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Question Number 52891 by ajfour last updated on 14/Jan/19

Commented by ajfour last updated on 15/Jan/19

Find tensions in the two strings.  let for example,  M=5kg, m= 3kg  R= 20cm , p = 40cm , r= 10cm,  q = 30cm  (to match our answers).

$${Find}\:{tensions}\:{in}\:{the}\:{two}\:{strings}. \\ $$$${let}\:{for}\:{example},\:\:{M}=\mathrm{5}{kg},\:{m}=\:\mathrm{3}{kg} \\ $$$${R}=\:\mathrm{20}{cm}\:,\:{p}\:=\:\mathrm{40}{cm}\:,\:{r}=\:\mathrm{10}{cm}, \\ $$$${q}\:=\:\mathrm{30}{cm}\:\:\left({to}\:{match}\:{our}\:{answers}\right). \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

T_a cosα−Mg−N_m sinθ=0.....eqn1  T_a sinα+N_m cosθ=N_w     eqn   2  sinα=(R/(a+R))  T_b sinβ−mg−N_m sinθ=0 .....eqn3  T_b cosβ=N_m cosθ....eqn4  cosγ=(((a+R)^2 +(b+r)^2 −(r+R)^2 )/(2(a+R)(b+r)))←known value  β=90^o −(α+γ)  sinθ=(h/(r+R))  h=(a+R)cosα−(b+r)sinβ  to solve...four eqn...pls wait...  pls check upto this....  [after checking ]  (T_b /(cosθ))=(N_m /(cosβ))=k  T_b =kcosθ      N_m =kcosβ  T_b sinβ−N_m sinθ=mg  kcosθsinβ−kcosβsinθ=mg  k=((mg)/(sin(β−θ)))  T_b =((mgcosθ)/(sin(β−θ)))    N_m =((mgcosβ)/(sin(β−θ)))      T_a =((Mg+N_m sinθ)/(cosα))     =((Mg+((mgcosβ)/(sin(β−θ)))sinθ)/(cosα))  =((Mgsin(β−θ)+mgcosβsinθ)/(cosαsin(β−θ)))  now something to say...  1)T_a =depends on  M,m,α,β,θ  2)T_b =depends on m,θ,β  sinα=(R/(a+R)) ←known value    angle γ also known  cosγ=(((a+R)^2 +(b+r)^2 −(R+r)^2 )/(2(a+R)(b+r)))    β=90^o −(α+γ) ←so β is known    sinθ=(h/(r+R))  h=(a+R)cosα−(b+r)sinβ  sinθ=(((a+R)cosα−(b+r)sinβ)/(r+R))←known value  T_a =((Mgsin(β−θ)+mgcosβsinθ)/(cos∝sin(β−θ)))  T_b =((mgcosθ)/(sin(β−θ)))  comments...since T_a  and T_b  depends on α,β,θ  but the value of α,β,θ are already calculated  by the given parameters(a,b,r,R)  so i think problem is solved...  if we put value of (α,β,θ) in expression of T_a   and T_b ...it will be more lengthy and zargon

$${T}_{{a}} {cos}\alpha−{Mg}−{N}_{{m}} {sin}\theta=\mathrm{0}.....{eqn}\mathrm{1} \\ $$$${T}_{{a}} {sin}\alpha+{N}_{{m}} {cos}\theta={N}_{{w}} \:\:\:\:{eqn}\:\:\:\mathrm{2} \\ $$$${sin}\alpha=\frac{{R}}{{a}+{R}} \\ $$$${T}_{{b}} {sin}\beta−{mg}−{N}_{{m}} {sin}\theta=\mathrm{0}\:.....{eqn}\mathrm{3} \\ $$$${T}_{{b}} {cos}\beta={N}_{{m}} {cos}\theta....{eqn}\mathrm{4} \\ $$$${cos}\gamma=\frac{\left({a}+{R}\right)^{\mathrm{2}} +\left({b}+{r}\right)^{\mathrm{2}} −\left({r}+{R}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{R}\right)\left({b}+{r}\right)}\leftarrow{known}\:{value} \\ $$$$\beta=\mathrm{90}^{{o}} −\left(\alpha+\gamma\right) \\ $$$${sin}\theta=\frac{{h}}{{r}+{R}} \\ $$$${h}=\left({a}+{R}\right){cos}\alpha−\left({b}+{r}\right){sin}\beta \\ $$$${to}\:{solve}...{four}\:{eqn}...{pls}\:{wait}... \\ $$$${pls}\:{check}\:{upto}\:{this}.... \\ $$$$\left[{after}\:{checking}\:\right] \\ $$$$\frac{{T}_{{b}} }{{cos}\theta}=\frac{{N}_{{m}} }{{cos}\beta}={k} \\ $$$${T}_{{b}} ={kcos}\theta\:\:\:\:\:\:{N}_{{m}} ={kcos}\beta \\ $$$${T}_{{b}} {sin}\beta−{N}_{{m}} {sin}\theta={mg} \\ $$$${kcos}\theta{sin}\beta−{kcos}\beta{sin}\theta={mg} \\ $$$${k}=\frac{{mg}}{{sin}\left(\beta−\theta\right)} \\ $$$${T}_{{b}} =\frac{{mgcos}\theta}{{sin}\left(\beta−\theta\right)}\:\:\:\:{N}_{{m}} =\frac{{mgcos}\beta}{{sin}\left(\beta−\theta\right)} \\ $$$$ \\ $$$$ \\ $$$${T}_{{a}} =\frac{{Mg}+{N}_{{m}} {sin}\theta}{{cos}\alpha} \\ $$$$\:\:\:=\frac{{Mg}+\frac{{mgcos}\beta}{{sin}\left(\beta−\theta\right)}{sin}\theta}{{cos}\alpha} \\ $$$$=\frac{{Mgsin}\left(\beta−\theta\right)+{mgcos}\beta{sin}\theta}{{cos}\alpha{sin}\left(\beta−\theta\right)} \\ $$$${now}\:{something}\:{to}\:{say}... \\ $$$$\left.\mathrm{1}\right){T}_{{a}} ={depends}\:{on}\:\:{M},{m},\alpha,\beta,\theta \\ $$$$\left.\mathrm{2}\right){T}_{{b}} ={depends}\:{on}\:{m},\theta,\beta \\ $$$${sin}\alpha=\frac{{R}}{{a}+{R}}\:\leftarrow{known}\:{value} \\ $$$$ \\ $$$${angle}\:\gamma\:{also}\:{known} \\ $$$${cos}\gamma=\frac{\left({a}+{R}\right)^{\mathrm{2}} +\left({b}+{r}\right)^{\mathrm{2}} −\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{R}\right)\left({b}+{r}\right)} \\ $$$$ \\ $$$$\beta=\mathrm{90}^{{o}} −\left(\alpha+\gamma\right)\:\leftarrow{so}\:\beta\:{is}\:{known} \\ $$$$ \\ $$$${sin}\theta=\frac{{h}}{{r}+{R}} \\ $$$${h}=\left({a}+{R}\right){cos}\alpha−\left({b}+{r}\right){sin}\beta \\ $$$${sin}\theta=\frac{\left({a}+{R}\right){cos}\alpha−\left({b}+{r}\right){sin}\beta}{{r}+{R}}\leftarrow{known}\:{value} \\ $$$$\boldsymbol{{T}}_{{a}} =\frac{{Mgsin}\left(\beta−\theta\right)+{mgcos}\beta{sin}\theta}{{cos}\propto{sin}\left(\beta−\theta\right)} \\ $$$${T}_{{b}} =\frac{{mgcos}\theta}{{sin}\left(\beta−\theta\right)} \\ $$$${comments}...{since}\:{T}_{{a}} \:{and}\:{T}_{{b}} \:{depends}\:{on}\:\alpha,\beta,\theta \\ $$$${but}\:{the}\:{value}\:{of}\:\alpha,\beta,\theta\:{are}\:{already}\:{calculated} \\ $$$${by}\:{the}\:{given}\:{parameters}\left({a},{b},{r},{R}\right) \\ $$$${so}\:{i}\:{think}\:{problem}\:{is}\:{solved}... \\ $$$${if}\:{we}\:{put}\:{value}\:{of}\:\left(\alpha,\beta,\theta\right)\:{in}\:{expression}\:{of}\:{T}_{{a}} \\ $$$${and}\:{T}_{{b}} ...\boldsymbol{{it}}\:\boldsymbol{{will}}\:\boldsymbol{{be}}\:\boldsymbol{{more}}\:\boldsymbol{{lengthy}}\:\boldsymbol{{and}}\:\boldsymbol{{zargon}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by ajfour last updated on 15/Jan/19

remove the intermediaries sir,  seek help from torque equations,  may be..

$${remove}\:{the}\:{intermediaries}\:{sir}, \\ $$$${seek}\:{help}\:{from}\:{torque}\:{equations}, \\ $$$${may}\:{be}.. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

thank you...

$${thank}\:{you}... \\ $$

Commented by ajfour last updated on 15/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

how you draw this diagram...

$${how}\:{you}\:{draw}\:{this}\:{diagram}... \\ $$

Commented by ajfour last updated on 15/Jan/19

sin α = (R/(a+R))  cos β = (((a+R)^2 +(b+r)^2 −(R+r)^2 )/(2(a+R)(b+r)))  tan θ = (((a+R)cos α−(b+r)cos (α+β))/((b+r)sin (α+β)−(a+R)sin α))  T_a cos α+T_b cos (α+β)=(M+m)g  taking moment about center of  bigger body:  T_b {[cos (α+β)](R+r)cos θ     +[sin (α+β)](R+r)sin θ}               = mg(R+r)cos θ  ⇒  T_b  = ((mgcos θ)/(cos (α+β−θ)))         T_a  = (((M+m)g−((mgcos θcos (α+β))/(cos (α+β−θ))))/(cos α))  I obtain:     α≈ 19.4712°  , β ≈ 26.384°,    θ≈ 73.135° , and with  g=10m/s^2 ,     T_b ≈ 9.7926N     T_a ≈ 77.6189N .

$$\mathrm{sin}\:\alpha\:=\:\frac{{R}}{{a}+{R}} \\ $$$$\mathrm{cos}\:\beta\:=\:\frac{\left({a}+{R}\right)^{\mathrm{2}} +\left({b}+{r}\right)^{\mathrm{2}} −\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{R}\right)\left({b}+{r}\right)} \\ $$$$\mathrm{tan}\:\theta\:=\:\frac{\left({a}+{R}\right)\mathrm{cos}\:\alpha−\left({b}+{r}\right)\mathrm{cos}\:\left(\alpha+\beta\right)}{\left({b}+{r}\right)\mathrm{sin}\:\left(\alpha+\beta\right)−\left({a}+{R}\right)\mathrm{sin}\:\alpha} \\ $$$${T}_{{a}} \mathrm{cos}\:\alpha+{T}_{{b}} \mathrm{cos}\:\left(\alpha+\beta\right)=\left({M}+{m}\right){g} \\ $$$${taking}\:{moment}\:{about}\:{center}\:{of} \\ $$$${bigger}\:{body}: \\ $$$${T}_{{b}} \left\{\left[\mathrm{cos}\:\left(\alpha+\beta\right)\right]\left({R}+{r}\right)\mathrm{cos}\:\theta\right. \\ $$$$\left.\:\:\:+\left[\mathrm{sin}\:\left(\alpha+\beta\right)\right]\left({R}+{r}\right)\mathrm{sin}\:\theta\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{mg}\left({R}+{r}\right)\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\:{T}_{{b}} \:=\:\frac{{mg}\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\alpha+\beta−\theta\right)} \\ $$$$\:\:\:\:\:\:\:{T}_{{a}} \:=\:\frac{\left({M}+{m}\right){g}−\frac{{mg}\mathrm{cos}\:\theta\mathrm{cos}\:\left(\alpha+\beta\right)}{\mathrm{cos}\:\left(\alpha+\beta−\theta\right)}}{\mathrm{cos}\:\alpha} \\ $$$${I}\:{obtain}: \\ $$$$\:\:\:\alpha\approx\:\mathrm{19}.\mathrm{4712}°\:\:,\:\beta\:\approx\:\mathrm{26}.\mathrm{384}°, \\ $$$$\:\:\theta\approx\:\mathrm{73}.\mathrm{135}°\:,\:{and}\:{with}\:\:{g}=\mathrm{10}{m}/{s}^{\mathrm{2}} , \\ $$$$\:\:\:\boldsymbol{{T}}_{\boldsymbol{{b}}} \approx\:\mathrm{9}.\mathrm{7926}{N} \\ $$$$\:\:\:\boldsymbol{{T}}_{\boldsymbol{{a}}} \approx\:\mathrm{77}.\mathrm{6189}{N}\:. \\ $$

Commented by ajfour last updated on 15/Jan/19

draw with an app  Lekh Diagram,  Tanmay Sir.

$${draw}\:{with}\:{an}\:{app}\:\:{Lekh}\:{Diagram}, \\ $$$${Tanmay}\:{Sir}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

ok i shall download it...

$${ok}\:{i}\:{shall}\:{download}\:{it}... \\ $$

Answered by mr W last updated on 15/Jan/19

Commented by mr W last updated on 15/Jan/19

α=sin^(−1) (R/(a+R))  γ=cos^(−1) (((a+R)^2 +(b+r)^2 −(R+r)^2 )/(2(a+R)(b+r)))  β=α+γ=sin^(−1) (R/(a+R))+cos^(−1) (((a+R)^2 +(b+r)^2 −(R+r)^2 )/(2(a+R)(b+r)))    ΣM_A =0:  T_b  (a+R) sin γ=mg [(b+r) sin β−R]  ⇒T_a =(((b+r) sin β−R)/((a+R) sin γ))×mg  ΣM_B =0:  T_a [(((a+R) cos α sin γ)/(cos β))]=Mg[(a+R) cos α tan β−R]+mg[(a+R) cos α tan β−(b+r) sin β]  ⇒T_a =(({Mg[(a+R) cos α tan β−R]+mg[(a+R) cos α tan β−(b+r) sin β]}cos β)/((a+R) cos α sin γ))

$$\alpha=\mathrm{sin}^{−\mathrm{1}} \frac{{R}}{{a}+{R}} \\ $$$$\gamma=\mathrm{cos}^{−\mathrm{1}} \frac{\left({a}+{R}\right)^{\mathrm{2}} +\left({b}+{r}\right)^{\mathrm{2}} −\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{R}\right)\left({b}+{r}\right)} \\ $$$$\beta=\alpha+\gamma=\mathrm{sin}^{−\mathrm{1}} \frac{{R}}{{a}+{R}}+\mathrm{cos}^{−\mathrm{1}} \frac{\left({a}+{R}\right)^{\mathrm{2}} +\left({b}+{r}\right)^{\mathrm{2}} −\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{R}\right)\left({b}+{r}\right)} \\ $$$$ \\ $$$$\Sigma{M}_{{A}} =\mathrm{0}: \\ $$$${T}_{{b}} \:\left({a}+{R}\right)\:\mathrm{sin}\:\gamma={mg}\:\left[\left({b}+{r}\right)\:\mathrm{sin}\:\beta−{R}\right] \\ $$$$\Rightarrow{T}_{{a}} =\frac{\left({b}+{r}\right)\:\mathrm{sin}\:\beta−{R}}{\left({a}+{R}\right)\:\mathrm{sin}\:\gamma}×{mg} \\ $$$$\Sigma{M}_{{B}} =\mathrm{0}: \\ $$$${T}_{{a}} \left[\frac{\left({a}+{R}\right)\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\gamma}{\mathrm{cos}\:\beta}\right]={Mg}\left[\left({a}+{R}\right)\:\mathrm{cos}\:\alpha\:\mathrm{tan}\:\beta−{R}\right]+{mg}\left[\left({a}+{R}\right)\:\mathrm{cos}\:\alpha\:\mathrm{tan}\:\beta−\left({b}+{r}\right)\:\mathrm{sin}\:\beta\right] \\ $$$$\Rightarrow{T}_{{a}} =\frac{\left\{{Mg}\left[\left({a}+{R}\right)\:\mathrm{cos}\:\alpha\:\mathrm{tan}\:\beta−{R}\right]+{mg}\left[\left({a}+{R}\right)\:\mathrm{cos}\:\alpha\:\mathrm{tan}\:\beta−\left({b}+{r}\right)\:\mathrm{sin}\:\beta\right]\right\}\mathrm{cos}\:\beta}{\left({a}+{R}\right)\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\gamma} \\ $$

Commented by ajfour last updated on 15/Jan/19

Great way Sir!

$${Great}\:{way}\:{Sir}! \\ $$

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