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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.

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.

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
Problem 1.1MA
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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. 

XA W 0 = 50° XB/A B Fig. 4. Position vectors for dependent motion. ANALYSIS: You can obtain a scalar equation by applying Newton's second law to each of these systems. a. System 1: ΕΣΕ. =mдaАx +1EF, = mAday b. Block B: +\ ΣFx = mBª Bx NA - 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 BaBy NB - WB cos 40° = −mª cos 40° (4) You now have four equations and five unknowns (T, NÃ, Nâ, ªд, and aBIA), so you need one more equation. The motions of A and B are related because they are connected by a cable. (1) (2) Constraint Equations. Define position vectors as shown in Fig. 4. Note that the positive directions for the position vectors for A and B are defined from the kinetic diagrams in Figs. 2 and 3. Assuming the cable is inextensible, you can write the lengths in terms of the coordinates and then differentiate. Constraint equation for the cable: XA + 2XB/A = constant Differentiating this twice gives aε = -2a BIA (5) You now have five equations and five unknowns, so all that remains is to substitute the known values and solve for the unknowns. The results are N₁ = -0.1281 lb, NB 0.869 lb, T = 0.281 lb, a = −13.46 ft/s², and aBIA = 6.73 ft/s². = T= 0.281 lb
Transcribed Image Text:XA W 0 = 50° XB/A B Fig. 4. Position vectors for dependent motion. ANALYSIS: You can obtain a scalar equation by applying Newton's second law to each of these systems. a. System 1: ΕΣΕ. =mдaАx +1EF, = mAday b. Block B: +\ ΣFx = mBª Bx NA - 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 BaBy NB - WB cos 40° = −mª cos 40° (4) You now have four equations and five unknowns (T, NÃ, Nâ, ªд, and aBIA), so you need one more equation. The motions of A and B are related because they are connected by a cable. (1) (2) Constraint Equations. Define position vectors as shown in Fig. 4. Note that the positive directions for the position vectors for A and B are defined from the kinetic diagrams in Figs. 2 and 3. Assuming the cable is inextensible, you can write the lengths in terms of the coordinates and then differentiate. Constraint equation for the cable: XA + 2XB/A = constant Differentiating this twice gives aε = -2a BIA (5) You now have five equations and five unknowns, so all that remains is to substitute the known values and solve for the unknowns. The results are N₁ = -0.1281 lb, NB 0.869 lb, T = 0.281 lb, a = −13.46 ft/s², and aBIA = 6.73 ft/s². = T= 0.281 lb
P A W 0 = 50° 0 = 50° B System 1 B -System 2 Fig. 1. System boundaries. Sample Problem 12.4 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 fric- tion, determine the tension in the cable. STRATEGY: The principle you need to use is Newton's second law. Because a block is sliding down an incline and a cable is connecting A and B, you also need to use relative motion and dependent motion. MODELING: Model A and B as particles and assume all surfaces are smooth. As usual, start by choosing a system and then drawing a free-body diagram and a kinetic diagram. This problem has two systems, and you need to be careful with how you define them. The easiest systems to use are (a) collar A with its pulley and the ramp welded to it (system 1) and (b) block B and the pulley attached to it (system 2), as shown in Fig. 1. The free-body and kinetic diagrams for system 1 are shown in Fig. 2. The free-body and kinetic diagrams for 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 sum of the acceleration of A and the acceleration of B relative to A. Hence, aB=aA + aBIA Here aBIA is directed along the inclined surface of the wedge. Now you can draw the appropriate diagrams (Fig. 3). Note that you do not need to use the same x-y coordinate system for each mass, because these directions are simply used for obtaining the scalar equations. W T T T 0 = 50° NB T Fig. 2. Free-body diagram and kinetic diagram for system 1. B y L X mдªA mBªB/A mBa A Fig. 3. Free-body diagram and kinetic diagram for B. 0 = 50°
Transcribed Image Text:P A W 0 = 50° 0 = 50° B System 1 B -System 2 Fig. 1. System boundaries. Sample Problem 12.4 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 fric- tion, determine the tension in the cable. STRATEGY: The principle you need to use is Newton's second law. Because a block is sliding down an incline and a cable is connecting A and B, you also need to use relative motion and dependent motion. MODELING: Model A and B as particles and assume all surfaces are smooth. As usual, start by choosing a system and then drawing a free-body diagram and a kinetic diagram. This problem has two systems, and you need to be careful with how you define them. The easiest systems to use are (a) collar A with its pulley and the ramp welded to it (system 1) and (b) block B and the pulley attached to it (system 2), as shown in Fig. 1. The free-body and kinetic diagrams for system 1 are shown in Fig. 2. The free-body and kinetic diagrams for 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 sum of the acceleration of A and the acceleration of B relative to A. Hence, aB=aA + aBIA Here aBIA is directed along the inclined surface of the wedge. Now you can draw the appropriate diagrams (Fig. 3). Note that you do not need to use the same x-y coordinate system for each mass, because these directions are simply used for obtaining the scalar equations. W T T T 0 = 50° NB T Fig. 2. Free-body diagram and kinetic diagram for system 1. B y L X mдªA mBªB/A mBa A Fig. 3. Free-body diagram and kinetic diagram for B. 0 = 50°
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