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Question Number 21759 by Tinkutara last updated on 03/Oct/17
A block of mass 1 kg is pushed against  a rough vertical wall with a force of 20  N, coefficient of static friction being (1/4).  Another horizontal force of 10 N is  applied on the block in a direction  parallel to the wall. Will the block move?  If yes, with what acceleration? If no,  find the frictional force exerted by wall  on the block. (g = 10 m/s^2 )
$$\mathrm{A}\:\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{1}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{pushed}\:\mathrm{against} \\ $$$$\mathrm{a}\:\mathrm{rough}\:\mathrm{vertical}\:\mathrm{wall}\:\mathrm{with}\:\mathrm{a}\:\mathrm{force}\:\mathrm{of}\:\mathrm{20} \\ $$$$\mathrm{N},\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{static}\:\mathrm{friction}\:\mathrm{being}\:\frac{\mathrm{1}}{\mathrm{4}}. \\ $$$$\mathrm{Another}\:\mathrm{horizontal}\:\mathrm{force}\:\mathrm{of}\:\mathrm{10}\:\mathrm{N}\:\mathrm{is} \\ $$$$\mathrm{applied}\:\mathrm{on}\:\mathrm{the}\:\mathrm{block}\:\mathrm{in}\:\mathrm{a}\:\mathrm{direction} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{wall}.\:\mathrm{Will}\:\mathrm{the}\:\mathrm{block}\:\mathrm{move}? \\ $$$$\mathrm{If}\:\mathrm{yes},\:\mathrm{with}\:\mathrm{what}\:\mathrm{acceleration}?\:\mathrm{If}\:\mathrm{no}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{frictional}\:\mathrm{force}\:\mathrm{exerted}\:\mathrm{by}\:\mathrm{wall} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{block}.\:\left({g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$
Answered by ajfour last updated on 03/Oct/17
Commented by ajfour last updated on 03/Oct/17
block′s acceleration be a.  Normal from wall=20N  (not in figure)  since net applied force along  the wall/surface is greater than  maximum value of static friction  kinetic friction = μ_k N comes into  play.  if μ_k =μ_s   =(1/4) , then  f=μ_k N= (1/4)×20=5  So  10(√2)−5=1×a  a=14.14−5 = 9.14m/s^2  .
$${block}'{s}\:{acceleration}\:{be}\:{a}. \\ $$$${Normal}\:{from}\:{wall}=\mathrm{20}{N} \\ $$$$\left({not}\:{in}\:{figure}\right) \\ $$$${since}\:{net}\:{applied}\:{force}\:{along} \\ $$$${the}\:{wall}/{surface}\:{is}\:{greater}\:{than} \\ $$$${maximum}\:{value}\:{of}\:{static}\:{friction} \\ $$$${kinetic}\:{friction}\:=\:\mu_{{k}} {N}\:{comes}\:{into} \\ $$$${play}. \\ $$$${if}\:\mu_{{k}} =\mu_{{s}} \:\:=\frac{\mathrm{1}}{\mathrm{4}}\:,\:{then} \\ $$$${f}=\mu_{{k}} {N}=\:\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{20}=\mathrm{5} \\ $$$${So} \\ $$$$\mathrm{10}\sqrt{\mathrm{2}}−\mathrm{5}=\mathrm{1}×{a} \\ $$$${a}=\mathrm{14}.\mathrm{14}−\mathrm{5}\:=\:\mathrm{9}.\mathrm{14}{m}/{s}^{\mathrm{2}} \:. \\ $$
Commented by Tinkutara last updated on 04/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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