#$&* course PHY121 7/26 11 025. More Forces
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Given Solution: The pendulum string makes an angle of 15 degrees with vertical. Since we have assumed that the pendulum is pulled in the positive x direction, the direction of the tension in the string will be upward and to the left at an angle of 15 degrees with vertical. The tension force will therefore be directed at 90 degrees + 15 degrees = 105 degrees as measured counterclockwise from the positive x axis. The tension will therefore have x component T cos(105 degrees) and y component T sin (105 degrees). STUDENT QUESTION: why did you add 90deg to 15deg, i knew that x should have been Tcos(15) and y Tsin(15), i wasn't sure about the 90deg INSTRUCTOR RESPONSE 15 deg is the angle with the y axis. If you use T cos(theta) and T sin(theta) then the angles must be measured counterclockwise from the positive x axis. The pendulum is displaced in the positive x direction. So relative to the position of the pendulum mass, the string pulls up and to the left--into the second quadrant--at an angle of 90 degrees + 15 degrees = 105 degrees. Starting from the positive x axis you would have to rotate through 90 degrees to get to the y axis, then through the 15 degree angle the string makes with the y axis. STUDENT COMMENT: Ok that makes sense. Its like the vertical pendelum is just shifted so the pendelum tension will be like a vector. INSTRUCTOR RESPONSE Right. The tension exerts a force, and forces are characterized by magnitude and direction so they can be represented by vectors. STUDENT QUESTION I didn’t know that I had to add 90 degrees to the y-axis angle, but that explains my confusion earlier. INSTRUCTOR RESPONSE If you use the circular definition of angles, as is done in the given solution, the y axis is at the 90 degree position and an angle of 15 degrees with the y axis is at 90 deg + 15 deg or 90 deg - 15 deg; in this case the angle is 90 deg + 15 deg = 105 deg. STUDENT QUESTIONS (instructor responses in boldface) I had some trouble with this one, but it does make sense for me to see that y (vertical axis) should be at 105 degrees since we look at it as already having an angle of 90 degrees and then we just add. The y axis is at 90 degrees with the x axis, as always. The x axis is horizontal, as specified, and the y axis is therefore vertical. The pendulum string, not the y axis, is at 105 degrees with respect to the x axis. I am still a little confused on how the x axis is 105 degrees as well. I read the explanation and understand how the math works out but the part that confuses me is it sounded as though the string was completely horizontal so I don’t see how it was also pulled 15 degrees from the horizontal position. There are two strings, the pendulum string (at 15 degrees from vertical, 105 degrees from horizontal), and the second string, which is used to pull the pendulum away from its 'natural' vertical (i.e., 90 degree) position. It is the second string that is horizontal. The pendulum string exerts a tension force, which we call T, which makes angle 105 degrees with the positive x axis. A vector with magnitude T at 105 degrees (as measured counterclockwise from the positive x axis) has components T_x = T cos(105 deg), and T_y = T sin(105 deg). (If this last statement isn't clear you should review Introductory Problem Set 5, where these rules are stated). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand why the 90 had to be added to the 15 before finding x and y. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q002. Continuing the preceding problem, we see that we have a vertical force of T sin(105 deg) from the tension. What other vertical force(s) act on the mass? What is the magnitude and direction of each of these forces? What therefore must be the magnitude of T sin(105 deg). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There's always gravity. I'm making a guess here, but looking at my drawing, it looks like the gravitational pull is all in the y direction. I'm going to guess that the angle will be 270 deg. For that x = 9.8 m/s^2 (cos)(270) = 0. Y is 9.8 m/s^2 (sin) (270), which is 9.8 m/s^2. That would made the tension 9.8 sin (105 deg) = 9.5. