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 154888 by physicstutes last updated on 22/Sep/21

Answered by mr W last updated on 23/Sep/21

Commented by physicstutes last updated on 23/Sep/21

brilliant sir thanks

$$\mathrm{brilliant}\:\mathrm{sir}\:\mathrm{thanks} \\ $$

Commented by mr W last updated on 23/Sep/21

mg=160 N Mg=740 N N=mg+Mg=900 N H=f (1) maximal friction f_(max) =μN=0.4×900=360 N (2) H×4=mg×(3/2)+Mg×((s/5))×3 H×4=160×(3/2)+740×((s/5))×3 H=60+111s at s=1 m: f=H=60+111×1=171 N (3) H=f≤f_(max) =360 60+111s≤360 ⇒s_(max) =((360−60)/(111))=((300)/(111))≈2.7 m

$${mg}=\mathrm{160}\:{N} \\ $$$${Mg}=\mathrm{740}\:{N} \\ $$$${N}={mg}+{Mg}=\mathrm{900}\:{N} \\ $$$${H}={f} \\ $$$$\left(\mathrm{1}\right) \\ $$$${maximal}\:{friction}\:{f}_{{max}} =\mu{N}=\mathrm{0}.\mathrm{4}×\mathrm{900}=\mathrm{360}\:{N} \\ $$$$\left(\mathrm{2}\right) \\ $$$${H}×\mathrm{4}={mg}×\frac{\mathrm{3}}{\mathrm{2}}+{Mg}×\left(\frac{{s}}{\mathrm{5}}\right)×\mathrm{3} \\ $$$${H}×\mathrm{4}=\mathrm{160}×\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{740}×\left(\frac{{s}}{\mathrm{5}}\right)×\mathrm{3} \\ $$$${H}=\mathrm{60}+\mathrm{111}{s} \\ $$$${at}\:{s}=\mathrm{1}\:{m}: \\ $$$${f}={H}=\mathrm{60}+\mathrm{111}×\mathrm{1}=\mathrm{171}\:{N} \\ $$$$ \\ $$$$\left(\mathrm{3}\right) \\ $$$${H}={f}\leqslant{f}_{{max}} =\mathrm{360} \\ $$$$\mathrm{60}+\mathrm{111}{s}\leqslant\mathrm{360} \\ $$$$\Rightarrow{s}_{{max}} =\frac{\mathrm{360}−\mathrm{60}}{\mathrm{111}}=\frac{\mathrm{300}}{\mathrm{111}}\approx\mathrm{2}.\mathrm{7}\:{m} \\ $$

Commented by Tawa11 last updated on 23/Sep/21

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com