A-particle-of-mass-1-5kg-rests-on-a-rough-plane-inclined-at-45-to-the-horizontal-It-is-maintained-in-equilibrium-by-a-horizontal-force-of-p-newtons-Given-that-the-coefficient-of-friction-between-

Question Number 54578 by pieroo last updated on 07/Feb/19

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{1}.\mathrm{5kg}\:\mathrm{rests}\:\mathrm{on}\:\mathrm{a}\:\mathrm{rough}\: \\ $$$$\mathrm{plane}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{45}°\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{maintained}\:\mathrm{in}\:\mathrm{equilibrium}\:\mathrm{by}\:\mathrm{a}\: \\ $$$$\mathrm{horizontal}\:\mathrm{force}\:\mathrm{of}\:{p}\:\mathrm{newtons}.\:\mathrm{Given} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{particle}\:\mathrm{and}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{4}},\:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{p}\:\mathrm{when}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{moving} \\ $$$$\mathrm{i}.\:\mathrm{down}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{ii}.\:\mathrm{up}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\left[\mathrm{take}\:\mathrm{g}=\mathrm{10ms}^{−\mathrm{2}} \right]. \\ $$

Commented by pieroo last updated on 07/Feb/19

$$\mathrm{Please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help}. \\ $$

Answered by mr W last updated on 07/Feb/19

N=mg cos α+p sin α (i) mg sin α−p cos α=μN=μ(mg cos α+p sin α) ⇒p=((mg(sin α−μ cos α))/(μ sin α+cos α)) =((1.5×10(1−(1/4)))/((1/4)+1))=9 N (ii) p cos α−mg sin α=μN=μ(mg cos α+p sin α) ⇒p=((mg(sin α+μ cos α))/(cos α−μ sin α)) =((1.5×10(1+(1/4)))/(1−(1/4)))=25 N

$${N}={mg}\:\mathrm{cos}\:\alpha+{p}\:\mathrm{sin}\:\alpha \\ $$$$\left({i}\right) \\ $$$${mg}\:\mathrm{sin}\:\alpha−{p}\:\mathrm{cos}\:\alpha=\mu{N}=\mu\left({mg}\:\mathrm{cos}\:\alpha+{p}\:\mathrm{sin}\:\alpha\right) \\ $$$$\Rightarrow{p}=\frac{{mg}\left(\mathrm{sin}\:\alpha−\mu\:\mathrm{cos}\:\alpha\right)}{\mu\:\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha} \\ $$$$=\frac{\mathrm{1}.\mathrm{5}×\mathrm{10}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)}{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}}=\mathrm{9}\:{N} \\ $$$$\left({ii}\right) \\ $$$${p}\:\mathrm{cos}\:\alpha−{mg}\:\mathrm{sin}\:\alpha=\mu{N}=\mu\left({mg}\:\mathrm{cos}\:\alpha+{p}\:\mathrm{sin}\:\alpha\right) \\ $$$$\Rightarrow{p}=\frac{{mg}\left(\mathrm{sin}\:\alpha+\mu\:\mathrm{cos}\:\alpha\right)}{\mathrm{cos}\:\alpha−\mu\:\mathrm{sin}\:\alpha} \\ $$$$=\frac{\mathrm{1}.\mathrm{5}×\mathrm{10}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\mathrm{25}\:{N} \\ $$

Commented by mr W last updated on 07/Feb/19