# Differences

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 phy131studiof15:lectures:chapter14 [2015/10/16 09:22]mdawber [Elastic Moduli] phy131studiof15:lectures:chapter14 [2015/10/19 08:32]mdawber [Hanging sign problem] Both sides previous revision Previous revision 2015/10/19 08:33 mdawber [Hanging sign solution] 2015/10/19 08:32 mdawber [Hanging sign problem] 2015/10/16 09:27 mdawber [14.P.042] 2015/10/16 09:22 mdawber [Elastic Moduli] 2015/10/16 09:20 mdawber [Will the ladder slip?] 2015/10/16 09:19 mdawber [Crane boom with arm and rope at angle] 2015/07/22 10:47 mdawber created Next revision Previous revision 2015/10/19 08:33 mdawber [Hanging sign solution] 2015/10/19 08:32 mdawber [Hanging sign problem] 2015/10/16 09:27 mdawber [14.P.042] 2015/10/16 09:22 mdawber [Elastic Moduli] 2015/10/16 09:20 mdawber [Will the ladder slip?] 2015/10/16 09:19 mdawber [Crane boom with arm and rope at angle] 2015/07/22 10:47 mdawber created Last revision Both sides next revision Line 116: Line 116: What are the horizontal and vertical components of the force exerted by the hinge on the rod? Take the positive directions to be up and to the right. What are the horizontal and vertical components of the force exerted by the hinge on the rod? Take the positive directions to be up and to the right. + ===== Hanging sign solution ===== + + $\theta=\tan^{-1}\frac{30}{60}=26.57^{o}$ + + Vertical forces, up is positive $T\sin\theta+Fh_{y}=m_{sign}g+m_{rod}g$ + + Horizontal forces, right is positive $-T\cos\theta+Fh_{x}=0$ + + Torques $0.6T\sin\theta=0.8m_{sign}g+0.4m_{rod}g$ + + $T=\frac{0.8\times3\times9.8+0.4\times1\times9.8}{0.6\times\sin26.57^{o}}=102.24\mathrm{N}$ + + $Fh_{y}=4.981-102.24\sin26.57^{o}=-6.53\mathrm{N}$ ​ + + So force points down. + + $Fh_{x}=102.24\cos26.57^{o}=91.4\mathrm{N}$ + + Force points to the right ===== Will the ladder slip? ===== ===== Will the ladder slip? ===== Line 252: Line 271: ===== 14.P.042 ===== ===== 14.P.042 ===== + ===== 14.P.051 ===== ===== Strength of materials ===== ===== Strength of materials ===== 