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Question Number 223591 by Tawa11 last updated on 31/Jul/25
Commented by ajfour last updated on 31/Jul/25
Very nice Question!
$${Very}\:{nice}\:{Question}! \\ $$
Commented by Tawa11 last updated on 31/Jul/25
Really? Please help sir.
$$\mathrm{Really}? \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 01/Aug/25
have you tried by yourself and what did you get?
$${have}\:{you}\:{tried}\:{by}\:{yourself}\:{and}\:{what} \\ $$$${did}\:{you}\:{get}? \\ $$
Commented by mahdipoor last updated on 01/Aug/25
i get 2467 N
$${i}\:{get}\:\mathrm{2467}\:{N} \\ $$
Commented by Tawa11 last updated on 01/Aug/25
No sir. Textbook answer says 1730.5N
$$\mathrm{No}\:\mathrm{sir}.\: \\ $$$$\mathrm{Textbook}\:\mathrm{answer}\:\mathrm{says}\:\:\:\mathrm{1730}.\mathrm{5N} \\ $$
Commented by ajfour last updated on 01/Aug/25
Nsin α−μNcos α=f(=μR) Ncos α+μNsin α+mg=R (normal from ground) dividing ((tan α−μ)/(1+μtan α+((mg)/(Ncos α))))=μ ...(i) Mg(L/2)cos α=N(H/sin α) N=(((100 ^5 g)15 ^3 ×3×4)/(4×5×4×5))=45g mg=Ncos α{((tan α)/μ)−1−1−μtan α} w=45g×(3/5){((4×5)/3)−2−(4/(5×5))} w=27g(((338)/(75)))=1192.5 newtons g=9.8m/s^2
$$ \\ $$$${N}\mathrm{sin}\:\alpha−\mu{N}\mathrm{cos}\:\alpha={f}\left(=\mu{R}\right) \\ $$$${N}\mathrm{cos}\:\alpha+\mu{N}\mathrm{sin}\:\alpha+{mg}={R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({normal}\:{from}\:{ground}\right) \\ $$$${dividing} \\ $$$$\frac{\mathrm{tan}\:\alpha−\mu}{\mathrm{1}+\mu\mathrm{tan}\:\alpha+\frac{{mg}}{{N}\mathrm{cos}\:\alpha}}=\mu\:\:\:\:\:…\left({i}\right) \\ $$$${Mg}\left({L}/\mathrm{2}\right)\mathrm{cos}\:\alpha={N}\left({H}/\mathrm{sin}\:\alpha\right) \\ $$$${N}=\frac{\left(\cancel{\mathrm{100}}\:\overset{\mathrm{5}} {\:}{g}\right)\cancel{\mathrm{15}}\overset{\mathrm{3}} {\:}×\mathrm{3}×\cancel{\mathrm{4}}}{\cancel{\mathrm{4}×\mathrm{5}}×\cancel{\mathrm{4}}×\cancel{\mathrm{5}}}=\mathrm{45}{g} \\ $$$${mg}={N}\mathrm{cos}\:\alpha\left\{\frac{\mathrm{tan}\:\alpha}{\mu}−\mathrm{1}−\mathrm{1}−\mu\mathrm{tan}\:\alpha\right\} \\ $$$$\:\:\:\:{w}=\mathrm{45}{g}×\frac{\mathrm{3}}{\mathrm{5}}\left\{\frac{\mathrm{4}×\mathrm{5}}{\mathrm{3}}−\mathrm{2}−\frac{\mathrm{4}}{\mathrm{5}×\mathrm{5}}\right\} \\ $$$$\:\:{w}=\mathrm{27}{g}\left(\frac{\mathrm{338}}{\mathrm{75}}\right)=\mathrm{1192}.\mathrm{5}\:{newtons} \\ $$$$\:\:{g}=\mathrm{9}.\mathrm{8}{m}/{s}^{\mathrm{2}} \\ $$
Commented by ajfour last updated on 01/Aug/25
Commented by mr W last updated on 01/Aug/25
Very nice Question! i also think.
$${Very}\:{nice}\:{Question}! \\ $$$${i}\:{also}\:{think}. \\ $$
Commented by mr W last updated on 01/Aug/25
ajfour sir: what′s your solution/answer?
$${ajfour}\:{sir}:\: \\ $$$${what}'{s}\:{your}\:{solution}/{answer}? \\ $$
Commented by mr W last updated on 02/Aug/25
i guess your solution has only considered that the block doesn′t slide (no slipping). this is ok, only if the block has large enough size such as like following:
$${i}\:{guess}\:{your}\:{solution}\:{has}\:{only}\: \\ $$$${considered}\:{that}\:{the}\:{block}\:{doesn}'{t} \\ $$$${slide}\:\left({no}\:{slipping}\right).\:{this}\:{is}\:{ok}, \\ $$$${only}\:{if}\:{the}\:{block}\:{has}\:{large}\:{enough} \\ $$$${size}\:{such}\:{as}\:{like}\:{following}: \\ $$
Commented by mr W last updated on 02/Aug/25
Commented by mr W last updated on 02/Aug/25
in current case the size of the block is limited. we must also ensure that the block doesn′t rotate (no tipping). considering this the block may need a larger weight.
