#$&* course Phy 231 8/3/11 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aIf the crate is moving then the force exerted by friction is .30 times the normal force between it and the floor. If the push is horizontal, then the only horizontal forces acting on the crate are the downward force of gravity and the upward force exerted by the floor. Since the crate is not accelerating in the vertical direction, these forces are equal and opposite so the normal force is equal to the weight of the crate. The weight of the crate is 35 kg * 9.8 m/s^2 = 340 N, approx. The frictional force is therefore f = .30 * 340 N = 100 N, approx.. If the crate moves at constant speed, then its acceleration is zero, so the net force acting on it is zero. The floor exerts its normal force upward, which counters the gravitational force (i.e., the weight). The frictional force acts in the direction opposite motion; if net force is zero an equal and opposite force is required, so you must push the box with a force of 100 N in the direction of motion. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qgen phy 4.55 18 kg box down 37 deg incline from rest, accel .27 m/s^2. what is the friction force and the coefficient of friction? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fx = 18 kg * 9.8 m/sec^2 * sin(37 deg) Fx = 106.16 N Fy = 18 kg * 9.8 m/sec^2 * cos(37 deg) Fy = -140.879 N Fnet = m * a Fnet = 18 kg * 0.27 m/sec^2 Fnet = 4.86 N 4.86 = 106.16 - mu * 140.879 coefficient = 0.72 f = 0.72 * 140.879 N f = -101.43 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The coordinate system is rotated from its traditional vertical-horizontal orientation in such a way that the positive x axis points down the incline, below the horizontal. As the system is rotated, the negative y axis 'swings away' from a vector which remains in the vertical downward direction, and that vector ends up in the fourth quadrant. The result will be something like the figure below, though the rotation in this figure is only 20 degrees and won't completely agree with the rotation in this problem: The weight vector being vertical in the downward direction is therefore in the fourth quadrant, at an angle of 37 degrees with respect to the negative y axis. Thus the weight vector makes angle 270 deg + 37 deg = 307 deg with the positive x axis and its x and y components are wtx = 18 kg * 9.8 m/s^2 * cos(307 deg) = 106 N and wty = 18 kg * 9.8 m/s^2 * sin(307 deg) = -141 N. The figure below depicts a fourth-quadrant vector and its components. As before the angle of rotation in this figure is only 20 degrees, as opposed to the 37 degrees of this problem, so the x in the problem component is relatively smaller and the y component relatively larger than depicted below: The indented discussion of the right-angle trigonometry of the situation is in a font different from the rest of the solution, and may be skipped without loss of continuity. You get the same results using the sin and cos of the 37 deg angle. The following assumes some familiarity with right-triangle trigonometry, which is assumed for General College Physics students and should be very familiar to University Physics students. In the figure below we have constructed a right triangle whose hypotenuse coincides with the weight vector, with one of the legs along the negative y axis. If theta is the angle at the vertex located at the origin, then theta is the angle through which the axes are rotated. Theta therefore coincides with the angle of the original incline. The x and y components of the vector are congruent with the legs of the triangle, the x component with the leg opposite the angle theta and the y component with the leg adjacent to theta. Thus the 'opposite' leg is hypotenuse * sin(theta), and the 'adjacent' leg is hypotenuse * cos(theta). The vector in this example represents the weight of the object, so the hypotenuse represents the weight. The x component is positive and the y component is negative. We conclude that x component = + weight * sin(theta) and y component = - weight * cos(theta), where again theta is the angle of the incline. For the mass of the object in this problem, the weight is 18 kg * 9.8 m/s^2, and we have x component = 18 kg * 9.8 m/s^2 * sin(theta) and y component = 18 kg * 9.8 m/s^2 * cos(theta) Common point of confusion: The x component in one model is given by the cosine, and in the other by the sine. The y components are also given by two different functions. The key idea is that the circular-model definitions lead us to an x-y coordinate system in which the weight vector is at angle 307 degrees relative to the positive x axis, while the right-triangle analysis can be done in terms of a triangle containing the 37-degree incline. The sine of one angle is the cosine of the other, and vice versa, except for + and - signs. The + and - signs arise naturally from the circular model, which is the advantage of this model. The triangular model also has its advantages. Detailed resolution of point of confusion: The angles used in the two