course Phy 231 7/10 10:15 025. More Forces Question: `q001. Note that this assignment contains 5 questions.
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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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Assessment: OK ********************************************* 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). Your solution: Gravity is the other vertical force: the magnitude is 9.8m/s^2* .15 kg = 1.47N. The direction is 270 degrees. Y component = 1.47N sin 270 = -1.47N Since the pendulum is not moving (no acceleration = 0 net force) the magnitude of the y component has to be equal and opposite to the gravity, so T sin 105 = 1.47N confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The only other vertical force acting on the mass will be the gravitational force, which is .150 kg * 9.8 meters/second ^ 2 = 1.47 Newtons. The direction of this force is vertically downward. Since the mass is in equilibrium, i.e., not accelerating, the net force in the y direction must be zero. Thus T sin(105 deg) - 1.47 Newtons = 0 and T sin(105 deg) = 1.47 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Assessment: OK ********************************************* Question: `q003. Continuing the preceding two problems, what therefore must be the tension T, and how much tension is there in the horizontal string which is holding the pendulum back? Your solution: T sin 105 = 1.47 N T = 1.52 N X component = 1.52N cos 105 = -.393 N The other string is horizontal and the mass is in equilibrium, so the net force in the x direction is 0. -.393N + Tension in the other string = 0 Tension in the horizontal string = .393 N confidence rating #$&*: 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. Self-critique (if necessary): OK Self-critique Assessment: OK ********************************************* 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? Your solution: X component = T cos 110 Y component = T sin 110 Weight = 2kg * 9.8m/s^2 = 19.6N Y component = 19.6 sin 270 = -19.6 19.6N = T sin 110 T = 20.9N Force in the string: 20.9N cos 110 = -7.15N Horizontal force = 7.15N confidence rating #$&*: 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): OK Self-critique Assessment: OK ********************************************* 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? Your solution: The diagram shows the mass being pulled 20 degrees from the vertical and then held in place by a chain going up to the positive x axis. The angle between a horizontal line coming off the mass and the chain is 40 degrees. The weight of the mass is pulling down and the tension of the string is going up and to the left at 110 degrees. The tension of the chain is going up and to the right at 40 degrees. Tension of the string = TS X component = TS cos 110 => 7.15N Y component = TS sin 110 => -19.6N Tension in the chain = TC X = TC cos 40 Y = TC sin 40 Weight = 2 kg * 9.8m/s^2 = 19.6 N X = 0N Y = -19.6N TScos110 + TCcos40 = 0 => -.342 TS + .776 TC = 0 TC = .441 TS TSsin110 + TCsin40 – 19.6N = 0 .94 TS + .643(.441 TS) -19.6 = 0 .94 TS +.284 TS -19.6 = 0 1.224 TS = 19.6 TS = 16 N = tension in the string TC = .441 * 16N = 7.1 N = tension in the chain confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*ent: OK "