The second photo has background info to help solve the practice problem Page 111 Practice Problem 4.8: If we double the mass of the chain, what are the new values forthe 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 massNow suppose that the mass of the chain in Example 4.7 is not negligible, but is 10.0 kg. Again, let's find theforces at the ends of the chain.SOLUTIONSET UP Againwe draw separate free-body diagrams for the re-flector and the chain (Figure 4.27); each is in equilibrium. We takethe y axis to be vertically upward. The weight wc of the chain iswc = mc8 = (10.0 kg) (9.8 m/s²) = 98 N. The free-body diagramsare different from those in Figure 4.26 because the magnitudes of theforces 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 forcesT₁ and T₂.SOLVE Note that the two forces labeled T₂ form an action-reactionpair; that's how we know that they have the same magnitude. The equi-librium condition ΣΕ = 0 for the reflector isT₂ + (-1960 N) = 0,SO T₂ = 1960 N.There are now three forces acting on the chain: its weight and the forcesat the two ends. The equilibrium condition EF, = 0 for the chain isT₁ + (-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 findthatT₁ = 2058 N.Alternative Solution: An alternative procedure is to draw a free-bodydiagram for the composite object consisting of the reflector and thechain together (Figure 4.27b). The two forces on this composite objectare 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 that100TYondePas d'andatWR5010WC(a) Free-body diagrams forreflector and chainASI 202Video Tutor SolutionReflector +ChainWe + WC(b) Free-body diagram for reflectorand chain treated as a unitA FIGURE 4.27 Two alternative diagrams we could draw forthis problem.T₁= 2058 N. Note that we cannot find T2 by this method; we still needa separate free-body diagram for one of the objects.REFLECT When a composite object can be divided into two or morecomponent 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 aconsistency check that's useful in finding errors.Practice Problem: If we double the mass of the chain, what are thenew values for the magnitude of the tension forces at both ends of thechain? Answer: T₁ = 2156 N; T₂ = 1960 N.

Given Initial Mass, m1 = 10 kg Final mass (after doubled), m2 = 20 kg…

Answered: age 111 Practice Problem 4.8: If we…

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

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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