models are different. The angle used in the 'circular' model is 307 deg, the angle of the weight vector with the positive x axis of our coordinate system. The angle used in the 'triangle' model is 37 deg, the angle of the incline with horizontal. The sines and cosines of these angles are related: cos(307 deg) = sin(37 deg) sin(307 deg) = - cos(37 deg) The only difference is that the circular model, if set up correctly, automatically gives you the correct signs of the components (i.e., x component is positive, y component negative). If you use the triangle you have to think a little bit more, and put the correct signs on the angles. In this respect the circular model is simpler. However it's very important in advanced applications to 'see' the triangles involved, so the triangle model is also essential for General College Physics and especially for University Physics students. Principals of Physics students who are not familiar with trigonometry generally do better with the circular model. The given solution continues below Having found the components of the weight we find the normal force: The only other force in the y direction, other than the y component of the weight, is the normal force (which is the elastic or compressive force exerted by the incline in response to the y component of the weight). Since the mass does not accelerate in the y direction, we know that the net force in the y direction is zero. It follows that normal force + y component of weight = 0, so that normal force = - y component of weight = - ( -141 N ) = 141 N. The frictional force is f_Frict = coefficient of friction * normal force, or in symbols f_Frict = mu * F_normal. The acceleration of the system is .27 m/s^2 down the incline. The component of the gravitational force down the incline is 106 N. The net force along the incline is therefore 106 N - mu * F_normal = 106 N - mu * 141 N. The net force is also F_net = m a = 18 kg * .27 m/s^2 = 5 N, approx.. Thus we have the equation F_net = wt_x - mu * F_normal. Solving for mu: mu = -(F_net - wt_x) / (F_normal) = - (5 N - 106 N) / (141 N) = 101 N / (141 N) = .7, approx. It's best to write and solve the equation for mu before substituting, but it doesn't hurt to represent the relationship using the actual quantities F_net = wt_x - mu * F_normal. Substituting the quantities we have already calculated this becomes 5 N = 106 N - mu * 141 N so that mu = (5 N - 106 N) / (-141 N) = .7. STUDENT QUESTION What does mu stand for? INSTRUCTOR RESPONSE mu is the coefficient of friction. Generally the lowercase Greek letter mu is the symbol used to represent the coefficient of friction. The symbol looks like this: m The meaning is implicit in the lines below, quoted from the given solution: The frictional force is f_Frict = coefficient of friction * normal force, or in symbols f_Frict = mu * F_normal. `aGOOD STUDENT SOLUTION USING RIGHT-ANGLE TRIGONOMETRY: (I don't know why, but I was hoping you would pick an odd numbered problem here)Here goes.....For an 18kg box on an incline of 37 degrees with an acceleration of .270 m/s/s, I first drew out a diagram showing the forces involved. Next the forces had to be derived. First, to find the force associated with the weight component parrallel to the inline moving the box downward....Fp=sin 37 deg(18kg)(9.8m/s/s)=106N. Next, the Normal force that is counter acting the mg of the box is found by.. Fn=cos 37 deg. (18kg)(9.8 m/s/s) = 141N. The frictional force can be found by using F=(mass)(acceleration) where (Net Force)-(frictional coeffecient*Normal Force)=(m)(a) so that... 106N - (141N * Friction Coeff.) = (18kg)(.270 m/s/s) where by rearranging, the frictional coeffecient is seen to be .717. INSTRUCTOR COMMENT: Note that you should specify the direction of your positive x axis. The simplest choice will be an x axis oriented down the incline, so that the acceleration will be positive. You either used triangle trigonometry or quoted formulas from the text (hopefully the former, so you will understand the picture that goes with this problem). Either way you got the correct result and did a good job. My solutions generally formulate vectors in the context of the circular definition of trigonometric functions. The triangle-based solution is probably more standard, but requires more understanding of right-angle trigonometry, and is more prone to confusion regarding signs. STUDENT COMMENTS I don’t know if I know and was trying the circular method or the triangular method. I read through the Given Solution and I am still confused. I don’t understand in the student’s solution why the net force was 106.2 N. That was determined by the sin product, and isn’t that the y-component?? And I thought I remember reading that friction works on the y-component. And I have no idea why your diagram puts this incline in the 4th quadrant. By picturing a 37 degree ramp with it ascending up towards the right, it’s clear that the incline should be in the 1st quadrant. Please don’t tell me that there are times when we switch the sin and cos when