$${in}\:{current}\:{case}\:{the}\:{size}\:{of}\:{the}\:{block} \\ $$$${is}\:{limited}.\:{we}\:{must}\:{also}\:{ensure}\:{that} \\ $$$${the}\:{block}\:{doesn}'{t}\:{rotate}\:\left({no}\:{tipping}\right). \\ $$$${considering}\:{this}\:{the}\:{block}\:{may}\:{need} \\ $$$${a}\:{larger}\:{weight}. \\ $$
Answered by mr W last updated on 02/Aug/25
Commented by mr W last updated on 02/Aug/25
the rod can only rotate about point A. the block can both slide and rotate in the tendency as shown. we must ensure that the block has no slipping and no tipping.
$${the}\:{rod}\:{can}\:{only}\:{rotate}\:{about}\:{point}\:{A}. \\ $$$${the}\:{block}\:{can}\:{both}\:{slide}\:{and}\:{rotate} \\ $$$${in}\:{the}\:{tendency}\:{as}\:{shown}.\:{we}\:{must} \\ $$$${ensure}\:{that}\:{the}\:{block}\:{has}\:{no}\:{slipping} \\ $$$$\underline{{and}}\:{no}\:{tipping}. \\ $$
Commented by mr W last updated on 04/Aug/25
Commented by mr W last updated on 02/Aug/25
tan φ=μ=0.2 tan θ=(4/3) ((0.5)/(tan α))+((0.5)/(tan (θ−β)))=4 ⇒tan α=(1/(8−(1/(tan (θ−β))))) for β=φ: tan α=(1/(8−((19)/(17))))=((17)/(117)) <0.2 ✓ ((1.25×sin ((π/2)+β))/(sin (θ−β)))=((3.75×sin ((π/2)−ϕ−θ))/(sin ϕ))=SG ((cos β)/(sin (θ−β)))=((3 cos (θ+ϕ))/(sin ϕ)) ⇒tan ϕ=((cos θ)/(((cos β)/(3 sin (θ−β)))+sin θ)) =((cos θ)/((1/(3(sin θ−cos θ tan β)))+sin θ)) for β=φ: tan ϕ=((0.6)/((1/(3(0.8−0.6×0.2)))+0.8))=((153)/(329)) ((mg sin ϕ)/(sin (θ−β+ϕ)))=((Mg sin α)/(sin (θ−β−α)))=R_3 ⇒(M/m)=((sin ϕ sin (θ−β−α))/(sin α sin (θ−β+ϕ))) =((sin ϕ[ sin (θ−β) cos α−cos (θ−β) sin α])/(sin α[sin (θ−β) cos ϕ+cos (θ−β) sin ϕ])) =((((tan (θ−β))/(tan α))−1)/(((tan (θ−β))/(tan ϕ))+1)) for β=φ: tan (θ−β)=((tan θ−tan β)/(1+tan θ tan β))=(((4/3)−0.2)/(1+(4/3)×0.2))=((17)/(19)) since β≤φ, (M/m) is minimum at β=φ: ((M/m))_(min) =((((17)/(19))×((117)/(17))−1)/(((17)/(19))×((329)/(153))+1))=((441)/(250))=1.764 i.e. (Mg)_(min) =1.764 mg =1.764×100×9.81≈1730.5 N ✓
$$\mathrm{tan}\:\phi=\mu=\mathrm{0}.\mathrm{2} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\frac{\mathrm{0}.\mathrm{5}}{\mathrm{tan}\:\alpha}+\frac{\mathrm{0}.\mathrm{5}}{\mathrm{tan}\:\left(\theta−\beta\right)}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{8}−\frac{\mathrm{1}}{\mathrm{tan}\:\left(\theta−\beta\right)}} \\ $$$${for}\:\beta=\phi:\: \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{8}−\frac{\mathrm{19}}{\mathrm{17}}}=\frac{\mathrm{17}}{\mathrm{117}}\:<\mathrm{0}.\mathrm{2}\:\checkmark \\ $$$$\frac{\mathrm{1}.\mathrm{25}×\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}+\beta\right)}{\mathrm{sin}\:\left(\theta−\beta\right)}=\frac{\mathrm{3}.\mathrm{75}×\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\varphi−\theta\right)}{\mathrm{sin}\:\varphi}={SG} \\ $$$$\frac{\mathrm{cos}\:\beta}{\mathrm{sin}\:\left(\theta−\beta\right)}=\frac{\mathrm{3}\:\mathrm{cos}\:\left(\theta+\varphi\right)}{\mathrm{sin}\:\varphi} \\ $$$$\Rightarrow\mathrm{tan}\:\varphi=\frac{\mathrm{cos}\:\theta}{\frac{\mathrm{cos}\:\beta}{\mathrm{3}\:\mathrm{sin}\:\left(\theta−\beta\right)}+\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{cos}\:\theta}{\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\:\mathrm{tan}\:\beta\right)}+\mathrm{sin}\:\theta} \\ $$$${for}\:\beta=\phi: \\ $$$$\mathrm{tan}\:\varphi=\frac{\mathrm{0}.