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Sample Problem 7.3 Determine the maximum force P that can be applied to block A in Fig. (a) without causing either block to move. Surface O A - 02 W.- 100 Ib Solution The problem statement indicates that we are to find P that would cause impend- ing motion. However, there are two possible ways in which motion can impend: Surface O impending sliding at surface 1, or impending sliding at surface 2. Because impending sliding is specified but not its location, this is a Type III problem. The free-body diagrams of the entire system and cach block are shown in Figs. (b) and (c), respectively. Note that the equilibrium of cach block yields W= 200 Ib A = 0.1 (a)

Sample Problem 7.3 Determine the maximum force P that can be applied to block A in Fig. (a) without causing either block to move. Surface O A - 02 W.- 100 Ib Solution The problem statement indicates that we are to find P that would cause impend- ing motion. However, there are two possible ways in which motion can impend: Surface O impending sliding at surface 1, or impending sliding at surface 2. Because impending sliding is specified but not its location, this is a Type III problem. The free-body diagrams of the entire system and cach block are shown in Figs. (b) and (c), respectively. Note that the equilibrium of cach block yields W= 200 Ib A = 0.1 (a)

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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7.3)

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Read and Analyze the sample problems given. Give a thorough analysis stating what type of situation the author dealt with, how the problem was solved, the assumptions that the author considered in solving the problem, and why where these assumptions were made. It should have 5 sentences for each problem.

Sample Problem 7.3 Determine the maximum force P that can be applied to block A in Fig. (a) without causing either block to move. Surface O A W, = 100 Ib H, = 0.2 Solution The problem statement indicates that we are to find P that would cause impend- ing motion. However, there are two possible ways in which motion can impend: impending sliding at surface 1, or impending sliding at surface 2. Because impending sliding is specified but not its location, this is a Type III problem. The free-body diagrams of the entire system and cach block are shown in Figs. (b) and (c), respectively. Note that the equilibrium of each block yields Wy= 200 lb Surface O A = 0.1 (a) 348 W,- J00 Ib IN,= 100 Ib |W = 100 Ib P. N - 100 lb Ws = 200 lb W 200 Ib F2 F, IN = 300 lb N, = 300 lb (b) (c) N1 = 100 lb and N2 = 300 lb, as shown on the FBDS. Attention should be paid to the friction forces. The friction force F2 on the bottom of block B is directed to the left, opposite the direction in which sliding would impend. At surface 1, block A would tend to slide to the right, across the top of block B. Therefore, F, is directed to the left on block A, and to the right on block B. The tendency of F to slide B to the right is resisted by the friction force F2. Note that F, and N, do not appear in the FBD in Fig. (b), because they are internal to the system of both blocks. Two solutions are presented here to illustrate both methods of analysis described in Art. 7.3. Method of Analysis 1 First, assume impending sliding at surface 1. Under this assumption we have Fi = (Fi)max = (,) N = 0.2(100) = 20 lb The FBD of block A then gives EF, = 0 + P - F1 = 0 P = F, = 20 lb Nexl, assuic inupculing sliding al suilace 2, wliclı gives F2 = (F;)mas = (4,)2N2 = 0.1(300) = 30 lb From the FBD of the entire system, Fig. (b), we then obtain EF, =0 + ΣΕ0 P - F2 = 0 P = F = 30 lb 349
Transcribed Image Text:Sample Problem 7.3 Determine the maximum force P that can be applied to block A in Fig. (a) without causing either block to move. Surface O A W, = 100 Ib H, = 0.2 Solution The problem statement indicates that we are to find P that would cause impend- ing motion. However, there are two possible ways in which motion can impend: impending sliding at surface 1, or impending sliding at surface 2. Because impending sliding is specified but not its location, this is a Type III problem. The free-body diagrams of the entire system and cach block are shown in Figs. (b) and (c), respectively. Note that the equilibrium of each block yields Wy= 200 lb Surface O A = 0.1 (a) 348 W,- J00 Ib IN,= 100 Ib |W = 100 Ib P. N - 100 lb Ws = 200 lb W 200 Ib F2 F, IN = 300 lb N, = 300 lb (b) (c) N1 = 100 lb and N2 = 300 lb, as shown on the FBDS. Attention should be paid to the friction forces. The friction force F2 on the bottom of block B is directed to the left, opposite the direction in which sliding would impend. At surface 1, block A would tend to slide to the right, across the top of block B. Therefore, F, is directed to the left on block A, and to the right on block B. The tendency of F to slide B to the right is resisted by the friction force F2. Note that F, and N, do not appear in the FBD in Fig. (b), because they are internal to the system of both blocks. Two solutions are presented here to illustrate both methods of analysis described in Art. 7.3. Method of Analysis 1 First, assume impending sliding at surface 1. Under this assumption we have Fi = (Fi)max = (,) N = 0.2(100) = 20 lb The FBD of block A then gives EF, = 0 + P - F1 = 0 P = F, = 20 lb Nexl, assuic inupculing sliding al suilace 2, wliclı gives F2 = (F;)mas = (4,)2N2 = 0.1(300) = 30 lb From the FBD of the entire system, Fig. (b), we then obtain EF, =0 + ΣΕ0 P - F2 = 0 P = F = 30 lb 349
So far, we have determined that P= 20 lb will cause motion to impend at surface 1 and that P= 30 lb will cause motion to impend at surface 2. Therefore, the largest force that can be applied without causing either block to move is P = 20 Ib Answer with sliding impending at surface 1. Be sure you understand that the largest force that can be applied is the smaller of the two values determined in the preceding calculations. If sliding impends when P = 20 lb, then the system would not be at rest when P= 30 lb. Method of Analysis 2 Assume impending motion at surface 1. We would then obtain P = (Fi)max = 20 lb, as determined in Method of Analysis 1. Next, we check the assumption. Check The assumption of impending motion at surface 1 is checked by comparing the friction force F2 with (F2)mar, its maximum possible value. Using the FBD of block B, we obtain EF, =0 F - F2 = 0 F1 = F2 = 20 lb Because (F2)max =(4,)2N2=0.1(300) = 30 Ib, we have F2 < (F2)max- Con- sequently, we conclude that impending motion at surface 1 is the correct assumption, so that the answer is P= 201b. Had F, turned out to be greater than (F2)max, we would know that sliding would first impend at surface 2, and the problem would have to be solved again making use of this fact. Comment There are five unknowns in this problem: P, N1, F1, N2, and F2. There are four independent equilibrium equations: two for each block. The assumption of impending motion at one surface provides the fifth equation, F = µ,N, making the problem statically determinate. In our solution, we have considered two possible modes of impending motion-impending sliding at surface 1 and impending sliding at surface 2. Impending sliding at both surfaces at the same time is obviously a third possibility, but it need not be examined independently. Both of the foregoing analyses would determine if simultancous impending sliding is indeed the case. In Method of Analysis 1 the two computed values of P would be equal. In Method of Analysis 2 the check would reveal that F= Fmax at both surfaces. Caution A mistake that is often made in the analysis of Type III problems is to assume that motion impends at the surface with the smallest coefficient of static friction. The solution to this problem illustrates that this need not be the case. 350
Transcribed Image Text:So far, we have determined that P= 20 lb will cause motion to impend at surface 1 and that P= 30 lb will cause motion to impend at surface 2. Therefore, the largest force that can be applied without causing either block to move is P = 20 Ib Answer with sliding impending at surface 1. Be sure you understand that the largest force that can be applied is the smaller of the two values determined in the preceding calculations. If sliding impends when P = 20 lb, then the system would not be at rest when P= 30 lb. Method of Analysis 2 Assume impending motion at surface 1. We would then obtain P = (Fi)max = 20 lb, as determined in Method of Analysis 1. Next, we check the assumption. Check The assumption of impending motion at surface 1 is checked by comparing the friction force F2 with (F2)mar, its maximum possible value. Using the FBD of block B, we obtain EF, =0 F - F2 = 0 F1 = F2 = 20 lb Because (F2)max =(4,)2N2=0.1(300) = 30 Ib, we have F2 < (F2)max- Con- sequently, we conclude that impending motion at surface 1 is the correct assumption, so that the answer is P= 201b. Had F, turned out to be greater than (F2)max, we would know that sliding would first impend at surface 2, and the problem would have to be solved again making use of this fact. Comment There are five unknowns in this problem: P, N1, F1, N2, and F2. There are four independent equilibrium equations: two for each block. The assumption of impending motion at one surface provides the fifth equation, F = µ,N, making the problem statically determinate. In our solution, we have considered two possible modes of impending motion-impending sliding at surface 1 and impending sliding at surface 2. Impending sliding at both surfaces at the same time is obviously a third possibility, but it need not be examined independently. Both of the foregoing analyses would determine if simultancous impending sliding is indeed the case. In Method of Analysis 1 the two computed values of P would be equal. In Method of Analysis 2 the check would reveal that F= Fmax at both surfaces. Caution A mistake that is often made in the analysis of Type III problems is to assume that motion impends at the surface with the smallest coefficient of static friction. The solution to this problem illustrates that this need not be the case. 350
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