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Question-29880




Question Number 29880 by tawa tawa last updated on 13/Feb/18
Commented by ajfour last updated on 13/Feb/18
Commented by ajfour last updated on 13/Feb/18
Fcos α−f−mgsin α=ma  ⇒  f=Fcos α−ma−mgsin α   =200cos 20°−1.5×0.4−15sin 20°    ⇒ f = 182.2N  μN=μ(mgcos α+Fsin α) =f  μ =(f/(mgcos α+Fsin α))     μ =((200cos 20°−0.6−15sin 20°)/(15cos 20°+200sin 20°))    ⇒  𝛍 = 2.208
$${F}\mathrm{cos}\:\alpha−{f}−{mg}\mathrm{sin}\:\alpha={ma} \\ $$$$\Rightarrow\:\:{f}={F}\mathrm{cos}\:\alpha−{ma}−{mg}\mathrm{sin}\:\alpha \\ $$$$\:=\mathrm{200cos}\:\mathrm{20}°−\mathrm{1}.\mathrm{5}×\mathrm{0}.\mathrm{4}−\mathrm{15sin}\:\mathrm{20}° \\ $$$$\:\:\Rightarrow\:\boldsymbol{{f}}\:=\:\mathrm{182}.\mathrm{2}\boldsymbol{{N}} \\ $$$$\mu{N}=\mu\left({mg}\mathrm{cos}\:\alpha+{F}\mathrm{sin}\:\alpha\right)\:={f} \\ $$$$\mu\:=\frac{{f}}{{mg}\mathrm{cos}\:\alpha+{F}\mathrm{sin}\:\alpha} \\ $$$$\:\:\:\mu\:=\frac{\mathrm{200cos}\:\mathrm{20}°−\mathrm{0}.\mathrm{6}−\mathrm{15sin}\:\mathrm{20}°}{\mathrm{15cos}\:\mathrm{20}°+\mathrm{200sin}\:\mathrm{20}°} \\ $$$$\:\:\Rightarrow\:\:\boldsymbol{\mu}\:=\:\mathrm{2}.\mathrm{208}\: \\ $$
Commented by tawa tawa last updated on 13/Feb/18
God bless you sir. i really appreciate
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate} \\ $$

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