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Question Number 108619 by ajfour last updated on 18/Aug/20

Commented by ajfour last updated on 18/Aug/20

Find tension in rod connecting the blocks.

$${Find}\:{tension}\:{in}\:{rod}\:{connecting} \\ $$$${the}\:{blocks}. \\ $$

Commented by Dwaipayan Shikari last updated on 18/Aug/20

For upper block T+mgsinθ−f_1 =ma (f=μN=μmgcosθ) (θ=tan^(−1) (8/(15))) Lower block mgsinθ−T−f_2 =ma T+mgsinθ−f_1 =mgsinθ−T−f_2 2T=f_1 −f_2 T=((μ_1 mgcosθ−μ_2 mgcosθ)/2)=(1/2)mgcosθ(0.2)=(g/(10)).170.cosθ=g.17.((15)/(17))N =15gN

$${For}\:{upper}\:{block} \\ $$$${T}+{mgsin}\theta−{f}_{\mathrm{1}} ={ma}\:\:\:\:\:\:\:\left({f}=\mu{N}=\mu{mgcos}\theta\right)\:\:\left(\theta={tan}^{−\mathrm{1}} \frac{\mathrm{8}}{\mathrm{15}}\right) \\ $$$${Lower}\:{block} \\ $$$${mgsin}\theta−{T}−{f}_{\mathrm{2}} ={ma} \\ $$$${T}+{mgsin}\theta−{f}_{\mathrm{1}} ={mgsin}\theta−{T}−{f}_{\mathrm{2}} \\ $$$$\mathrm{2}{T}={f}_{\mathrm{1}} −{f}_{\mathrm{2}} \\ $$$${T}=\frac{\mu_{\mathrm{1}} {mgcos}\theta−\mu_{\mathrm{2}} {mgcos}\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{mgcos}\theta\left(\mathrm{0}.\mathrm{2}\right)=\frac{{g}}{\mathrm{10}}.\mathrm{170}.{cos}\theta={g}.\mathrm{17}.\frac{\mathrm{15}}{\mathrm{17}}{N} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{15}{gN} \\ $$$$ \\ $$$$ \\ $$

Commented by ajfour last updated on 18/Aug/20

Nice way, thanks Sir.

$${Nice}\:{way},\:{thanks}\:{Sir}. \\ $$

Answered by mr W last updated on 18/Aug/20

a=acceleration a=((2mg sin θ−mg(μ_1 +μ_2 )cos θ)/(2m)) =[sin θ−(((μ_1 +μ_2 )cos θ)/2)]g >0 T=mg sin θ−μ_2 mg cos θ−ma =mg[sin θ−μ_2 cos θ−((2sin θ−(μ_1 +μ_2 )cos θ)/2)] =((mg(μ_1 −μ_2 )cos θ)/2)

$${a}={acceleration} \\ $$$${a}=\frac{\mathrm{2}{mg}\:\mathrm{sin}\:\theta−{mg}\left(\mu_{\mathrm{1}} +\mu_{\mathrm{2}} \right)\mathrm{cos}\:\theta}{\mathrm{2}{m}} \\ $$$$=\left[\mathrm{sin}\:\theta−\frac{\left(\mu_{\mathrm{1}} +\mu_{\mathrm{2}} \right)\mathrm{cos}\:\theta}{\mathrm{2}}\right]{g}\:>\mathrm{0} \\ $$$$ \\ $$$${T}={mg}\:\mathrm{sin}\:\theta−\mu_{\mathrm{2}} {mg}\:\mathrm{cos}\:\theta−{ma} \\ $$$$={mg}\left[\mathrm{sin}\:\theta−\mu_{\mathrm{2}} \mathrm{cos}\:\theta−\frac{\mathrm{2sin}\:\theta−\left(\mu_{\mathrm{1}} +\mu_{\mathrm{2}} \right)\mathrm{cos}\:\theta}{\mathrm{2}}\right] \\ $$$$=\frac{{mg}\left(\mu_{\mathrm{1}} −\mu_{\mathrm{2}} \right)\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$

Commented by ajfour last updated on 18/Aug/20

Thanks Sir, right answer; 150N.

$${Thanks}\:{Sir},\:{right}\:{answer};\:\mathrm{150}{N}. \\ $$

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