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Category: Mechanics

Question-40012

Question Number 40012 by rahul 19 last updated on 15/Jul/18 Answered by ajfour last updated on 15/Jul/18 $$\left({mg}\right)_{{max}} =\frac{\mathrm{3}{Mg}}{\mathrm{5}}+\frac{\mathrm{0}.\mathrm{6}\left(\mathrm{4}{Mg}\right)}{\mathrm{5}} \\ $$$$\:\:\:\:\Rightarrow\:{m}_{{max}} =\mathrm{30}{kg}+\mathrm{24}{kg}\:=\:\mathrm{54}{kg} \\ $$$$\left({mg}\right)_{{min}} =\frac{\mathrm{3}{Mg}}{\mathrm{5}}−\frac{\mathrm{0}.\mathrm{6}\left(\mathrm{4}{Mg}\right)}{\mathrm{5}}…

Question-39970

Question Number 39970 by rahul 19 last updated on 14/Jul/18 Answered by MrW3 last updated on 14/Jul/18 $${let}\:{p}_{\mathrm{1}} ,{p}_{\mathrm{2}} ,{p}_{\mathrm{3}} \:{be}\:{the}\:{acceleration}\:{of}\:{string} \\ $$$${about}\:{the}\:{pulley}\:\mathrm{1},\mathrm{2},\mathrm{3}\:{respectively}\:\left(\curvearrowright\:{positive}\right). \\ $$$$…

Question-105476

Question Number 105476 by ajfour last updated on 29/Jul/20 Commented by ajfour last updated on 29/Jul/20 $${Find}\:{the}\:{mass}\:{of}\:{the}\:{square}\:{plate} \\ $$$${OABC},\:{if}\:{mass}\:{density}\:{per}\:{unit} \\ $$$${area}\:{is}\:{proportional}\:{to}\:{the} \\ $$$${distance}\:{from}\:{corner}\:{O},\:\:{given} \\ $$$${by}\:\:\boldsymbol{\rho}=\boldsymbol{{kr}}/\boldsymbol{{a}}.…

Question-39929

Question Number 39929 by rahul 19 last updated on 13/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18 $$ \\ $$$$\mathrm{75}×\mathrm{10}=\mathrm{80}×\mathrm{10}−\mathrm{80}×{a}×{cos}\mathrm{60}^{{o}} \\ $$$$\mathrm{80}×{a}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{50} \\ $$$${a}=\frac{\mathrm{5}}{\mathrm{4}}\:{pls}\:{check}… \\…

Question-39776

Question Number 39776 by ajfour last updated on 10/Jul/18 Commented by ajfour last updated on 11/Jul/18 $${Find}\:{the}\:{acceleration}\:{of}\:{wedge}, \\ $$$${if}\:{ground}\:{is}\:{smooth}\:{while}\:{the} \\ $$$${discs}\:{purely}\:{rolls}\:{on}\:{the}\:{wedge}. \\ $$$${mass}\:{of}\:{wedge}\:{is}\:{M}\:'\:={M}+{m}\:. \\ $$$${The}\:{connecting}\:{rod}\:{is}\:{light}\:{and}…

Question-39755

Question Number 39755 by rahul 19 last updated on 10/Jul/18 Answered by ajfour last updated on 11/Jul/18 $${a}={g}\mathrm{cot}\:\theta \\ $$$${as}\:\:\:{a}=\frac{{M}\:'{g}}{{M}+{M}\:'}\:=\:{g}\mathrm{cot}\:\theta \\ $$$$\Rightarrow\:\:\:\mathrm{1}+\frac{{M}}{{M}\:'}\:=\mathrm{tan}\:\theta \\ $$$$\Rightarrow\:\:\:\:\:{M}\:'\:=\:\frac{{M}}{\mathrm{tan}\:\theta−\mathrm{1}}\:=\:\frac{{M}\mathrm{cot}\:\theta}{\mathrm{1}−\mathrm{cot}\:\theta} \\…

Question-39655

Question Number 39655 by rahul 19 last updated on 09/Jul/18 Commented by rahul 19 last updated on 09/Jul/18 $$\mathrm{Given}\:\mathrm{Mass}\:\mathrm{of}\:\mathrm{Block}\:\mathrm{A}\:=\:\mathrm{m}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Mass}\:\mathrm{of}\:\mathrm{Block}\:\mathrm{B}\:=\:\mathrm{3m}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{acc}^{\mathrm{n}} \:\mathrm{of}\:\mathrm{blocks}\:\mathrm{at}\:\mathrm{the}\:\mathrm{instant} \\…

Question-39649

Question Number 39649 by rahul 19 last updated on 09/Jul/18 Commented by rahul 19 last updated on 09/Jul/18 $$\mathrm{Given}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{block}\:\mathrm{A}\:\&\:\mathrm{C}\:=\:\mathrm{m} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{mass}\:\mathrm{of}\:\mathrm{block}\:\mathrm{B}\:=\:\mathrm{2m}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Spring}\:\mathrm{constant}\:=\:\mathrm{k}. \\ $$$$\mathrm{Find}\:\mathrm{acc}^{\mathrm{n}}…