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

Commented by ajfour last updated on 18/Aug/20

$$Findtensioninrodconnecting$$$$theblocks.$$

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

$$Forupperblock$$$$T+mgsin\theta -f_1=ma(f=\mu N=\mu mgcos\theta )(\theta ={tan}^{-1}\frac{8}{15})$$$$Lowerblock$$$$mgsin\theta -T-f_2=ma$$$$T+mgsin\theta -f_1=mgsin\theta -T-f_2$$$$2T=f_1-f_2$$$$T=\frac{{\mu }_{1}mgcos\theta -{\mu }_{2}mgcos\theta }{2}=\frac{1}{2}mgcos\theta (0.2)=\frac{g}{10}.170.cos\theta =g.17.\frac{15}{17}N$$$$=15gN$$$$$$$$$$

Commented by ajfour last updated on 18/Aug/20

$$Niceway,thanksSir.$$

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

$$a=acceleration$$$$a=\frac{2mg\sin \theta -mg({\mu }_1+{\mu }_2)\cos \theta }{2m}$$$$=[\sin \theta -\frac{({\mu }_1+{\mu }_2)\cos \theta }{2}]g>0$$$$$$$$T=mg\sin \theta -{\mu }_{2}mg\cos \theta -ma$$$$=mg[\sin \theta -{\mu }_2\cos \theta -\frac{2\sin \theta -({\mu }_1+{\mu }_2)\cos \theta }{2}]$$$$=\frac{mg({\mu }_1-{\mu }_2)\cos \theta }{2}$$

Commented by ajfour last updated on 18/Aug/20

$$ThanksSir,rightanswer;150N.$$

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