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age 111 Practice Problem 4.8: If we double the mass of the chain, what are the new values for e magnitude of the tension forces at both ends of the chain? Answer: T₁= 2156 N; T₂- 1960 N.

age 111 Practice Problem 4.8: If we double the mass of the chain, what are the new values for e magnitude of the tension forces at both ends of the chain? Answer: T₁= 2156 N; T₂- 1960 N.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Page 111 Practice Problem 4.8: If we double the mass of the chain, what are the new values for the magnitude of the tension forces at both ends of the chain? Answer: T₁= 2156 N; T₂= 1960 N.
Transcribed Image Text:Page 111 Practice Problem 4.8: If we double the mass of the chain, what are the new values for the magnitude of the tension forces at both ends of the chain? Answer: T₁= 2156 N; T₂= 1960 N.
EXAMPLE 4.8 Tension in a chain with mass Now suppose that the mass of the chain in Example 4.7 is not negligible, but is 10.0 kg. Again, let's find the forces at the ends of the chain. SOLUTION SET UP Again we draw separate free-body diagrams for the re- flector and the chain (Figure 4.27); each is in equilibrium. We take the y axis to be vertically upward. The weight wc of the chain is wc = mc8 = (10.0 kg) (9.8 m/s²) = 98 N. The free-body diagrams are different from those in Figure 4.26 because the magnitudes of the forces acting on the two ends of the chain are no longer equal. (Other- wise, the chain couldn't be in equilibrium.) We label these two forces T₁ and T₂. SOLVE Note that the two forces labeled T₂ form an action-reaction pair; that's how we know that they have the same magnitude. The equi- librium condition ΣΕ = 0 for the reflector is T₂ + (-1960 N) = 0, SO T₂ = 1960 N. There are now three forces acting on the chain: its weight and the forces at the two ends. The equilibrium condition EF, = 0 for the chain is T₁ + (-1₂) + (-98 N) = 0. Note that the y component of T₁ is positive because it points in the +y direction, but the y components of both T2 and 98 N are negative. When we substitute the value T₂ = 1960 N and solve for T₁, we find that T₁ = 2058 N. Alternative Solution: An alternative procedure is to draw a free-body diagram for the composite object consisting of the reflector and the chain together (Figure 4.27b). The two forces on this composite object are the upward force T₁ at the top of the chain and the total weight, with magnitude 1960 N +98 N = 2058 N. Again, we find that 100 T Yonde Pas d'andat WR 5010 WC (a) Free-body diagrams for reflector and chain ASI 202 Video Tutor Solution Reflector + Chain We + WC (b) Free-body diagram for reflector and chain treated as a unit A FIGURE 4.27 Two alternative diagrams we could draw for this problem. T₁= 2058 N. Note that we cannot find T2 by this method; we still need a separate free-body diagram for one of the objects. REFLECT When a composite object can be divided into two or more component parts, there are usually several alternative solutions. Often, the calculations can be simplified by a clever choice of subsystems. Also, solving a problem in two or more alternative ways provides a consistency check that's useful in finding errors. Practice Problem: If we double the mass of the chain, what are the new values for the magnitude of the tension forces at both ends of the chain? Answer: T₁ = 2156 N; T₂ = 1960 N.
Transcribed Image Text:EXAMPLE 4.8 Tension in a chain with mass Now suppose that the mass of the chain in Example 4.7 is not negligible, but is 10.0 kg. Again, let's find the forces at the ends of the chain. SOLUTION SET UP Again we draw separate free-body diagrams for the re- flector and the chain (Figure 4.27); each is in equilibrium. We take the y axis to be vertically upward. The weight wc of the chain is wc = mc8 = (10.0 kg) (9.8 m/s²) = 98 N. The free-body diagrams are different from those in Figure 4.26 because the magnitudes of the forces acting on the two ends of the chain are no longer equal. (Other- wise, the chain couldn't be in equilibrium.) We label these two forces T₁ and T₂. SOLVE Note that the two forces labeled T₂ form an action-reaction pair; that's how we know that they have the same magnitude. The equi- librium condition ΣΕ = 0 for the reflector is T₂ + (-1960 N) = 0, SO T₂ = 1960 N. There are now three forces acting on the chain: its weight and the forces at the two ends. The equilibrium condition EF, = 0 for the chain is T₁ + (-1₂) + (-98 N) = 0. Note that the y component of T₁ is positive because it points in the +y direction, but the y components of both T2 and 98 N are negative. When we substitute the value T₂ = 1960 N and solve for T₁, we find that T₁ = 2058 N. Alternative Solution: An alternative procedure is to draw a free-body diagram for the composite object consisting of the reflector and the chain together (Figure 4.27b). The two forces on this composite object are the upward force T₁ at the top of the chain and the total weight, with magnitude 1960 N +98 N = 2058 N. Again, we find that 100 T Yonde Pas d'andat WR 5010 WC (a) Free-body diagrams for reflector and chain ASI 202 Video Tutor Solution Reflector + Chain We + WC (b) Free-body diagram for reflector and chain treated as a unit A FIGURE 4.27 Two alternative diagrams we could draw for this problem. T₁= 2058 N. Note that we cannot find T2 by this method; we still need a separate free-body diagram for one of the objects. REFLECT When a composite object can be divided into two or more component parts, there are usually several alternative solutions. Often, the calculations can be simplified by a clever choice of subsystems. Also, solving a problem in two or more alternative ways provides a consistency check that's useful in finding errors. Practice Problem: If we double the mass of the chain, what are the new values for the magnitude of the tension forces at both ends of the chain? Answer: T₁ = 2156 N; T₂ = 1960 N.
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