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

A particle of mass 1 . 5 kg 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

the particle and the plane is \frac{1}{4}, calculate

the value of p when the particle is on

the point of moving

i . down the plane

ii . up the plane

[take g = {10 ms}^{- 2}] .

Commented by pieroo last updated on 07/Feb/19

Please i need your help .

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

N = m g \cos α + p \sin α

(i)

m g \sin α - p \cos α = μ N = μ (m g \cos α + p \sin α)

\Rightarrow p = \frac{m g (\sin α - μ \cos α)}{μ \sin α + \cos α}

= \frac{1.5 \times 10 (1 - \frac{1}{4})}{\frac{1}{4} + 1} = 9 N

(i i)

p \cos α - m g \sin α = μ N = μ (m g \cos α + p \sin α)

\Rightarrow p = \frac{m g (\sin α + μ \cos α)}{\cos α - μ \sin α}

= \frac{1.5 \times 10 (1 + \frac{1}{4})}{1 - \frac{1}{4}} = 25 N

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

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

f i g u r e a b o v e s h o w s c a s e (i) .

Commented by pieroo last updated on 08/Feb/19

thanks sir

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