Question

A force P is applied at B, in a regular hexagon, as shown. Determine

 (a) the moment of the force P about E, 

 (b) the magnitude and sense of the horizontal force applied at A that creates the same moment about E, 

 (c) the (i) magnitude and (ii) direction (from horizontal) of the smallest force applied at A that creates the same moment about E. (Hint: Internal angle of regular hexagon is 120°) 

 

 If P=500 N, a=40°, a=300 mm

Answer 1

600 - (a) from diagram, X+60+ 8 = 180 B = 120-2 Nos, perb. seberation between fosce P & point E A LE = X = BE Sim B → X=2a Sin (120-) Neco, moment of force, T = Force x peop. Seperation - P X LE = Pxx - Px za Sin (120-x) = (SON) (@x300 mm) Sin(120-40) TT-- - 500 x 2 x 300 x Sin (120-40) Istoo = 300 sin. 8o - 300 x 0.9848 T = 995.44 N-m

horizontally. (b). let force (P) on A acts LCDE = 1200 LL DE = 60° - DE Perpendicularo seberation betworen / force P & point of rotation at LE = de from a LDE, Sin 60-LET x= a Sin 6o e Ba/2 Moment of force, Taxexp = va pa but ascending to Roblem, 2 Pa = 295.44 EXP x(300) = 295.44 A P= 295-4+ x20 loco --1/37.18 N-m 313 & *P = 1137-2 NM

(C) tet force () on A acts at angle o from horizontal. from diagram, AE: 3 a A A 28-'12 T 20 BC f LLAE = (@+30) Nag In A ALE, Sin (0+30) - Le 5 Sin(@+30) Now, Coreple of force, : I = PX LE * T = Px 53a sin(+30) but accerding to problem, 0995.44 =13*(30)x P x Sin (0+30) loco

2954.4 2954.4 *** = 568.59 N-m P Sin (0+30) = - 353 5.196 nax. or min, p Sin CO +30) - 568.59 a pe 568.59 Corec (0+30) - 0 for p to be max. or min, de = 0 = 500:59. f- corec (ora) torad} 7 Corec (0+30). Cot(0+30) = 0 Here, fon Corec (0+30) to Cot (0+30) = 0 Cot (0+3.) - Got 90 | 0430-96 Jo=6ool .:64 0=60°, force will be minimum, from0, Pmin = 568.59 Corec (60 +30) = 568.59 Cobec 90 -568.59 N - P min = 568.59 NL at 0= 60 - -

Answer 2

Not sure if the forces around A are what this question is asking for, but are they correct for what they are?  How does the moment around E look?  Thanks for any advice.  :)