PROBLEM 8-23 (page 422, 14th edition) The beam is supported in a horizontal position by a pin at A and a roller at B. The weight P of the beam is 60 N. The weight of the roller may be neglected. The coefficients of static friction at B and Care 0.4 and 0.5. The value of 0 is increased until impending motion occurs. · Determine the values of all friction and normal forces at impending motion. · Determine the value of 0 at impending motion. P 2 m 2 m A B C 08_PROB_023 Copyright 0 2016 Pearson fiducation, Az Rights Reserunit ALWAYS LEARNING Statics, Fourteenth Edition R.C. Hibbeler Copyright @2016 by Pearson Education, Inc. All rights reserved. PEARSON
8-23 (page 422, 14th edition) COMPLETION OF PROBLEM · Data P= 60N l = 2m W roller =0 HB = 0.4 · Cartesian component equations of equilibrium MC = 0.5 Beam : IF = 0: `X (1) `x A + F = 0 T'B IF = 0: y (2) y A + N IP = 0 I (M2)A = 0: z'A B 2IN_ 0 1P = 0 (3) Roller: IF =0: N sin/ I F cos! IFp = 0 X (4) or IF = 0: IF IF cos / + N sin = 0 (4*) IF = 0: NC Cose + Fsin/ INp = 0 y (5) or IF =0: N IF sim/ IN cos/ = 0 (5*) I (M2)% =0: rFp = IF c (6) Restrictions: B Z B U Eqs. (1) to (6) contain 7 unknowns: X y AANENE We need another equation to solve the problem.
8-23 (page 422, 14th edition) COMPLETION OF PROBLEM page 2 · Data P = 60N l = 2m W = 0 HB= 0.4 uc=0.5 roller · It follows from eqs. (2) and (3) that A = N = P /2 = 30N It follows from eqs. (4), (4*) and (6) or (5), (5*) and (6) that N = N B SO N = 30N · Impending sliding occurs at B since UB < Hlc. `B F = UN 'B (7) Then F =F. = 12N and A = 8-12N X · Angle / Write F = N tan/ where tan/ = 1B B B Eq. (4*): I tan /, I cos / tan/ + sin/ = 0 SO sin/ p = sin/ cos/ I cos/ sin/ B = sin( [ [B) That is, @ = 20 p = 2tan11 up = 43.6 (8)