course Phy 201 ?????L????:????assignment #025
......!!!!!!!!...................................
19:07:10 `q001. Note that this assignment contains 5 questions. . A pendulum consists of a 150 g mass suspended from a light string. Another light string is attached to the mass, which is then pulled back from its equilibrium position by that string until the first string makes an angle of 15 degrees with vertical. The second string remains horizontal. Let the x axis be horizontal and the y axis vertical. Assume that the mass is pulled in the positive x direction. If T stands for the tension in the pendulum string, then in terms of the variable T what are the x and y components of the tension?
......!!!!!!!!...................................
RESPONSE --> If we cart this on a graph we have the first string at 15deg in the second quadrant and T will be in the positive x direction we have an angle now of 90+15=105 degrees.Now to find the x-component of tension T cos(105deg) and y-exponent T sin(105deg) confidence assessment: 2
.................................................
......!!!!!!!!...................................
19:07:53 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).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:13:26 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.
......!!!!!!!!...................................
RESPONSE --> You have the gravitational force which is 9.8m/s^2*.15kg=1.47N and it would be directed downward. self critique assessment: 1
.................................................
......!!!!!!!!...................................
19:32:10 `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?
......!!!!!!!!...................................
RESPONSE --> If we start with the last question we know that T sin(105deg)=1.47 N so we can divide both sides by sin(105deg) to fin T, T=1.47N/sin(105deg),T=1.52N. Since we know that T cos(105deg) and T=1.52N we get 1.52N*cos(105deg)=-.39N confidence assessment: 2
.................................................
......!!!!!!!!...................................
19:34:13 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:39:24 `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?
......!!!!!!!!...................................
RESPONSE --> We will do this problem as we did the preceding problems. Since we have 20 deg and x is in the psoitive direction we have a total of 110deg. The vertical force is 2kg*9.8m/s^2=19.6N. We know T=19.6N/sin(110deg),T=20.9N so the horizontal component is 20.9N*cos(110deg)=-7.1N and since it is in equilibrium we will have another 7.1N of horizontal force to keep it there. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:44:42 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:11:26 `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?
......!!!!!!!!...................................
RESPONSE --> The pendulum string is in the upper left quadrant at 110deg and the chain is in the upper right quadrant at 40deg. If we break this down into two T we get T1 and T2. Since it is in equilibrium we can say T1 cos + T2 cos will =0 and T1 sin + T2 sin-gravity =0. With this we can say, T1 cos(110deg)=-34,T2 cos(40deg)=.77. T1 sin(110deg)=.94,T2 sin(40deg)=.64.The force by gravity will be 9.8m/s^2*2kg=19.6N. We can find T1 by taking T1 -34+T2 .77=0, we subtract T2 .77 from both sides and then divide bothe sides by -.34 we get T1=T2 .77/-.34,T1=T2 -2.3. Now that we know T1 we can put this into the other equation, .95*T2 -2.3+T2 .64-19.6N=0,we first take .95*T2-2.3=T2 -2.2+T2 .64-19.6N=0, we add 19.6N to both sides and then add the T2 we get,T2 -1.56=19.6N, now divide bothe sides by -1.56 we get T2=-12.7N, and since T1=T2 -2.3 we get T1=-12.7N*-2.3N,T1=29.21N confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:15:35 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 -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.
......!!!!!!!!...................................
RESPONSE --> Do we make the x component positive since the othe horizontal force is positive that keeps it at equilibrium self critique assessment: 2