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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

$${wow}...\:{Thanks} \\ $$

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

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

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

Commented by NECx last updated on 25/Oct/17

buddy your explanations are  vivid.Thanks

$${buddy}\:{your}\:{explanations}\:{are} \\ $$$${vivid}.{Thanks} \\ $$$$ \\ $$$$ \\ $$

Commented by Joel577 last updated on 25/Oct/17

very nice

$${very}\:{nice} \\ $$

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

lol thanks and ya r welcome.

$${lol}\:{thanks}\:{and}\:{ya}\:{r}\:{welcome}. \\ $$

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