solving a problem.... INSTRUCTOR RESPONSE The student's solution used the angle of the incline rather than the angle with the positive x axis. The given solution then put the problem in the context of the circular definition. An incline at 37 deg from horizontal could be depicted rising from left to right, or as descending from left to right. It has to be one or the other, and either choice is equally valid. The choice illustrated in my solution is the latter. When using the circular definition, you set your coordinate system, and the sines and cosines follow the definitions. They don't switch. The frictional force is proportional to the normal force, which is often (as it is in the given solution) in the y direction. The normal force is in the direction perpendicular to the surface, whatever that direction is (again, in this problem the y axis is perpendicular to the incline, so the normal force is in the y direction). When using the trigonometric definitions, the choice of sine or cosine for a given quantity is also clear in terms of those definitions. STUDENT SOLUTION (COMMON ERROR: CONFUSING ANGLE OF INCLINE WITH ANGLE OF WEIGHT VECTOR) Weight = 18kg * 9.8m/s^2 = 176.4N Parallel = 176.4 cos 37 = 141N Perpendicular = 176.4 sin 37 = 106N = normal force INSTRUCTOR RESPONSE The magnitude of the parallel component is of the weight clearly less than that of the perpendicular component for an angle less than 45 degrees. The weight vector does not make a 37 degree angle with the positive x axis. 37 degrees is the angle of elevation of the incline. If you use cos and sin, the angle has to be between the positive x axis and the weight vector. In this case that angle would be either 233 deg or 307 deg, depending on whether the incline is drawn sloping upward or downward as you move from left to right. In the 233 deg case you would get parallel component = 176.4 cos(233 deg) = -106 N, perpendicular component 176.4 sin(233 deg) = -141 N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qUniv. 5.90 (5.86 10th edition). 4 kg and 8 kg blocks, 30 deg plane, coeff .25 and .35 resp. Connected by string. Accel of each, tension in string. What if reversed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F1x = 4 kg * 9.8 m/sec^2 * sin(30 deg) F1x = 19.6 N F1y = 4 kg * 9.8 m/sec^2 * cos(30 deg) F1y = 33.95 N F2x = 8 kg * 9.8 m/sec^2 * sin(30 deg) F2x = 39.20 N F2y = 8 kg * 9.8 m/sec^2 * cos(30 deg) F2y = 67.90 N f1 = 0.25 * 33.95 N f1 = 8.49 N f2 = 0.35 * 67.90 N f2 = 23.77 N Block 1 19.6 N - 8.49 N - T = 4 kg * a a = -0.25(T - 11.11) Block 2 39.20 N - 23.77 N - T = 8 kg * a Now plug into Block 2 39.20 N - 23.77 N + T = 8 kg * -0.25(T - 11.11 N) T = 2.26 N sub T into Block 1 a = -0.25(2.26 N - 11.11 N) a = 2.21 m/sec^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We will use the direction down the incline as the positive direction in all the following: The normal forces on the two blocks are 4 kg * 9.8 m/s^2 * cos(30 deg) = 34 N, approx., and 8 kg * 9.8 m/s^2 * cos(30 deg) = 68 N, approx. If sliding the 4 kg block will therefore experience frictional resistance .25 * 34 N = 8.5 N, approx. and the 8 kg block a frictional resistance .35 * 68 N = 24 N, approx. The gravitational components down the incline are 4 kg * 9.8 m/s^2 * sin(30 deg) = 19.6 N and 8 kg * 9.8 m/s^2 * sin(30 deg) = 39.2 N. If the blocks were separate the 4 kg block would experience net force 19.6 N - 8.5 N = 11.1 N down the incline, and the 8 kg block a net force of 39.2 N - 24 N = 15.2 N down the incline. The accelerations would be 11.1 N / (4 kg) = 2.8 m/s^2, approx., and 15.2 N / (8 kg) = 1.9 m/s^2, approx. If the 4 kg block is higher on the incline than the 8 kg block then the 4 kg block will tend to accelerate faster than the 8 kg block and the string will be unable to resist this tendency, so the blocks will have the indicated accelerations (at least until they collide). If the 4 kg block is lower on the incline than the 8 kg block it will tend to accelerate away from the block but the string will restrain it, and the two blocks will move as a system with total mass 12 kg and net force 15.2 N + 11.1 N = 26.3 N down the incline. The acceleration of the system will therefore be 26.3 N / (12 kg) = 2.2 m/s^2, approx.. In this case the net force on the 8 kg block will be 8 kg * 2.2 m/s^2 = 17.6 N, approx.. This net force is the sum of the tension T, the gravitational component m g sin(theta) down the incline and the frictional resistance mu * N: Fnet = T + m g sin(theta) - mu * N so that T = Fnet - m g sin(theta) + mu * N = 17.6 N - 39.2 N + 24 N = 2.4 N approx., or about 2.4 N directed down the incline. The relationship for the 4 kg mass, noting that for this mass T 'pulls' back up the incline, is Fnet = m g sin(theta) - T - mu * N so that T = -Fnet + m g sin(theta) - mu * N = -8.8 N + 19.6 N - 8.5 N = -2.3 N. equal within the accuracy of the mental approximations used here to the result obtained by considering the 8 kg block and confirming that calculation. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!