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Question Number 22319 by Tinkutara last updated on 15/Oct/17

A string of negligible mass going over a clamped pulley of mass m supports a block of mass M. The force on the pulley by the clamp is given by

$$\mathrm{A}\:\mathrm{string}\:\mathrm{of}\:\mathrm{negligible}\:\mathrm{mass}\:\mathrm{going}\:\mathrm{over}\:\mathrm{a} \\ $$$$\mathrm{clamped}\:\mathrm{pulley}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{supports}\:\mathrm{a} \\ $$$$\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:{M}.\:\mathrm{The}\:\mathrm{force}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{pulley}\:\mathrm{by}\:\mathrm{the}\:\mathrm{clamp}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$

Commented by Tinkutara last updated on 15/Oct/17

Commented by mrW1 last updated on 15/Oct/17

F_x =Mg F_y =(M+m)g F=(√(M^2 +(M+m)^2 )) g =(√(2M(M+m)+m^2 )) g

$$\mathrm{F}_{\mathrm{x}} =\mathrm{Mg} \\ $$$$\mathrm{F}_{\mathrm{y}} =\left(\mathrm{M}+\mathrm{m}\right)\mathrm{g} \\ $$$$\mathrm{F}=\sqrt{\mathrm{M}^{\mathrm{2}} +\left(\mathrm{M}+\mathrm{m}\right)^{\mathrm{2}} }\:\mathrm{g} \\ $$$$=\sqrt{\mathrm{2M}\left(\mathrm{M}+\mathrm{m}\right)+\mathrm{m}^{\mathrm{2}} }\:\mathrm{g} \\ $$

Commented by Tinkutara last updated on 15/Oct/17

But answer is [(√((M + m)^2 + m^2 ))]g

$${But}\:{answer}\:{is}\:\left[\sqrt{\left({M}\:+\:{m}\right)^{\mathrm{2}} \:+\:{m}^{\mathrm{2}} }\right]{g} \\ $$

Commented by ajfour last updated on 15/Oct/17

If m=0 , which answer agrees ?

$${If}\:{m}=\mathrm{0}\:,\:{which}\:{answer}\:{agrees}\:? \\ $$

Commented by mrW1 last updated on 15/Oct/17

If M=0, the force should be mg. If m=0, the force should be (√2)Mg.

$$\mathrm{If}\:\mathrm{M}=\mathrm{0},\:\mathrm{the}\:\mathrm{force}\:\mathrm{should}\:\mathrm{be}\:\mathrm{mg}. \\ $$$$\mathrm{If}\:\mathrm{m}=\mathrm{0},\:\mathrm{the}\:\mathrm{force}\:\mathrm{should}\:\mathrm{be}\:\sqrt{\mathrm{2}}\mathrm{Mg}. \\ $$

Commented by mrW1 last updated on 16/Oct/17

Commented by ajfour last updated on 16/Oct/17

yes sir, good agreement.

$${yes}\:{sir},\:{good}\:{agreement}. \\ $$

Commented by Tinkutara last updated on 18/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Answered by math solver last updated on 16/Oct/17

answer by mrw1 is absolutely correct! P.S : it is an old jee problem

$${answer}\:{by}\:{mrw}\mathrm{1}\:{is}\:{absolutely}\:{correct}! \\ $$$${P}.{S}\::\:{it}\:{is}\:{an}\:{old}\:{jee}\:{problem} \\ $$

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