Engineering Mechanics - Statics Episode 3 Part 7 pptx

... Engineering Mechanics - Statics Chapter 11Problem 1 1-1 7 Each member of the pin-connected mechanism has amass m1. If the spring is unstretched ... publisher. Engineering Mechanics - Statics Chapter 11Vγπa2h22W3a8−⎛⎜⎝⎞⎟⎠cosθ()=θVddγπa2h22W3a8−⎛⎜⎝⎞⎟⎠− sinθ()=2θVdd2γπa2h22W3a8−⎛⎜⎝⎞⎟⎠− ... publisher. Engineering Mechanics - Statics Chapter 11 abc60(+ 60 (- 60(Solution:asin 60 deg()bsin 60 degθ−()= basin 60 degθ−()sin 60 deg()=VW2 3 dsin 60 deg( ) cosθ()bcos 30 ...

... publisher. Engineering Mechanics - Statics Chapter 10the x and y axes.Given:a 30 mm=b 170 mm=c 30 mm=d 140 mm=e 30 mm=f 30 mm=g 70 mm=Solution:Ix1 3 ac d+ e+() 3 1 3 bc 3 +112ge 3 + ... publisher. Engineering Mechanics - Statics Chapter 10Solution:Iyab 3 31 36 ac 3 +12ac bc 3 +⎛⎜⎝⎞⎟⎠2+1 36 db c+() 3 +12db c+()2 bc+() 3 ⎡⎢⎣⎤⎥⎦2+=Iy1 971 in4=Problem 1 0-4 5Locate ... publisher. Engineering Mechanics - Statics Chapter 10Solution:Ix'1 36 bh 3 =xc2 3 a12ha aba− 3 +⎛⎜⎝⎞⎟⎠12hb a−()+12ha12hb a−()+=ba+ 3 =Iy'1 36 ha 3 12haba+ 3 2 3 a−⎛⎜⎝⎞⎟⎠2+1 36 hb...

... the publisher. Engineering Mechanics - Statics Chapter 9Units Used:Mg 10 3 kg=kN 10 3 N=Given:L 8m=ρw1.0Mgm 3 =a 3m=b 2m=g 9.81ms2=Solution:F 3 ρwgabL= F 3 470 .88 kN=F2ρwgabL= ... writing from the publisher. Engineering Mechanics - Statics Chapter 9F 7. 62 10 3 ×= Nxc1F05xx240−x 1+ 34 0+⎛⎜⎝⎞⎟⎠6⌠⎮⎮⌡d=xc2 .74 = m992© 20 07 R. C. Hibbeler. Published ... Engineering Mechanics - Statics Chapter 9Units Used:kip 10 3 lb=Given:γ56lbft 3 =c 8ft=w 6ft= d 4ft=a 10 ft= e 3ft=b 2ft= f 4ft=Solution:wBwγb= wB 672 lbft= wCwγbc+()=...

... publisher. Engineering Mechanics - Statics Chapter 9Solution:V2 3 πa 3 π 3 a2a−=π 3 a 3 =zc1V5a82 3 πa 3 ⎛⎜⎝⎞⎟⎠ 3 4aπ 3 a 3 ⎜⎝⎞⎟⎠−⎡⎢⎣⎤⎥⎦=zca2=Problem 9 -7 6Determine ... Engineering Mechanics - Statics Chapter 9Solution:A 2a 3 2a22π=A 3 πa2=V 212a2⎛⎜⎝⎞⎟⎠ 3 2a⎛⎜⎝⎞⎟⎠ 3 6a2π⎛⎜⎝⎞⎟⎠=Vπ4a 3 =Problem 9-8 9Sand is ... cone.Given: a 80 mm=b 60 mm=c 30 mm=d 30 mm=Solution:V1 3 πd2aπd2b+2 3 πd 3 +=xc1V1 3 πd2a3a4⎛⎜⎝⎞⎟⎠πd2bab2+⎛⎜⎝⎞⎟⎠+2 3 πd 3 ab+3c8+⎛⎜⎝⎞⎟⎠+⎡⎢⎣⎤⎥⎦=xc105.2...

... publisher. Engineering Mechanics - Statics Chapter 9Given: a 2ft=b 2ft=Solution:V0bzπa2zb⌠⎮⎮⌡d=V 12.566 ft 3 =zc1V0bzzπa2zb⌠⎮⎮⌡d=zc1 .33 3 ft=Problem 9 -3 5Locate ... writing from the publisher. Engineering Mechanics - Statics Chapter 9Problem 9 -3 6Locate the centroid of the quarter-cone.Solution:rahhz−()= zcz= xcyc=4r 3 π=V0hzπ4ahhz−()⎡⎢⎣⎤⎥⎦2⌠⎮⎮⌡d=112ha2π=zc12ha2π0hzzπ4ahhz−()⎡⎢⎣⎤⎥⎦2⌠⎮⎮⌡d=14h=xc12ha2π0hz4 3 πahhz−()⎡⎢⎣⎤⎥⎦π4ahh ... z2a21y2b2−⎛⎜⎜⎝⎞⎟⎟⎠=V0byπa21y2b2−⎛⎜⎜⎝⎞⎟⎟⎠⌠⎮⎮⎮⌡d=1 3 b 3 b2b2−b2a2π=yc 3 2ba2π0byyπa21y2b2−⎛⎜⎜⎝⎞⎟⎟⎠⌠⎮⎮⎮⌡d= 3 8 bb2= yc3b8=By symmetryxczc= 0=Problem 9 -3 3 Locate the centroid...

... permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8Given:M 75 kg=a 70 0 mm=b 25 mm=c 30 0 mm=d 200 mm=e160 mm=μs0 .3= e 2 .71 8=Solution:Initial guesses:T11N= ... publisher. Engineering Mechanics - Statics Chapter 8FEPcbb2c2+ cosβ()b sinβ()a+()= FE 72 .7 N=The equilibrium of clamped block requires that FDFE= FD 72 .7 N=Problem 8-8 2The ... publisher. Engineering Mechanics - Statics Chapter 8a 70 0 mm=b 25 mm=c 30 0 mm=d 200 mm=e160 mm=μs0 .3= e 2 .71 8=Solution:Initial guesses:T11N= T21N= M 1kg=GivenDrum:T2T1eμs 3 π2⎛⎜⎝⎞⎟⎠=T2−...

... x,()=PNBFBTNDFD⎛⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎠ 13. 33 33. 33 6. 67 6. 67 30 .006. 67 ⎛⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎠lb= x 0. 67 ft=Now checke the assumptionsFDmaxμDND=Since FD6. 67 lb= < FDmax9.00 ... publisher. Engineering Mechanics - Statics Chapter 8Since x 0. 67 ft= < b20 .75 ft= then the block does not tip.So our original assumption is correct.P 13. 33 lb=Problem 8-4 6Determine ... mm=μC0.2= c 30 0 mm=w 800Nm= d 3= g 9.81ms2= e 4=Solution: Member AB:ΣMA = 0; 12wa⎛⎜⎝⎞⎟⎠−2a 3 ⎛⎜⎝⎞⎟⎠NBa+ 0= NB1 3 wa= NB 533 .33 N=806© 20 07 R. C. Hibbeler....