course PHY 121 1/7 025. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:ent: ---- ********************************************* 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: Gravity is also a vertical force acting on the mass. The force = m * a = .150kg *9.8m/s^2 = 1.47N. I know that the force can be influenced by the mag/direction but am unsure where to go from here. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see now how, since the vertical direction is not accelerating, the Fnet must be 0, so it must be the opposite force to -1.47N. ------------------------------------------------ Self-critique Rating:ent: ---- ********************************************* 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: If T*sin(105) = 1.47 Newtons, then 1.47/ 0.97 = T = 1.52. In the horizontal (X) direction, x = tcos(105), so x = (1.52)cos105 = -.4N confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:ent: ---- ********************************************* Question: `q004. If a 2 kg pendulum is held back at an angle of 20 degrees from vertical by a horizontal force, what is the magnitude of that horizontal force? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m = 2kg, angle = 20degrees. We have to adjust our axis, so 90+20 = 110 degrees. For the vertical, Fnet = m*a = 2kg*9.8m/s = 19.6N. The tension in the vertical direction must be opposite of this, so 19.6N + tsin(110) = 0 . tsin(110) = -19.6 t(.94) = -19.6 t = -21N. x = tcos(110) = -21Ncos(110) = 7.2N. confidence rating: 2/3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I somehow ended up with the correct final answer but messed up my positive/negative signs about halfway through the problem. I just got confused. ------------------------------------------------ Self-critique Rating:ent: ---- ********************************************* 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: my graph kind of looks like an inverted triangle, with the location of the pendulum weight being at some (0,y). There are two different angles so we have to do an x and y component for both. x = T2cos(110) y = T2sin(110) x = T1cos(40) y = T1sin(40) The pendulum isn’t moving so everything must be at 0, though gravity is still working in the vertical direction at -19.6N on the weight of the pendulum. for y, T1sin(40) + T2sin(110) = 0 .62(T1) + .94(T2) -19.6 = 0 for x, T1cos(40) + T2cos(110) = 0 .77(T1) + -.34(T2) = 0 We have to solve for one of the T’s. T1 = .34T2/.77 = .44T2. Plugging that into the y equation, .62(.44T2) + .94T2 -19.6 = 0. 0.27T2 + .94T2 -19.6 = 0 1.21T2 = 19.6 T2 = 16.3 tension in the string at 110 degrees. .77(T1) + -.34(16.3) = 0 .77T1 = 5.5 T1 = 7.1 tension in the string at 40 degrees. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 obtai .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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Whew. That was rough. My graph reasoning was only half correct. My answers were also quite a few tenths off.. not sure if it was due to calculation error by me or if you just did approximations in the given answer.