having two components, one parallel to the incline and one perpendicular to the incline. The incline exerts a force perpendicular to itself, and thereby balances the weight component perpendicular to the incline. The weight component parallel to the incline is not balanced, and tends to accelerate the object down the incline. Frictional forces tend to resist this parallel component of the weight and reduce or eliminate the acceleration. A complete analysis of these forces is best done using the techniques of vectors, which will be encountered later in the course. For now you can safely assume that for small slopes (less than .1) the component of the gravitational force parallel to the incline is very close to the product of the slope and the weight of the object. [If you remember your trigonometry you might note that the exact value of the parallel weight component is the product of the weight and the sine of the angle of the incline, that for small angles the sine of the angle is equal to the tangent of the angle, and that the tangent of the angle of the incline is the slope. The product of slope and weight is therefore a good approximation for small angles or small slopes.] What therefore would be the component of the gravitational force acting parallel to an incline with slope .07 on a cart of mass 3 kg? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: slope * m * a = F_grav [ if slope is < .1] = .07 * 3 kg * 9.81 = 2.060 N ** it is slightly more accurate to find the angle = .0699 rad Then the sin (theta) = the displacement of vehicle with respect to a vertical y-axis. => sin(.0699rad)*3kg*9.81m/s^2= 2.0551 N In this case there is about a .15 percent error which is very small error confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The gravitational force on a 3 kg object is its weight and is equal to 3 kg * 9.8 m/s^2 = 29.4 Newtons. The weight component parallel to the incline is found approximately as (parallel weight component) = slope * weight = .07 * 29.4 Newtons = 2.1 Newtons (approx.).. STUDENT COMMENT: It's hard to think of using the acceleration of 9.8 m/s^2 in a situation where the object is not free falling. Is weight known as a force measured in Newtons? Once again I'm not used to using mass and weight differently. If I set a 3 kg object on a scale, it looks to me like the weight is 3 kg.INSTRUCTOR RESPONSE: kg is commonly used as if it is a unit of force, but it's not. Mass indicates resistance to acceleration, as in F = m a. {}{}eight is the force exerted by gravity. {}{}The weight of a given object changes as you move away from Earth and as you move into the proximity of other planets, stars, galaxies, etc.. As long as the object remains intact its mass remains the same, meaning it will require the same net force to give it a specified acceleration wherever it is.{}{}An object in free fall is subjected to the force of gravity and accelerates at 9.8 m/s^2. This tells us how much force gravity exerts on a given mass: F = mass * accel = mass * 9.8 m/s^2. This is the weight of the object. ** STUDENT NOTE: i didn't think to use the acceleration of gravity for this one, it said the object was paralell INSTRUCTOR RESPONSE: The situation talks about the weight having two components, one being parallel to the incline, and the instruction tells you how to find that parallel component when the slope of the incline is small. We know from experience that an object will pick up speed along the incline, as opposed to the direction perpendicular to the incline (to move in the perpendicular direction the object would have to leave the incline, either burrowing down into the incline or levitating up off the incline). The direction along the incline is parallel to the incline. So its acceleration is parallel to the incline, and the net force must be parallel to the incline. In the absence of other forces, only gravity has a component parallel to the incline. Therefore in this ideal case the gravitational component parallel to the incline is the net force. In reality there are other forces present (e.g., friction) but the parallel gravitational component is nevertheless present, and contributes to the net force in the direction of motion. STUDENT QUESTION If we are finding the force at 29.4 newtons is that the perpendicular, and the slope * weight is parallel, why is this considered the weight instead of force in the calculation, is this because the weight is the force that is moving the object INSTRUCTOR RESPONSE The weight of an object is the force exerted on it by gravity. Since objects near the surface of the Earth accelerates downward, if free to do so, at 9.8 m/s^2, the weight of an object is 9.8 m/s^2 multiplied by its mass. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q009. What will be the acceleration of the cart in the previous example, assuming that it is free to accelerate down the incline and that frictional forces are negligible? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We caltulaed the F_grav to be 2.0551 N F/m = a => 2.0551 N / 3 kg = about .685 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The weight component perpendicular to the incline is balanced by the perpendicular force exerted by the incline. The only remaining force is the parallel component of the weight, which is therefore the net force. The acceleration will therefore be a = F / m = 2.1 Newtons / (3 kg) = .7 m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ self-critique rating #$&*: + ********************************************* Question: `q010. What would be the acceleration of the cart in the previous example if friction exerted a force equal to 2% of the weight of the cart, assuming that the cart is moving down the incline? [Note that friction is in fact a percent of the perpendicular force exerted by the incline; however for small slopes the perpendicular force is very close to the total weight of the object]. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Cos (theta) * 9.81 m/s^2 * .02 * 3 kg = x` component fric = .587 N F_net = F_grav F_fric= 1.468 N F / mt = a = 1.47N / 3 kg = .4893 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: The weight of the cart was found to be 29.4 Newtons. The frictional force will therefore be .02 * 29.4 Newtons = .59 Newtons approx.. This frictional force will oppose the motion of the cart, which is down the incline. If the downward direction along the incline is taken as positive, the frictional force will be negative and the 2.1 Newton parallel component of the weight will be positive. The net force on the object will therefore be net force = 2.1 Newtons - .59 Newtons = 1.5 Newtons (approx.). This will result in an acceleration of a = Fnet / m = 1.5 Newtons / 3 kg = .5 m/s^2. STUDENT QUESTION (instructor comments in bold) ok, so the weight is equal to the force that is making the onject go down the incline (because force and mass are equal here????) the force is equal to the mass times the acceleration of gravity the mass is 3 kg, the weight is 29.4 Newtons the mass or weight was 29.4newtons. 2 percent of this is .02 * 29.4 = .59 newtons which is the frictional force, and that should be -.59 since friction works in the opposite direction of the system. To find the net force of the system with add -.59 + 2.1Newtons = 1.5 newtons So the mass of the systme = 3kg a= f/m = 1.5/3kg = .5m/s^2 I am a little confused, I am not sure about the weight (being the gravity 9.8m/s^2 * the mass 3kg= 29.4newtons) is related to the force of the system slope .07 * weight 29.4????? That goes back to the statement in the preceding problem, about how the gravitational force splits into two components, one being parallel to the incline and equal to slope * weight (provided slope is small). We will see soon, in terms of vectors, why this is so, and also how to deal with the situation where the slope isn't small. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: K-should have done this a while ago! ********************************************* Question: `q011. Given the conditions of the previous question, what would be the acceleration of the cart if it was moving up the incline? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F_grav and F_fric are working aginst motion. => F_net = F_grav + F_fric = 2.0551N + .587 N = 2.64 N against motion. => -2.64N / 3kg = -.8807m/s^2 with motion confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In this case the frictional force would still have magnitude .59 Newtons, but would be directed opposite to the motion, or down the incline. If the direction down the incline is still taken as positive, the net force must be net force = 2.1 Newtons + .59 Newtons = 2.7 Newtons (approx). The cart would then have acceleration a = Fnet / m = 2.7 Newtons / 3 kg = .9 m/s^2. STUDENT QUESTION ?????????????????????????????????if the cart is moving downhill and this is considered in the positive direction parallel to the weight of the object which is also considered to be positve, why if it is going in the opposite direction would that not be considered negative since the systme is moving against gravity and the force of it own weight ( which made it go down the incline in the previous problem) ?????????????????????????? INSTRUCTOR RESPONSE The direction of gravity does not determine the positive direction; the positive direction is simply declared in the solution, and you would be free to use either direction as positive. Once the positive direction is declared, all forces, displacements, velocities and accelerations will be positive or negative depending on whether they are in the positive direction, or opposite to it. The positive direction as chosen in the given solution is down the incline. The displacement is up the incline, as is the velocity, so both displacement and velocity are negative. The frictional force and the component of the gravitational force along the incline are both positive, according to the choice of positive direction. So the object has a negative velocity and a positive acceleration, meaning that it is moving in the negative direction but slowing. It takes some thinking to get used to this idea; the idea is far from trivial. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q012. Assuming a very long incline, describe the motion of the cart which is given an initial velocity up the incline from a point a few meters up from the lower end of the incline. Be sure to include any acceleration experienced by the cart. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The car would travel slower and slower due to such a F_g + F_fric = After it comes to rest the vehicle will descend at a slower rate and increase in speed but increase at a slower rate than it would decelerate due to a lesser acceleration. In fact its initial speed might be close to its finial speed coming off the ramp. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The cart begins with a velocity up the incline, which we still taken to be the negative direction, and an acceleration of +.9 m/s^2. This positive acceleration tends to slow the cart while it is moving in the negative direction, and the cart slows by .9 m/s every second it spends moving up the incline. Eventually its velocity will be 0 for an instant, immediately after which it begins moving down the incline as result of the acceleration provided by the weight component parallel to the incline. As soon as it starts moving down the incline its acceleration decreases to +.5 m/s^2, but since the acceleration and velocity are now parallel the cart speeds up, increasing its velocity by .5 m/s every second, until it reaches the lower end of the incline. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This makes sense I could visualize it; though I did not use the analysis of the lab results. ------------------------------------------------ self-critique rating #$&*: OK "