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The only other vertical force acting on the mass will be the gravitational force, which is .150 kg * 9.8 meters/second ^ 2 = 1.47 Newtons. The direction of this force is vertically downward. Since the mass is in equilibrium, i.e., not accelerating, the net force in the y direction must be zero. Thus T sin(105 deg) - 1.47 Newtons = 0 and T sin(105 deg) = 1.47 Newtons. STUDENT QUESTION Ok. So what I basically do is set the gravitational force equal to that of T sin(105). And the 1.47 Newtons of gravitational force now becomes the magnitude of the x component….??? INSTRUCTOR RESPONSE The gravitational force remains what it is--just the gravitational force. We assume the gravitational force to be in the downward vertical direction. Since the mass is assumed to be in equilibrium, with the only other force in the y direction being the y component of the string tension (which is T sin(105 deg) ), it follows that these two forces are in equilibrium so that T sin(105 deg) = 1.47 N. Nothing in the solution to this question addresses the x component of the force, but that's coming up in the next question. STUDENT QUESTION T sin(105 deg) - 1.47 Newtons = 0 and T sin(105 deg) = 1.47 Newtons. So this means that the only force acting on the object is the force of gravity, since there is no acceleration y is = to zero, correct? INSTRUCTOR RESPONSE The reason this quantity is zero is that the net force in the y direction is zero. We know this because the object accelerates only in the x direction (y acceleration is negligible). In the y direction we have both the tension and the force of gravity, which must be equal and opposite. STUDENT QUESTIONS (instructor responses in boldface) I feel like I kind of know how to solve these and then when I read the given solution I am really confused as to why problems have been solved a particular way. I determined the net force correctly, but struggled when solving for the magnitude. You correctly found the 1.47 Newton weight of the mass. However this is not the net force. The net force is the sum of all forces acting on the mass, including the tensions of any strings that might be pulling on it. The present question was related to the y component of the tension. Since the mass doesn't accelerate significantly in the y direction (the vertical direction), the net force in this direction must be zero. The forces acting in the vertical direction are the weight, which acts downward, and the y component of the tension, which acts upward. It follows that the y component of the tension is equal and opposite to the weight. We therefore know that the y component of the tension is 1.47 Newtons upward. I thought that the net force was multiplied by Tsin (120 degrees0. Instead it should have been written as follows: T sin(105 deg) - 1.47 Newtons = 0 and T sin(105 deg) = 1.47 Newtons. 120 degrees is not involved in this problem. The tension force is at 105 degrees, and its y component is 1.47 Newtons. It follows that T sin(105 deg) = 1.47 Newtons. From this we can calculate the tension T, but that calculation wasn't requested in this question. It's coming up soon, though. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): First of all, I forgot to multiply by the mass. Also, I did not think of the mass as being in equilibrium because of the 15 degree pull. That's in the x direction. I guess it is in equlibrium in the y direction because it's being held up by a string. I'm still processing what in this I understand and don't understand. I started getting confused when you have to swing the 15 degrees around the y axis to get the angle. I have done this on my picture and I can see how it will work and why, but as far as application...I'm not there yet. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q003. Continuing the preceding two problems, what therefore must be the tension T, and how much tension is there in the horizontal string which is holding the pendulum back? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I know that there must be a way to work backward from the information I now have, but I don't know what it is. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If T sin(105 deg) = 1.47 Newtons then T = 1.47 Newtons / (sin(15 deg)) = 1.47 Newtons/.97 = 1.52 Newtons. Thus the horizontal component of the tension will be T cos(105 deg) = 1.52 Newtons * cos(105 deg) = 1.52 Newtons * (-.26) = -.39 Newtons, approximately. Since the mass is in equilibrium, the net force in the x direction must be zero. The only forces acting in the x direction are the x component of the tension, to which we just found to be -.39 Newtons, and the tension in the second string, which for the moment will call T2. Thus T2 + (-.39 N) = 0 and T2 = .39 N. That is, the tension in the second string is .39 Newtons. STUDENT COMMENT: I'm really confused now. If we started out with a .15 kg mass that is equal to 1.47 Newtons. How did we create more weight to get 1.52 Newtons? Is the horizontal string not helping support the weight or is it puling on the weight adding more force? INSTRUCTOR RESPONSE: A horizontal force has no vertical component and cannot help to support an object against a vertical force. The vertical component of the tension is what supports the weight, so the tension has to be a bit greater than the weight. The tension in the string is resisting the downward weight vector as well as the horizontal pull, so by the Pythagorean Theorem it must be greater than either. STUDENT COMMENT: Thats odd how that works that the tension is negative INSTRUCTOR RESPONSE: The tension isn't negative, but in this case, where the displacement from equilibrium is positive, the string pulls back (upward and to the left) so the horizontal component of the tension is negative. STUDENT QUESTION Is where I seemed to get off is I assumed I could set the X component equal to the force in Newtons found from the Y direction (.15kg*9.8m/s^2). From what I can tell now, the tension in X is equal to the X component being set equal to the force in the Y direction * the displacement from the Y axis…I THINK that is a way of looking at it. INSTRUCTOR COMMENT: Here's an overview of what we know and how we use it. The given solution will fill in the details at the end. Let T be the unknown magnitude of the tension vector. We know that T is at 105 degrees. The components of T are therefore T_y = T sin(105 deg) and T_x = T cos(105 deg). The y component of the tension is what supports the mass of the pendulum in opposition to the force exerted on it by gravity. Setting T sin(105 deg) equal to m g we find T, as shown in the give solution. Then we can find the x component T_x, as shown in the given solution. STUDENT QUESTION I’m not sure where the -.26 came from. INSTRUCTOR RESPONSE -.26 is roughly equal to the cosine of 105 degrees STUDENT QUESTION The solution says....... If T sin(105 deg) = 1.47 Newtons then T = 1.47 Newtons / (sin (105 deg)) = 1.47 Newtons/.97 = 1.52 Newtons. Shouldnt T= 1.47 Newtons / (cos(15degrees)) = 1.47N/ .97 = 1.52N in the horizontal direction? There's only one string with its tension. T has only one value, and that's 1.52 N. So vertically gravity is the only force action on the weight which is creating tension on the string vertically so Tsin(105)= 1.47, vertically. I think I understand this. Gravity isn't the only force; the other force is tension. The net force is the gravitational force added (as a vector) to the tension force. INSTRUCTOR RESPONSE You've just about got it sorted out. Hopefully the following will help: The tension is a vector of magnitude T at angle 105 degrees. Its components are T_x = T * cos(105 deg) and T_y = T * cos(105 deg). Its vertical component must be 1.47 Newtons in order to keep the mass from accelerating in the y direction (i.e., to keep it from falling). In order to have 1.47 Newtons in the vertical direction the tension must be 1.52 N (shown in preceding problem). Once we know this it's easy to find T_x, which is just T * cos(105 deg) = 1.52 N * cos (105 deg) = =.39 N, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): When I put in sin (15), I get .26. When I put in sin (105), I get .97. The answer says to put in sin (15) to get .97. I am confused about whether to use the 15 or the 105 for the y. I was able to walk through the given solution and get a matching answer for the cosine. This is still a work in progress. It's been a long time since I took trig. ------------------------------------------------ Self-critique rating: 1
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Given Solution: At the 20 degree angle the tension in the pendulum string will have a vertical component equal and opposite to the force exerted by gravity. The tension with therefore have a horizontal component. To achieve equilibrium by exerting the horizontal force, this horizontal force must balance the horizontal component of the tension. We therefore begin by letting T stand for the tension in the pendulum string. We also assumed that the pendulum is displaced in the positive x, so that the direction of the string as measured counterclockwise from the positive x axis will be 90 degrees + 20 degrees = 110 degrees. Thus the x component of the tension will be T cos(110 deg) and the y component of the tension will be T sin(110 deg). The weight of the 2 kg pendulum is 2 kg * 9.8 meters/second ^ 2 = 19.6 Newtons, directed in the negative vertical direction. Since the pendulum are in equilibrium, the net vertical force is zero: T sin(110 deg) + (-19.6 N) = 0 This equation is easily solved for the tension: T = 19.6 N / (sin(110 deg) ) = 19.6 N / (.94) = 20.8 Newtons, approximately. The horizontal component of the tension is therefore T cos(110 deg) = 20.8 N * cos(110 deg) = 20.8 N * (-.34) = -7 N, approx.. To achieve equilibrium, the additional horizontal force needed will be + 7 Newtons. STUDENT QUESTION F_grav = 2kg * 9.8m/s^2= 19.6N Net vertical force = 0 T sin(105 deg) - 19.6 N = 0 so T = 19.6N/ (sin(110)) = 19.6/ .94 = 20.8N So is this the force vertically that acts on the horizontal? INSTRUCTOR RESPONSE The 20.8 N is the tension in the string, which acts along the direction of the string, with components in both the x and y directions. The horizontal component of this tension is 20.8 N * cos(110 deg). