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Question Number 102880 by Dwaipayan Shikari last updated on 11/Jul/20

Commented by Dwaipayan Shikari last updated on 11/Jul/20

$$Sorrysir.Ididnotincludeit.Theweightrecorderismass$$$$less.Themanstandsonit.ifRecorderrecords72kgasthe$$$$weightofthemanthe.Heretheweightrecorderisatrianguler$$$$platform.Anditslides.Frictionisalongtheslope$$$$$$

Commented by mr W last updated on 11/Jul/20

$$doyouunderstandbasedonthetext$$$$howthisrecorderworksandwhat$$$$thesystemdoes?doestheman$$$$standontherecorderwhichslideo$$$$alongthewedge?canthewedgemove$$$$ontheground?ittalksabout$$$$friction,whichfrictionismeant?$$$$betweentherecorderandthewedge$$$$orelsewhere?$$

Commented by Dwaipayan Shikari last updated on 11/Jul/20

Commented by prakash jain last updated on 11/Jul/20

$$\mathrm{Can}\mathrm{you}\mathrm{use}\mathrm{smaller}\mathrm{lines}.\mathrm{You}\mathrm{may}$$$$\mathrm{be}\mathrm{using}\mathrm{a}\mathrm{tablet}\mathrm{but}\mathrm{most}\mathrm{of}\mathrm{us}\mathrm{use}$$$$\mathrm{phones}.$$

Commented by Dwaipayan Shikari last updated on 11/Jul/20

$$Iwillmaintainitfromthenexttime$$

Answered by mr W last updated on 11/Jul/20

Commented by mr W last updated on 11/Jul/20

$$N=mg\cos \theta$$$$f=\mu N=\mu mg\cos \theta$$$$ma=mg\sin \theta -f=mg\sin \theta -\mu mg\cos \theta$$$$a=g(\sin \theta -\mu \cos \theta )$$$$\Rightarrow \mu =(\sin \theta -\frac{a}{g})\frac{1}{\cos \theta }$$$$mg-N_1=ma\sin \theta$$$$\Rightarrow a=\frac{mg-N_1}{m\sin \theta }=\frac{80\times 10-72\times 10}{80\times 0.6}=\frac{5}{3}m/s^2$$$$\Rightarrow \mu =(\frac{3}{5}-\frac{5}{3\times 10})\frac{1}{\frac{4}{5}}=\frac{13}{24}=0.54$$

Commented by mr W last updated on 11/Jul/20

Commented by Dwaipayan Shikari last updated on 11/Jul/20

$$Thankingyou.Great!$$

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