what-are-the-conditions-necessary-for-a-body-of-mass-mkg-on-an-incline-plane-at-angle-to-the-horizontal-to-remain-at-rest-

Question Number 23022 by NECx last updated on 25/Oct/17

what are the conditions necessary for a body of mass mkg on an incline plane at angle θ to the horizontal to remain at rest?

$${what}\:{are}\:{the}\:{conditions}\:{necessary} \\ $$$${for}\:{a}\:{body}\:{of}\:{mass}\:{mkg}\:{on}\:{an}\:{incline}\:{plane} \\ $$$${at}\:{angle}\:\theta\:{to}\:{the}\:{horizontal}\:{to} \\ $$$${remain}\:{at}\:{rest}? \\ $$

Commented by NECx last updated on 25/Oct/17

please state all necessary conditions with diagrams where necessary. Thanks

$${please}\:{state}\:{all}\:{necessary} \\ $$$${conditions}\:{with}\:{diagrams}\:{where} \\ $$$${necessary}.\:\:\boldsymbol{\mathrm{T}}{hanks} \\ $$$$ \\ $$

Commented by mrW1 last updated on 25/Oct/17

Commented by mrW1 last updated on 25/Oct/17

1) mg sin θ≤μmg cos θ i.e. tan θ≤μ 2) COM of the body lies behind the out most points of the contact area, i.e. e≥0

$$\left.\mathrm{1}\right) \\ $$$$\mathrm{mg}\:\mathrm{sin}\:\theta\leqslant\mu\mathrm{mg}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{tan}\:\theta\leqslant\mu \\ $$$$\left.\mathrm{2}\right) \\ $$$$\mathrm{COM}\:\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{lies}\:\mathrm{behind}\:\mathrm{the}\:\mathrm{out} \\ $$$$\mathrm{most}\:\mathrm{points}\:\mathrm{of}\:\mathrm{the}\:\mathrm{contact}\:\mathrm{area}, \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{e}\geqslant\mathrm{0} \\ $$

Commented by NECx last updated on 25/Oct/17

$${wow}…\:{Thanks} \\ $$

Commented by ajfour last updated on 25/Oct/17

N^→ +f^(→) +(F^(→) )_(ext) =−mg^(→) and initial velocity u^(→_ ) =0 , and 𝛕_N ^→ +𝛕_f ^(→) =0 about centre of mass with 𝛚_0 =0 .

$$\overset{\rightarrow} {\boldsymbol{{N}}}+\overset{\rightarrow} {\boldsymbol{{f}}}+\left(\overset{\rightarrow} {\boldsymbol{{F}}}\right)_{{ext}} =−\boldsymbol{{m}}\overset{\rightarrow} {\boldsymbol{{g}}}\:\: \\ $$$${and}\:{initial}\:{velocity}\:\overset{\rightarrow_{} } {\boldsymbol{{u}}}=\mathrm{0}\:,\:{and} \\ $$$$\overset{\rightarrow} {\boldsymbol{\tau}}_{{N}} +\overset{\rightarrow} {\boldsymbol{\tau}_{{f}} }\:=\mathrm{0}\:\:{about}\:{centre}\:{of}\:{mass} \\ $$$${with}\:\boldsymbol{\omega}_{\mathrm{0}} =\mathrm{0}\:. \\ $$

Answered by Physics lover last updated on 25/Oct/17

there can be three possibilities : 1. force for friction can balance the component of gravity. { μ ≥ tan θ } 2. there can be an external force on the body { f_(ext ) ≥ mg .Sin θ } 3. the inclined plane/wedge is accelerating and the pseudo force balances the component of gravity. { ma.Cos θ ≥ mg .Sin θ } In All the three cases the body will be at rest w.r.t the surface of the inclined .

$${there}\:{can}\:{be}\:{three}\:{possibilities}\:: \\ $$$$\mathrm{1}.\:\:{force}\:{for}\:{friction}\:{can}\:{balance} \\ $$$${the}\:{component}\:{of}\:{gravity}. \\ $$$$\left\{\:\mu\:\geqslant\:{tan}\:\theta\:\right\} \\ $$$$\mathrm{2}.\:{there}\:{can}\:{be}\:{an}\:{external}\:{force} \\ $$$${on}\:{the}\:{body} \\ $$$$\left\{\:{f}_{{ext}\:} \geqslant\:{mg}\:.{Sin}\:\theta\:\right\} \\ $$$$\mathrm{3}.\:{the}\:{inclined}\:{plane}/{wedge} \\ $$$${is}\:{accelerating}\:{and}\:{the}\:{pseudo} \\ $$$${force}\:{balances}\:{the}\:{component} \\ $$$${of}\:{gravity}.\: \\ $$$$\left\{\:{ma}.{Cos}\:\theta\:\geqslant\:{mg}\:.{Sin}\:\theta\:\right\} \\ $$$${In}\:{All}\:{the}\:{three}\:{cases}\:{the}\:{body} \\ $$$${will}\:{be}\:{at}\:{rest}\:{w}.{r}.{t}\:{the}\:{surface} \\ $$$${of}\:{the}\:{inclined}\:. \\ $$$$ \\ $$

Commented by Physics lover last updated on 25/Oct/17