7.3)instructions: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 7.3Determine the maximum force P that can be applied to block A in Fig. (a) withoutcausing either block to move.Surface OAW, = 100 IbH, = 0.2SolutionThe 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. Becauseimpending 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 inFigs. (b) and (c), respectively. Note that the equilibrium of each block yieldsWy= 200 lbSurface OA = 0.1(a)348W,- J00 IbIN,= 100 Ib|W = 100 IbP.N - 100 lbWs = 200 lbW 200 IbF2F,IN = 300 lbN, = 300 lb(b)(c)N1 = 100 lb and N2 = 300 lb, as shown on the FBDS. Attention should be paidto the friction forces. The friction force F2 on the bottom of block B is directedto 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 Fto slide B to the right is resisted by the friction force F2. Note that F, and N, donot appear in the FBD in Fig. (b), because they are internal to the system of bothblocks.Two solutions are presented here to illustrate both methods of analysisdescribed in Art. 7.3.Method of Analysis 1First, assume impending sliding at surface 1. Under this assumption we haveFi = (Fi)max = (,) N = 0.2(100) = 20 lbThe FBD of block A then givesEF, = 0 +P - F1 = 0P = F, = 20 lbNexl, assuic inupculing sliding al suilace 2, wliclı givesF2 = (F;)mas = (4,)2N2 = 0.1(300) = 30 lbFrom the FBD of the entire system, Fig. (b), we then obtainEF, =0 +ΣΕ0P - F2 = 0P = F = 30 lb349 So far, we have determined that P= 20 lb will cause motion to impend atsurface 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 isP = 20 IbAnswerwith sliding impending at surface 1.Be sure you understand that the largest force that can be applied is the smallerof the two values determined in the preceding calculations. If sliding impendswhen P = 20 lb, then the system would not be at rest when P= 30 lb.Method of Analysis 2Assume 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.CheckThe assumption of impending motion at surface 1 is checked by comparing thefriction force F2 with (F2)mar, its maximum possible value. Using the FBD ofblock B, we obtainEF, =0F - F2 = 0F1 = F2 = 20 lbBecause (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 correctassumption, so that the answer is P= 201b.Had F, turned out to be greater than (F2)max, we would know that slidingwould first impend at surface 2, and the problem would have to be solved againmaking use of this fact.CommentThere are five unknowns in this problem: P, N1, F1, N2, and F2. There arefour independent equilibrium equations: two for each block. The assumption ofimpending motion at one surface provides the fifth equation, F = µ,N, makingthe problem statically determinate.In our solution, we have considered two possible modes of impendingmotion-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 woulddetermine if simultancous impending sliding is indeed the case. In Method ofAnalysis 1 the two computed values of P would be equal. In Method of Analysis 2the check would reveal that F= Fmax at both surfaces.Caution A mistake that is often made in the analysis of Type III problems is toassume that motion impends at the surface with the smallest coefficient of staticfriction. The solution to this problem illustrates that this need not be the case.350

In this problem, question asking what is the value of force P required to move either block we find…

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

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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