\mathrm{6}}{\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{0}.\mathrm{8}−\mathrm{0}.\mathrm{6}×\mathrm{0}.\mathrm{2}\right)}+\mathrm{0}.\mathrm{8}}=\frac{\mathrm{153}}{\mathrm{329}} \\ $$$$ \\ $$$$\frac{{mg}\:\mathrm{sin}\:\varphi}{\mathrm{sin}\:\left(\theta−\beta+\varphi\right)}=\frac{{Mg}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\theta−\beta−\alpha\right)}={R}_{\mathrm{3}} \\ $$$$\Rightarrow\frac{{M}}{{m}}=\frac{\mathrm{sin}\:\varphi\:\mathrm{sin}\:\left(\theta−\beta−\alpha\right)}{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\left(\theta−\beta+\varphi\right)} \\ $$$$\:\:\:\:\:=\frac{\mathrm{sin}\:\varphi\left[\:\mathrm{sin}\:\left(\theta−\beta\right)\:\mathrm{cos}\:\alpha−\mathrm{cos}\:\left(\theta−\beta\right)\:\mathrm{sin}\:\alpha\right]}{\mathrm{sin}\:\alpha\left[\mathrm{sin}\:\left(\theta−\beta\right)\:\mathrm{cos}\:\varphi+\mathrm{cos}\:\left(\theta−\beta\right)\:\mathrm{sin}\:\varphi\right]} \\ $$$$\:\:\:\:\:=\frac{\frac{\mathrm{tan}\:\left(\theta−\beta\right)}{\mathrm{tan}\:\alpha}−\mathrm{1}}{\frac{\mathrm{tan}\:\left(\theta−\beta\right)}{\mathrm{tan}\:\varphi}+\mathrm{1}} \\ $$$${for}\:\beta=\phi: \\ $$$$\mathrm{tan}\:\left(\theta−\beta\right)=\frac{\mathrm{tan}\:\theta−\mathrm{tan}\:\beta}{\mathrm{1}+\mathrm{tan}\:\theta\:\mathrm{tan}\:\beta}=\frac{\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{0}.\mathrm{2}}{\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}×\mathrm{0}.\mathrm{2}}=\frac{\mathrm{17}}{\mathrm{19}} \\ $$$${since}\:\beta\leqslant\phi,\:\frac{{M}}{{m}}\:{is}\:{minimum}\:{at}\:\beta=\phi: \\ $$$$\left(\frac{{M}}{{m}}\right)_{{min}} =\frac{\frac{\mathrm{17}}{\mathrm{19}}×\frac{\mathrm{117}}{\mathrm{17}}−\mathrm{1}}{\frac{\mathrm{17}}{\mathrm{19}}×\frac{\mathrm{329}}{\mathrm{153}}+\mathrm{1}}=\frac{\mathrm{441}}{\mathrm{250}}=\mathrm{1}.\mathrm{764} \\ $$$${i}.{e}. \\ $$$$\left({Mg}\right)_{{min}} =\mathrm{1}.\mathrm{764}\:{mg} \\ $$$$\:\:\:\:\:\:\:=\mathrm{1}.\mathrm{764}×\mathrm{100}×\mathrm{9}.\mathrm{81}\approx\mathrm{1730}.\mathrm{5}\:{N}\:\checkmark \\ $$
Commented by mr W last updated on 02/Aug/25
as usual i preferred again a pure geometrical path.
$${as}\:{usual}\:{i}\:{preferred}\:{again}\:{a}\:{pure}\: \\ $$$${geometrical}\:{path}. \\ $$
Commented by ajfour last updated on 02/Aug/25
wow! congratulations! I was next going to consider this csse too...
$${wow}!\:{congratulations}!\:{I}\:{was}\:{next} \\ $$$${going}\:{to}\:{consider}\:{this}\:{csse}\:{too}… \\ $$
Commented by Tawa11 last updated on 02/Aug/25
Wow, thank you sir. I really appreciate.
$$\mathrm{Wow},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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