Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 22499 by Tinkutara last updated on 19/Oct/17

A cylinder of weight 200 N is supported  on a smooth horizontal plane by a light  cord AC and pulled with force of 400 N.  The normal reaction at B is equal to

$$\mathrm{A}\:\mathrm{cylinder}\:\mathrm{of}\:\mathrm{weight}\:\mathrm{200}\:\mathrm{N}\:\mathrm{is}\:\mathrm{supported} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{plane}\:\mathrm{by}\:\mathrm{a}\:\mathrm{light} \\ $$$$\mathrm{cord}\:{AC}\:\mathrm{and}\:\mathrm{pulled}\:\mathrm{with}\:\mathrm{force}\:\mathrm{of}\:\mathrm{400}\:\mathrm{N}. \\ $$$$\mathrm{The}\:\mathrm{normal}\:\mathrm{reaction}\:\mathrm{at}\:{B}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Commented by Tinkutara last updated on 19/Oct/17

Answered by ajfour last updated on 19/Oct/17

Tcos 37 °=400N  ⇒ T=500N  Let normal reaction at B is R.  R=Mg+Tsin 37 °     =200N+300N = 500N =(1/2)kN .

$${T}\mathrm{cos}\:\mathrm{37}\:°=\mathrm{400}{N} \\ $$$$\Rightarrow\:{T}=\mathrm{500}{N} \\ $$$${Let}\:{normal}\:{reaction}\:{at}\:{B}\:{is}\:{R}. \\ $$$${R}={Mg}+{T}\mathrm{sin}\:\mathrm{37}\:° \\ $$$$\:\:\:=\mathrm{200}{N}+\mathrm{300}{N}\:=\:\mathrm{500}{N}\:=\frac{\mathrm{1}}{\mathrm{2}}{kN}\:. \\ $$

Commented by Tinkutara last updated on 19/Oct/17

Why Tcos 37°=400N?

$${Why}\:{T}\mathrm{cos}\:\mathrm{37}°=\mathrm{400}{N}? \\ $$

Commented by ajfour last updated on 19/Oct/17

Cord is pivoted at A. cylinder  doesn′ t move forward.  ΣF_x =ma_x =0  400N−Tcos 37 °=0

$${Cord}\:{is}\:{pivoted}\:{at}\:{A}.\:{cylinder} \\ $$$${doesn}'\:{t}\:{move}\:{forward}. \\ $$$$\Sigma{F}_{{x}} ={ma}_{{x}} =\mathrm{0} \\ $$$$\mathrm{400}{N}−{T}\mathrm{cos}\:\mathrm{37}\:°=\mathrm{0} \\ $$

Commented by Tinkutara last updated on 19/Oct/17

Thank you very much Sir!

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com