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Category: Mechanics

Question-188062

Question Number 188062 by mr W last updated on 25/Feb/23 Commented by mr W last updated on 25/Feb/23 $${A}\:{and}\:{B}\:{are}\:{connected}\:{with}\:{a}\:{rope} \\ $$$${of}\:{length}\:{L}.\:{A}\:{mouse}\:{of}\:{mass}\:{m} \\ $$$${moves}\:{from}\:{A}\:{to}\:{B}\:{along}\:{the}\:{rope}. \\ $$$${Find}\:{the}\:{locus}\:{of}\:{the}\:{mouse},…

Question-56874

Question Number 56874 by Runzzy last updated on 25/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 26/Mar/19 $${F}_{{R}} =\sqrt{\left(\mathrm{0}.\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{0}.\mathrm{3}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{0}.\mathrm{2}×\mathrm{0}.\mathrm{3}{cos}\mathrm{95}^{{o}} }\: \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{0}.\mathrm{3457}\approx\mathrm{0}.\mathrm{346} \\ $$$${F}_{{x}}…

Question-187789

Question Number 187789 by ajfour last updated on 21/Feb/23 Answered by mahdipoor last updated on 22/Feb/23 $${F}_{{x}} =\mathrm{0}\Rightarrow{Tcos}\left(\theta\right)=\left({T}+{dT}\right){cos}\left(\theta+{d}\theta\right) \\ $$$${F}_{{y}} =\mathrm{0}\Rightarrow{T}\left({sin}\theta\right)+{dW}=\left({T}+{dT}\right){sin}\left(\theta+{d}\theta\right) \\ $$$$\Rightarrow{Lem}\:{I}\Rightarrow \\ $$$$\begin{cases}{\mathrm{0}=−{T}.{sin}\left(\theta\right).{d}\theta+{dT}.{cos}\left(\theta\right)}\\{{g}.{dm}={T}.{cos}\left(\theta\right).{d}\theta+{dT}.{sin}\left(\theta\right)}\end{cases}\Rightarrow…

Question-187675

Question Number 187675 by mr W last updated on 20/Feb/23 Commented by mr W last updated on 20/Feb/23 $${an}\:{uniform}\:{rope}\:{of}\:{length}\:{L}\:{and}\: \\ $$$${mass}\:{M}\:{is}\:{fixed}\:{on}\:{both}\:{ends}\:{at}\:{a} \\ $$$${distance}\:{d}\:{as}\:{shown}.\:{a}\:{small}\:{object} \\ $$$${of}\:{mass}\:{m}\:{moves}\:{very}\:{slowly}\:{along}…

Question-56584

Question Number 56584 by Tinkutara last updated on 18/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 19/Mar/19 $$\Gamma=\frac{{dL}}{{dt}}={A}×{L}\:\left({given}\right) \\ $$$$\frac{{dL}}{{dt}}.\left({A}×{L}\right)=\mathrm{0} \\ $$$${hence}\:\frac{{dL}}{{dt}}\bot\left({A}×{L}\right) \\ $$$$ \\ $$…