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Question Number 27202 by Tinkutara last updated on 03/Jan/18

Answered by mrW1 last updated on 03/Jan/18

N=normal force between m and M  T=tension in thread  N=mg cos θ  T=N sin θ=mg sin θ cos θ=3×10×(1/2)=15 N

$${N}={normal}\:{force}\:{between}\:{m}\:{and}\:{M} \\ $$$${T}={tension}\:{in}\:{thread} \\ $$$${N}={mg}\:\mathrm{cos}\:\theta \\ $$$${T}={N}\:\mathrm{sin}\:\theta={mg}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta=\mathrm{3}×\mathrm{10}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{15}\:{N} \\ $$

Commented by Tinkutara last updated on 03/Jan/18

Why we can't resolve tension along the inclined plane and normal to it and then balance the forces?

Commented by mrW1 last updated on 03/Jan/18

because the normal force between block  M and floor as well as Mg have also component  alone the inclined plane and normal  to it.

$${because}\:{the}\:{normal}\:{force}\:{between}\:{block} \\ $$$${M}\:{and}\:{floor}\:{as}\:{well}\:{as}\:{Mg}\:{have}\:{also}\:{component} \\ $$$${alone}\:{the}\:{inclined}\:{plane}\:{and}\:{normal} \\ $$$${to}\:{it}. \\ $$

Commented by Tinkutara last updated on 03/Jan/18

Thank you very much Sir! I got the answer.

Commented by mrW1 last updated on 03/Jan/18

we see T=((mg sin 2θ)/2)  so if the question is what is the  inclination such that the tension in  thread is maximum?  we know the answer is when θ=45°.

$${we}\:{see}\:{T}=\frac{{mg}\:\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}} \\ $$$${so}\:{if}\:{the}\:{question}\:{is}\:{what}\:{is}\:{the} \\ $$$${inclination}\:{such}\:{that}\:{the}\:{tension}\:{in} \\ $$$${thread}\:{is}\:{maximum}? \\ $$$${we}\:{know}\:{the}\:{answer}\:{is}\:{when}\:\theta=\mathrm{45}°. \\ $$

Commented by Tinkutara last updated on 03/Jan/18

Yes thanks Sir.

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