Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 129542 by mr W last updated on 16/Jan/21

Commented by mr W last updated on 16/Jan/21

$$whenthemanclimbstheladderfrom$$$$bottomtotop,howmuchdoesthe$$$$blockmoveandinwhichdirection?$$

Answered by ajfour last updated on 17/Jan/21

$$Ifmanclimbsladderintimet.$$$$F-mg=ma\uparrow (system:manonly)$$$$(system:ladderandblock)$$$$F-mg=(2M+m)A\downarrow$$$$(2M+m)A=ma$$$$\frac{1}{2}(A+a)t^2=L$$$$\frac{1}{2}(1+1+\frac{2M}{m}){At}^2=L$$$$\frac{1}{2}{At}^2=y$$$$dividing$$$$\frac{y}{L}=\frac{m}{2(M+m)}$$$$y_{block}=\frac{mL}{2(M+m)}\uparrow$$

Commented by mr W last updated on 16/Jan/21

$$great!thanks!$$

Commented by Emptyset last updated on 17/Jan/21

$$canyoupleaseelaborate$$

Answered by mr W last updated on 17/Jan/21

Commented by mr W last updated on 17/Jan/21

$$T=tensioninmasslessrope$$$$F=forcebetweenmanandladder$$$$a=accelerationofmanupwards$$$$A=accelerationofblockupwards$$$$=accelerationofladderdownwards$$$$$$$$systemman:$$$$F-mg=ma...\left(i\right)$$$$$$$$systemladder:$$$$F+Mg-T=MA...\left(ii\right)$$$$$$$$systemblock:$$$$T-(M+m)g=(M+m)A...\left(iii\right)$$$$$$$$\left(ii\right)+\left(iii\right):$$$$F-mg=(2M+m)A$$$$\Rightarrow F=2(M+m)A+mg$$$$putthisinto\left(i\right):$$$$(2M+m)A=ma$$$$\Rightarrow a=\frac{2M+m}{m}\times A$$$$relativeaccelerationbetweenman$$$$andladder{a}_R=a+A=\frac{2(M+m)}{m}\times A$$$$\Rightarrow \frac{A}{a_R}=\frac{m}{2(M+m)}$$$$saythedistancetheblockmovesisH,$$$$H=\frac{1}{2}{At}^2$$$$thedistancethemanmovesrelatively$$$$toladderisL,$$$$L=\frac{1}{2}{a}_R{t}^2$$$$\Rightarrow \frac{H}{L}=\frac{A}{a_R}=\frac{m}{2(M+m)}$$$$\Rightarrow H=\frac{mL}{2(M+m)}$$$$itsdirectionisupwards.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com