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Practice Problem: To set up for another run, you pull the glider back up the track at a constant speed with a horizontal force F. Determine the magnitude of F, in terms of the two masses in the problem. Answer m₂8.

Practice Problem: To set up for another run, you pull the glider back up the track at a constant speed with a horizontal force F. Determine the magnitude of F, in terms of the two masses in the problem. Answer m₂8.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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130 CHAPTER 5 Applications of Newton's Laws EXAMPLE 5.7 An air track in a physics lab Now we will work a problem involving two objects tied together by a string. Figure 5.9a shows a glider with mass m, that moves on a level, frictionless air track in a physics lab. It is connected by a string passing over a small frictionless pulley to a hanging weight with total mass m₂. The string is light and flexible, and it doesn't stretch. Find the acceleration of each object and the tension in the string. ►FIGURE 5.9 (a) Apparatus SOLUTION SET UP The two objects have different motions, so we need to draw a separate free-body diagram and coordinate system for each, as shown in Figure 5.9b and c. We are free to use different coordinate axes for the two objects; in this case, it's convenient to take the +x direction to the right for the glider and the +y direction downward for the hanging weight. Then the glider has only an x component of acceleration (i.e., ax = a₁, a₁ = 0), and the weight has only a y component (i.e., a2x = 0, a2y = a₂). There is no friction in the pulley, and we consider the string to be massless, so the tension T' in the string is the same throughout; it applies a force with magnitude 7 to each object. The weights are mig and m2g. SOLVE We apply Newton's second law, in component form, to each object in turn. For the glider on the track, Newton's second law gives ΣF₁ = T = max EF, = n + (-m₁g) = m₁a₁y = 0, and for the hanging weight, EF, = m28 + (-T) = m₂a2y- Now, because the string doesn't stretch, the two objects must move equal distances in equal times, and their speeds at any instant must be equal. When the speeds change, they change by equal amounts in a given time, so the accelerations of the two bodies must have the same magnitude a. With our choice of coordinate systems, if a₁x is positive, azy will also be positive, and if a₁, is negative, so will be azy. We can express this relationship as alx = a2y = a. (The directions of the two accelerations are different, of course.) The two Newton's second law equations are then T = mya, m28 + (-T) = m₂a. We'd like to get separate expressions for T and a in terms of the masses and g. m₂ m₁9 (b) Free-body diagram for glider DEXO (n F X m₂9 Y (c) Free-body diagram for weight An easy way to get an expression for a is to replace -T in the second equation by -m₁a and rearrange the result; when we do so, we obtain m2g + (-ma) = m₂a, m2g = (m₁ + m₂)a, and, finally, m1m2 T = 8 = m₁ + m₂ m₂ 8. a = m₁ + m₂ Video Tutor Solution Then, to get an expression for T, we substitute this expression for a back into the first equation. The result is m1 m₁ + m₂ m₂8. 200 REFLECT It's always a good idea to check general symbolic results such as this for particular cases where we can guess what the answer ought to be. For example, if the mass of the glider is zero (m₁ = 0), we expect that the hanging weight (mass m₂) will fall freely with accelera- tion g and there will be no tension in the string. When we substitute m₁ = 0 into the preceding expressions for T and a, they do give T = 0 and a = g, as expected. Also, if m₂ = 0, there is nothing to create ten- sion in the string or accelerate either mass. For this case, the equations give T = 0 and a = 0. Thus, in these two special cases, the results agree with our intuitive expectations. We also note that, in general, the tension T is not equal to the weight m2g of the hanging mass m2, but is less by a factor of m₁/(m₁ + m₂). If I were equal to m2g, then m₂ would be in equilibrium, but it isn't. Practice Problem: To set up for another run, you pull the glider back up the track at a constant speed with a horizontal force F. Determine the magnitude of F, in terms of the two masses in the problem. Answer m28. 18d) 2005 hile wary
