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Question Number 81843 by TawaTawa last updated on 15/Feb/20

Commented by mr W last updated on 16/Feb/20

Commented by mr W last updated on 16/Feb/20

(T_1 /(sin ((π/2)−θ_2 )))=(T_2 /(sin ((π/2)−θ_1 )))=(F_g /(sin (θ_1 +θ_2 ))) ⇒T_1 =((F_g cos θ_2 )/(sin (θ_1 +θ_2 ))) ⇒T_2 =((F_g cos θ_1 )/(sin (θ_1 +θ_2 )))

$$\frac{{T}_{\mathrm{1}} }{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta_{\mathrm{2}} \right)}=\frac{{T}_{\mathrm{2}} }{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta_{\mathrm{1}} \right)}=\frac{{F}_{{g}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\frac{{F}_{{g}} \mathrm{cos}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{{F}_{{g}} \mathrm{cos}\:\theta_{\mathrm{1}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$

Commented by TawaTawa last updated on 15/Feb/20

Please help me.

$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}. \\ $$

Commented by TawaTawa last updated on 15/Feb/20

Please help me.

$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}. \\ $$

Commented by TawaTawa last updated on 16/Feb/20

God bless you sir. I appreciate your time

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time} \\ $$

Answered by ajfour last updated on 15/Feb/20

Commented by ajfour last updated on 15/Feb/20

T_1 cos θ_1 =T_2 cos θ_2 T_1 sin θ_1 +T_2 sin θ_2 =F_g ⇒ T_1 sin θ_1 +((T_1 cos θ_1 sin θ_2 )/(cos θ_2 ))=F_g T_1 =((F_g cos θ_2 )/(sin (θ_1 +θ_2 ))) .

$${T}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} ={T}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{2}} \\ $$$${T}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} +{T}_{\mathrm{2}} \mathrm{sin}\:\theta_{\mathrm{2}} ={F}_{{g}} \\ $$$$\Rightarrow\:{T}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} +\frac{{T}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{2}} }{\mathrm{cos}\:\theta_{\mathrm{2}} }={F}_{{g}} \\ $$$${T}_{\mathrm{1}} =\frac{{F}_{{g}} \mathrm{cos}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)}\:. \\ $$

Commented by TawaTawa last updated on 16/Feb/20

God bless you sir. I appreciate your time

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time} \\ $$

Answered by mind is power last updated on 15/Feb/20

f_g ^→ +T_3 ^→ =0^→ ⇒F_g =T_3 ..1 by projection (yy′) T_3 ^→ +T_2 ^→ sin(θ_2 )+T_1 ^→ sin(θ_1 )=0..2 projection (xx′) ⇒T_1 cos(θ_1 )=T_2 cos(θ_2 )⇒T_2 =T_1 ((cos(θ_1 ))/(cos(θ_2 )))...3 (1&2&3⇒ T_1 sin(θ_1 )+T_1 .((sin(θ_2 )cos(θ_1 ))/(cos(θ_2 )))−F_g =0 ⇒T_1 (((sin(θ_1 )cos(θ_2 )+sin(θ_2 )cos(θ_1 ))/(cos(θ_2 ))))=F_g sin(θ_1 )cos(θ_2 )+cos(θ_1 )sin(θ_2 )=sin(θ_1 +θ_2 )⇒ ⇒T_1 =((F_g cos(θ_2 ))/(sin(θ_1 +θ_2 )))

$$\overset{\rightarrow} {{f}}_{{g}} +\overset{\rightarrow} {{T}}_{\mathrm{3}} =\overset{\rightarrow} {\mathrm{0}} \\ $$$$\Rightarrow{F}_{{g}} ={T}_{\mathrm{3}} ..\mathrm{1} \\ $$$${by}\:{projection}\:\:\left({yy}'\right) \\ $$$$\overset{\rightarrow} {{T}}_{\mathrm{3}} +\overset{\rightarrow} {{T}}_{\mathrm{2}} {sin}\left(\theta_{\mathrm{2}} \right)+\overset{\rightarrow} {{T}}_{\mathrm{1}} {sin}\left(\theta_{\mathrm{1}} \right)=\mathrm{0}..\mathrm{2} \\ $$$${projection}\:\left({xx}'\right) \\ $$$$\Rightarrow{T}_{\mathrm{1}} {cos}\left(\theta_{\mathrm{1}} \right)={T}_{\mathrm{2}} {cos}\left(\theta_{\mathrm{2}} \right)\Rightarrow{T}_{\mathrm{2}} ={T}_{\mathrm{1}} \frac{{cos}\left(\theta_{\mathrm{1}} \right)}{{cos}\left(\theta_{\mathrm{2}} \right)}...\mathrm{3} \\ $$$$\left(\mathrm{1\&2\&3}\Rightarrow\right. \\ $$$${T}_{\mathrm{1}} {sin}\left(\theta_{\mathrm{1}} \right)+{T}_{\mathrm{1}} .\frac{{sin}\left(\theta_{\mathrm{2}} \right){cos}\left(\theta_{\mathrm{1}} \right)}{{cos}\left(\theta_{\mathrm{2}} \right)}−{F}_{{g}} =\mathrm{0} \\ $$$$\Rightarrow{T}_{\mathrm{1}} \left(\frac{{sin}\left(\theta_{\mathrm{1}} \right){cos}\left(\theta_{\mathrm{2}} \right)+{sin}\left(\theta_{\mathrm{2}} \right){cos}\left(\theta_{\mathrm{1}} \right)}{{cos}\left(\theta_{\mathrm{2}} \right)}\right)={F}_{{g}} \\ $$$${sin}\left(\theta_{\mathrm{1}} \right){cos}\left(\theta_{\mathrm{2}} \right)+{cos}\left(\theta_{\mathrm{1}} \right){sin}\left(\theta_{\mathrm{2}} \right)={sin}\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)\Rightarrow \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\frac{{F}_{{g}} {cos}\left(\theta_{\mathrm{2}} \right)}{{sin}\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by TawaTawa last updated on 16/Feb/20

God bless you sir. I appreciate your time.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}. \\ $$

Commented by mind is power last updated on 16/Feb/20

withe Pleasur miss

$${withe}\:{Pleasur}\:{miss} \\ $$

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