This question does NOT need to be solved, but I would appreciate some clarification. --------PROBLEM: Collar A has a ramp that is welded to it and a force P = 5 lb applied as shown. Collar A and the ramp weigh 3 lb, and block B weighs 0.8 lb. Neglecting friction, determine the tension in the cable.-------Can you explain in detail how to do the constraint equations ( in general)? And then specifically explain in this case why it is XB/A and how the two tension from the cable with block B does not effect the equation (figure four)? I understand how to solve it but I am still having trouble with the constraint equations. XAW0 = 50°XB/ABFig. 4.Position vectors fordependent motion.ANALYSIS: You can obtain a scalar equation by applying Newton's secondlaw to each of these systems.a. System 1:ΕΣΕ. =mдaАx+1EF, = mAdayb. Block B:+\ ΣFx = mBª BxNA - NB COS 50° +27 cos 40° = 0-WA+P+T-27 sin 40° - NB sin 50° = -mдад−2T+ Wg sin 40° = mpa BIA + mua, sin 40°(3)+/ΣF₂= m BaByNB - WB cos 40° = −mª cos 40°(4)You now have four equations and five unknowns (T, NÃ, Nâ, ªд, and aBIA), soyou need one more equation. The motions of A and B are related because theyare connected by a cable.(1)(2)Constraint Equations. Define position vectors as shown in Fig. 4. Notethat the positive directions for the position vectors for A and B are defined fromthe kinetic diagrams in Figs. 2 and 3. Assuming the cable is inextensible, youcan write the lengths in terms of the coordinates and then differentiate.Constraint equation for the cable: XA + 2XB/A = constantDifferentiating this twice givesaε = -2a BIA(5)You now have five equations and five unknowns, so all that remains isto substitute the known values and solve for the unknowns. The resultsare N₁ = -0.1281 lb, NB 0.869 lb, T = 0.281 lb, a = −13.46 ft/s², andaBIA = 6.73 ft/s².=T= 0.281 lb PAW0 = 50°0 = 50°BSystem 1B-System 2Fig. 1. System boundaries.Sample Problem 12.4Collar A has a ramp that is welded to it and a force P = 5 lb applied as shown.Collar A and the ramp weigh 3 lb, and block B weighs 0.8 lb. Neglecting fric-tion, determine the tension in the cable.STRATEGY: The principle you need to use is Newton's second law. Becausea block is sliding down an incline and a cable is connecting A and B, you alsoneed to use relative motion and dependent motion.MODELING: Model A and B as particles and assume all surfaces aresmooth. As usual, start by choosing a system and then drawing a free-bodydiagram and a kinetic diagram. This problem has two systems, and you need tobe careful with how you define them. The easiest systems to use are (a) collarA with its pulley and the ramp welded to it (system 1) and (b) block B and thepulley attached to it (system 2), as shown in Fig. 1. The free-body and kineticdiagrams for system 1 are shown in Fig. 2. The free-body and kinetic diagramsfor B are a little trickier, because you don't know the direction of the accelera-tion of B.Kinematics for Block B. Express the acceleration as of block B as the sumof the acceleration of A and the acceleration of B relative to A. Hence,aB=aA + aBIAHereaBIAis directed along the inclined surface of the wedge. Now you can drawthe appropriate diagrams (Fig. 3). Note that you do not need to use the same x-ycoordinate system for each mass, because these directions are simply used forobtaining the scalar equations.WTTT0 = 50°NBTFig. 2. Free-body diagram and kinetic diagram for system 1.ByLXmдªAmBªB/AmBa AFig. 3. Free-body diagram and kinetic diagram for B.0 = 50°

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This question does NOT need to be solved, but I would appreciate some clarification. --------PROBLEM: Collar A has a ramp that is welded to it and a force P = 5 lb applied as shown. Collar A and the ramp weigh 3 lb, and block B weighs 0.8 lb. Neglecting friction, determine the tension in the cable.-------Can you explain in detail how to do the constraint equations ( in general)? And then specifically explain in this case why it is XB/A and how the two tension from the cable with block B does not effect the equation (figure four)?  I understand how to solve it but I am still having trouble with the constraint equations. XAW0 = 50°XB/ABFig. 4.Position vectors fordependent motion.ANALYSIS: You can obtain a scalar equation by applying Newton's secondlaw to each of these systems.a. System 1:ΕΣΕ. =mдaАx+1EF, = mAdayb. Block B:+\ ΣFx = mBª BxNA - NB COS 50° +27 cos 40° = 0-WA+P+T-27 sin 40° - NB sin 50° = -mдад−2T+ Wg sin 40° = mpa BIA + mua, sin 40°(3)+/ΣF₂= m BaByNB - WB cos 40° = −mª cos 40°(4)You now have four equations and five unknowns (T, NÃ, Nâ, ªд, and aBIA), soyou need one more equation. The motions of A and B are related because theyare connected by a cable.(1)(2)Constraint Equations. Define position vectors as shown in Fig. 4. Notethat the positive directions for the position vectors for A and B are defined fromthe kinetic diagrams in Figs. 2 and 3. Assuming the cable is inextensible, youcan write the lengths in terms of the coordinates and then differentiate.Constraint equation for the cable: XA + 2XB/A = constantDifferentiating this twice givesaε = -2a BIA(5)You now have five equations and five unknowns, so all that remains isto substitute the known values and solve for the unknowns. The resultsare N₁ = -0.1281 lb, NB 0.869 lb, T = 0.281 lb, a = −13.46 ft/s², andaBIA = 6.73 ft/s².=T= 0.281 lb PAW0 = 50°0 = 50°BSystem 1B-System 2Fig. 1. System boundaries.Sample Problem 12.4Collar A has a ramp that is welded to it and a force P = 5 lb applied as shown.Collar A and the ramp weigh 3 lb, and block B weighs 0.8 lb. Neglecting fric-tion, determine the tension in the cable.STRATEGY: The principle you need to use is Newton's second law. Becausea block is sliding down an incline and a cable is connecting A and B, you alsoneed to use relative motion and dependent motion.MODELING: Model A and B as particles and assume all surfaces aresmooth. As usual, start by choosing a system and then drawing a free-bodydiagram and a kinetic diagram. This problem has two systems, and you need tobe careful with how you define them. The easiest systems to use are (a) collarA with its pulley and the ramp welded to it (system 1) and (b) block B and thepulley attached to it (system 2), as shown in Fig. 1. The free-body and kineticdiagrams for system 1 are shown in Fig. 2. The free-body and kinetic diagramsfor B are a little trickier, because you don't know the direction of the accelera-tion of B.Kinematics for Block B. Express the acceleration as of block B as the sumof the acceleration of A and the acceleration of B relative to A. Hence,aB=aA + aBIAHereaBIAis directed along the inclined surface of the wedge. Now you can drawthe appropriate diagrams (Fig. 3). Note that you do not need to use the same x-ycoordinate system for each mass, because these directions are simply used forobtaining the scalar equations.WTTT0 = 50°NBTFig. 2. Free-body diagram and kinetic diagram for system 1.ByLXmдªAmBªB/AmBa AFig. 3. Free-body diagram and kinetic diagram for B.0 = 50°

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