Transcribed Image Text:130 CHAPTER 5 Applications of Newton's Laws EXAMPLE 5.7 An air track in a physics lab Now we will work a problem involving two objects tied together by a string. Figure 5.9a shows a glider with mass m, that moves on a level, frictionless air track in a physics lab. It is connected by a string passing over a small frictionless pulley to a hanging weight with total mass m₂. The string is light and flexible, and it doesn't stretch. Find the acceleration of each object and the tension in the string. ►FIGURE 5.9 (a) Apparatus SOLUTION SET UP The two objects have different motions, so we need to draw a separate free-body diagram and coordinate system for each, as shown in Figure 5.9b and c. We are free to use different coordinate axes for the two objects; in this case, it's convenient to take the +x direction to the right for the glider and the +y direction downward for the hanging weight. Then the glider has only an x component of acceleration (i.e., ax = a₁, a₁ = 0), and the weight has only a y component (i.e., a2x = 0, a2y = a₂). There is no friction in the pulley, and we consider the string to be massless, so the tension T' in the string is the same throughout; it applies a force with magnitude 7 to each object. The weights are mig and m2g. SOLVE We apply Newton's second law, in component form, to each object in turn. For the glider on the track, Newton's second law gives ΣF₁ = T = max EF, = n + (-m₁g) = m₁a₁y = 0, and for the hanging weight, EF, = m28 + (-T) = m₂a2y- Now, because the string doesn't stretch, the two objects must move equal distances in equal times, and their speeds at any instant must be equal. When the speeds change, they change by equal amounts in a given time, so the accelerations of the two bodies must have the same magnitude a. With our choice of coordinate systems, if a₁x is positive, azy will also be positive, and if a₁, is negative, so will be azy. We can express this relationship as alx = a2y = a. (The directions of the two accelerations are different, of course.) The two Newton's second law equations are then T = mya, m28 + (-T) = m₂a. We'd like to get separate expressions for T and a in terms of the masses and g. m₂ m₁9 (b) Free-body diagram for glider DEXO (n F X m₂9 Y (c) Free-body diagram for weight An easy way to get an expression for a is to replace -T in the second equation by -m₁a and rearrange the result; when we do so, we obtain m2g + (-ma) = m₂a, m2g = (m₁ + m₂)a, and, finally, m1m2 T = 8 = m₁ + m₂ m₂ 8. a = m₁ + m₂ Video Tutor Solution Then, to get an expression for T, we substitute this expression for a back into the first equation. The result is m1 m₁ + m₂ m₂8. 200 REFLECT It's always a good idea to check general symbolic results such as this for particular cases where we can guess what the answer ought to be. For example, if the mass of the glider is zero (m₁ = 0), we expect that the hanging weight (mass m₂) will fall freely with accelera- tion g and there will be no tension in the string. When we substitute m₁ = 0 into the preceding expressions for T and a, they do give T = 0 and a = g, as expected. Also, if m₂ = 0, there is nothing to create ten- sion in the string or accelerate either mass. For this case, the equations give T = 0 and a = 0. Thus, in these two special cases, the results agree with our intuitive expectations. We also note that, in general, the tension T is not equal to the weight m2g of the hanging mass m2, but is less by a factor of m₁/(m₁ + m₂). If I were equal to m2g, then m₂ would be in equilibrium, but it isn't. Practice Problem: To set up for another run, you pull the glider back up the track at a constant speed with a horizontal force F. Determine the magnitude of F, in terms of the two masses in the problem. Answer m28. 18d) 2005 hile wary
Page 129 Practice Problem 5.7: To set up for another run, you pull the glider back up the track at a constant speed with a horizontal force F. Determine the magnitude of F in terms of the two a a masses in the problem. Answer: m₂g.
Transcribed Image Text:Page 129 Practice Problem 5.7: To set up for another run, you pull the glider back up the track at a constant speed with a horizontal force F. Determine the magnitude of F in terms of the two a a masses in the problem. Answer: m₂g.
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