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Question Number 19511 by Tinkutara last updated on 12/Aug/17

In the arrangement shown, the wedge is smooth and has a mass M. The sphere has a mass m. The system is released from rest from the position shown. There is no friction anywhere. Find the contact force between the wall and the sphere.

$$\mathrm{In}\:\mathrm{the}\:\mathrm{arrangement}\:\mathrm{shown},\:\mathrm{the}\:\mathrm{wedge} \\ $$$$\mathrm{is}\:\mathrm{smooth}\:\mathrm{and}\:\mathrm{has}\:\mathrm{a}\:\mathrm{mass}\:{M}.\:\mathrm{The}\:\mathrm{sphere} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{mass}\:{m}.\:\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{released} \\ $$$$\mathrm{from}\:\mathrm{rest}\:\mathrm{from}\:\mathrm{the}\:\mathrm{position}\:\mathrm{shown}. \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{friction}\:\mathrm{anywhere}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{contact}\:\mathrm{force}\:\mathrm{between}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{sphere}. \\ $$

Commented by Tinkutara last updated on 12/Aug/17

Answered by ajfour last updated on 12/Aug/17

Commented by ajfour last updated on 12/Aug/17

Asin θ = acos θ ...(i) because along the direction perpendicular to incline of wedge sphere and wedge maintain contact. From diagram of wedge Nsin θ= MA .... (ii) from diagram of sphere N_w = Nsin θ .....(iii) and mg−Ncos θ=ma ....(iv) As N_w = Nsin θ=MA we have from (iv) mg−((MAcos θ)/(sin θ))=((mAsin θ)/(cos θ)) ⇒ A(mtan θ+Mcot θ)=mg So, N_w =MA=((mMg)/(mtan 𝛉+Mcot 𝛉)) .

$$\:\:\mathrm{Asin}\:\theta\:=\:\mathrm{acos}\:\theta\:\:\:\:\:\:...\left(\mathrm{i}\right) \\ $$$$\mathrm{because}\:\mathrm{along}\:\mathrm{the}\:\mathrm{direction}\: \\ $$$$\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{incline}\:\mathrm{of}\:\mathrm{wedge} \\ $$$$\mathrm{sphere}\:\mathrm{and}\:\mathrm{wedge}\:\mathrm{maintain}\:\mathrm{contact}. \\ $$$$\:\:\:\mathrm{From}\:\mathrm{diagram}\:\mathrm{of}\:\mathrm{wedge} \\ $$$$\:\:\:\:\:\mathrm{Nsin}\:\theta=\:\mathrm{MA}\:\:\:\:....\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\mathrm{from}\:\mathrm{diagram}\:\mathrm{of}\:\mathrm{sphere} \\ $$$$\:\:\:\:\:\:\:\mathrm{N}_{\mathrm{w}} =\:\mathrm{Nsin}\:\theta\:\:\:\:\:\:\:\:.....\left(\mathrm{iii}\right) \\ $$$$\:\:\mathrm{and}\:\:\:\:\mathrm{mg}−\mathrm{Ncos}\:\theta=\mathrm{ma}\:\:\:....\left(\mathrm{iv}\right) \\ $$$$\mathrm{As}\:\:\:\mathrm{N}_{\mathrm{w}} =\:\mathrm{Nsin}\:\theta=\mathrm{MA} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{from}\:\left(\mathrm{iv}\right) \\ $$$$\:\:\:\:\:\mathrm{mg}−\frac{\mathrm{MAcos}\:\theta}{\mathrm{sin}\:\theta}=\frac{\mathrm{mAsin}\:\theta}{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\:\:\:\:\mathrm{A}\left(\mathrm{mtan}\:\theta+\mathrm{Mcot}\:\theta\right)=\mathrm{mg} \\ $$$$\mathrm{So},\:\:\boldsymbol{\mathrm{N}}_{\mathrm{w}} =\boldsymbol{\mathrm{MA}}=\frac{\boldsymbol{\mathrm{mMg}}}{\boldsymbol{\mathrm{mtan}}\:\boldsymbol{\theta}+\boldsymbol{\mathrm{Mcot}}\:\boldsymbol{\theta}}\:. \\ $$

Commented by Tinkutara last updated on 12/Aug/17

Thank you very much Sir!

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

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