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It's all off as soon as I used .2 kg instead of 2 kg. I had the steps correct, just the decimal point in the wrong place. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q005. The 2 kg pendulum in the previous exercise is again pulled back to an angle of 20 degrees with vertical. This time it is held in that position by a chain of negligible mass which makes an angle of 40 degrees above horizontal. Describe your sketch of the forces acting on the mass of the pendulum. What must be the tension in the chain? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I made an x-y axis. The pendulum is hanging from the y axis at a 20 degree angle. Attached to the pendulum is a chain pulling at an upward angle of 40 degrees. The forces that I have shown on this sketch are: up and to the right at a 40 degree angle, vertical from gravity, and horizontal back to the origin because the pendulum wants to return to equilibrium. The mass of the pendulum, which is 2 kg, multiplied by 9.8 m/s^2 = 19.6 N 19.6 N cos 40 deg = 15 N in the x direction 19.6 N sin 40 deg = 12 N in the y direction (15 N)^2 + (12 N)^2 = 369 N^2 `sqrt 369 N^2 = 19.2 Newtons of force in the chain. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The weight of the pendulum is partially supported by the tension in the chain. Thus the tension in the pendulum string is not the same as before. The horizontal component of the tension in the chain will be equal and opposite to the horizontal component of the tension in the pendulum string. Your picture should show the weight vector acting straight downward, the tension in the pendulum string acting upward and to the left at an angle of 20 degrees to vertical and the tension in the chain should act upward into the right at an angle of 40 degrees above horizontal. The lengths of the vectors should be adjusted so that the horizontal components of the two tensions are equal and opposite, and so that the sum of the vertical components of the two tensions is equal of opposite to the weight vector. Since both tensions are unknown we will let T1 stand for the tension in the pendulum and T2 for the tension in the chain. Then T1, as in the preceding problem, acts at an angle of 110 degrees as measured counterclockwise from the positive x axis, and T2 acts at an angle of 40 degrees. At this point whether or not we know where we are going, we should realize that we need to break everything into x and y components. It is advisable to put this information into a table something like the following: x comp y comp T1 T1 * cos(110 deg) T1 * sin(110 deg) in T2 T2 * cos(40 deg) T2 * sin(40 deg) Weight 0 &nb sp; -19.6 N The pendulum is held in equilibrium, so the sum of all the x components must be 0, as must the sum of all y components. We thus obtain the two equations T1 * cos(110 deg) + T2 * cos(40 deg) = 0 and T1 * sin(110 deg) + T2 * sin(40 deg) - 19.6 N = 0. The values of the sines and cosines can be substituted into the equations obtain the equations -.33 T1 + .77 T2 = 0 .95 T1 + .64 T2 - 19.6 N = 0. We solve these two simultaneous equations for T1 and T2 using one of the usual methods. Here we will solve using the method of substitution. If we solve the first equation for T1 in terms of T2 we obtain T1 = .77 T2 / .33 = 2.3 T2. Substituting 2.3 T2 for T1 in the second equation we obtain .95 * 2.3 T2 + .64 T2 - 19.6 N = 0, which we easily rearrange to obtain 2.18 T2 + .64 T2 = 19.6 Newtons, or 2.82 T2 = 19.6 N, which has solution T2 = 19.6 Newtons/2.82 = 6.9 N, approximately. Since T1 = 2.3 T2, we have T1 = 2.3 * 6.9 N = 15.9 N, approximately. Thus the pendulum string has tension approximately 15.9 Newtons and the chain the tension of approximately 6.9 Newtons. STUDENT QUESTION: I thought we already found the tension for the original string so we just have to solve for the T components in the chain. How come the tension found earlier is not the same one we use here? INSTRUCTOR RESPONSE: Previously the 'pullback' used a horizontal string, which exerted no force in the vertical direction. The string is now at a 40 degree angle, so any tension must have both horizontal and vertical components. Since we need a horizontal component to pull the mass back, there will be a nonzero vertical component. This will have the effect of reducing the tension in the pendulum. STUDENT QUESTION I read through the Given Solution, and I think I understand. I’m just not very good at recognizing solutions where I have to use substitutions. But one question…. When you were solving for T1 * cos(110) + T2 * cos(40) = 0, I solution for T1 = -2.3 * T2. Why did you change the solution from negative to positive?? Does it have something to do with the tensions being equal and opposite, or what? INSTRUCTOR RESPONSE The given solution didn't change any signs. The equation T1 * cos(110) + T2 * cos(40) = 0, solved for T1, gives you T1 = + 2.3 * T2. You just have to be careful about the signs of your trig functions. T1 cos(110 deg) + T2 cos(40 deg) = 0, so that T1 = -T2 * cos(40 deg) / cos(110 deg). cos(110 deg) is negative so the relationship is T1 = -T2 * (.77 / (-.33) ) = +2.3 T2. The tension T1 is not -2.3 T2. STUDENT QUESTION Ok, after drawing this out I am rather comfortable with the forces present; however, seemed to just assume as in one of the previous student comments, that the only tension we needed to solve would be the added chain at 40 degrees. Still a bit unsure here how substitution method was utilized to solve this? INSTRUCTOR RESPONSE The first statement in the solution: 'If we solve the first equation for T1 in terms of T2 we obtain T1 = .77 T2 / .33 = 2.3 T2. ' More specifically, the first equation is -.33 T1 + .77 T2 = 0. Subtracting .77 T2 from both sides we get -.33 T1 = -.77 T2. Dividing both sides by .33 we get T1 = .77 / .33 T2. The second equation is .95 T1 + .64 T2 - 19.6 N = 0. If we substitute .77 / .33 T2 for T1 we get .95 * (.77 / .33 ) T2 + .64 T2 - 19.6 N = 0. Multiplying out the first term, and adding 19.6 N to both sides we get 2.18 T2 + .64 T2 = 19.6 N, so that 2.82 T2 = 19.6 N and T2 = 19.6 N / (2.82) = 6.9 N. Now since T1 = .77 / .33 T2, we have T1 = .77 / .33 * 6.9 N = 15.9 N. As usual the numbers are approximate, but may not be completely accurate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not have the tension in the pendulum string working upward. I did not split up the tensions between the chain and the string, but it makes sense to do it. I worked through the solution once I had the 2 basic equations for the two tensions and did manage to get the same answer. ------------------------------------------------ Self-critique rating: 2 If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q006. Sketch a force diagram for a simple pendulum of length 4 meters and mass 0.5 kg, which is held at a position 0.2 meters from equilibrium by the tension of a horizontal rubber band. Describe your diagram. What do you think the horizontal force is, as a percent of the pendulum's weight? How did you reason out your answer based on your diagram? (Note that the angle of the pendulum string with the positive x axis will be about 93 degrees. You should be able to determine this using vector methods, or if you prefer plain old trigonometry, and your solution should show how you got this. But if you didn't get it, don't let that stop you; just go ahead and use the 93 degrees.) If the pendulum was held at position 0.4 meters from equilibrium by the rubber band, would the rubber band's length change? What do you think the tension force would be at the 0.2 meter position, and at the 0.4 meter position? If the pendulum was released from the 0.4 meter position, how much kinetic energy do you think it would gain by the time it reached the 0.2 meter position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is an x-y axis. The pendulum is .2 m away from the origin. The pendulum is labeled ""4 m"". The forces are: equal and opposite forces that are horizontal, or close to horizontal, that show pulling out and a wish to return to equilibrium. There is a force going up and down the pendulum string pulling on the weight as the weight pulls on the string, and there is a vertical force, which is the pull of gravity. I also have a rubber band atached to the weight. It shows forces pulling on the weight and the weight pulling on the rubber band. The pendulum's weight is .5 kg * 9.8 m/s^2 = 4.9 N Horizontal force as a percent of the pendulum's weight? Well, since I'm getting lost here, I've gone back to the last problem. The horizontal force was about 35% of the pendulum's total weight. Since the angle is much smaller on this (which I couldn't figure out, I kept coming up with 89 degrees, which I knew was wrong), I would think that the horizontal force would be less and the vertical force would be stronger. No math--just educated guesses. If the position changed from 0.2 m to 0.4 m, the rubber band would stretch. Since I'm lost above, I guess I'll ask for some clues to get me jump started. Thanks! confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: