Question-212752

Question Number 212752 by mr W last updated on 25/Oct/24

Commented by mr W last updated on 25/Oct/24

the uniform rod with length l has mass m. find the tensions in the strings a and b.

$${the}\:{uniform}\:{rod}\:{with}\:{length}\:\boldsymbol{{l}}\:{has}\: \\ $$$${mass}\:\boldsymbol{{m}}.\:{find}\:{the}\:{tensions}\:{in}\:{the} \\ $$$${strings}\:{a}\:{and}\:{b}. \\ $$

Answered by ajfour last updated on 23/Oct/24

Commented by ajfour last updated on 23/Oct/24

On rod ΣF_x =0 say T_a =T & T_b =F ⇒ Tcos α=Fcos β ...(i) Tsin α+Fsin β=w ...(ii) Tsin (α−θ)=Fsin (β+θ) ...(iii) acos α+c+bcos β = Lcos θ ...(iv) asin α=Lsin θ+bsin β ....(v) we have the necessary eqs. now.

$${On}\:{rod}\:\Sigma{F}_{{x}} =\mathrm{0}\:\: \\ $$$$\:{say}\:{T}_{{a}} ={T}\:\:\:\:\:\:\&\:\:\:\:\:{T}_{{b}} ={F} \\ $$$$\Rightarrow\:{T}\mathrm{cos}\:\alpha={F}\mathrm{cos}\:\beta\:\:\:…\left({i}\right) \\ $$$${T}\mathrm{sin}\:\alpha+{F}\mathrm{sin}\:\beta=\mathrm{w}\:\:\:\:\:…\left({ii}\right) \\ $$$${T}\mathrm{sin}\:\left(\alpha−\theta\right)={F}\mathrm{sin}\:\left(\beta+\theta\right)\:\:\:…\left({iii}\right) \\ $$$${a}\mathrm{cos}\:\alpha+{c}+{b}\mathrm{cos}\:\beta\:=\:{L}\mathrm{cos}\:\theta\:\:\:…\left({iv}\right) \\ $$$${a}\mathrm{sin}\:\alpha={L}\mathrm{sin}\:\theta+{b}\mathrm{sin}\:\beta\:\:\:\:….\left({v}\right) \\ $$$${we}\:{have}\:{the}\:{necessary}\:{eqs}.\:{now}. \\ $$

Commented by ajfour last updated on 26/Oct/24

(T/F)=((cos β)/(cos α))=((sin (β+θ))/(sin (α−θ))) ⇒ cos β[sin αcos θ−cos αsin θ] =cos α[sin βcos θ+cos βsin θ] ⇒ cos θsin (α−β)=2sin θcos αcos β ⇒ tan α−tan β=2tan θ

$$\frac{{T}}{{F}}=\frac{\mathrm{cos}\:\beta}{\mathrm{cos}\:\alpha}=\frac{\mathrm{sin}\:\left(\beta+\theta\right)}{\mathrm{sin}\:\left(\alpha−\theta\right)} \\ $$$$\Rightarrow\:\mathrm{cos}\:\beta\left[\mathrm{sin}\:\alpha\mathrm{cos}\:\theta−\mathrm{cos}\:\alpha\mathrm{sin}\:\theta\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{cos}\:\alpha\left[\mathrm{sin}\:\beta\mathrm{cos}\:\theta+\mathrm{cos}\:\beta\mathrm{sin}\:\theta\right] \\ $$$$\Rightarrow\:\mathrm{cos}\:\theta\mathrm{sin}\:\left(\alpha−\beta\right)=\mathrm{2sin}\:\theta\mathrm{cos}\:\alpha\mathrm{cos}\:\beta \\ $$$$\Rightarrow\:\mathrm{tan}\:\alpha−\mathrm{tan}\:\beta=\mathrm{2tan}\:\theta \\ $$

Commented by ajfour last updated on 25/Oct/24

$${yes},\:{thanks}\:{sir}.\:{now}\:{only}\:{i}\:{can}\:{see}! \\ $$

Commented by mr W last updated on 25/Oct/24

⇏ cos θsin (α−β)=sin θcos (α−β) ⇒ cos θsin (α−β)=2 sin θ cos α cos β it should follow that tan α−tan β=2 tan θ

$$\nRightarrow\:\mathrm{cos}\:\theta\mathrm{sin}\:\left(\alpha−\beta\right)=\mathrm{sin}\:\theta\mathrm{cos}\:\left(\alpha−\beta\right) \\ $$$$\Rightarrow\:\mathrm{cos}\:\theta\mathrm{sin}\:\left(\alpha−\beta\right)=\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta \\ $$$$ \\ $$$${it}\:{should}\:{follow}\:{that} \\ $$$$\mathrm{tan}\:\alpha−\mathrm{tan}\:\beta=\mathrm{2}\:\mathrm{tan}\:\theta \\ $$

Commented by mr W last updated on 26/Oct/24

$${see}\:{a}\:{geometrical}\:{approach}\:{in} \\ $$$${Q}\mathrm{212867} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply