Question #123553

A particle of mass 2kg is placed on a rough plane which is inclined at 30°to the horizontal. A force of 20N acts on the particle in a direction parallel to and up the plane. If the particle is just about to move up the plane, analytically obtain the coefficient of friction. (g=10m/s^2)

Expert's answer


According to Newton's second law, the particle is in rest (which we can compare to "about to move") if all forces are in equilibrium:


Ox:−mg sinθ−f+F=0,Oy:N−mg cosθ=0,f=μN.Ox: -mg\text{ sin}\theta-f+F=0,\\ Oy: N-mg\text{ cos}\theta=0,\\ f=\mu N.

Therefore, we see that the force force of friction can be expressed as


f=μmg cosθ,f=\mu mg\text{ cos}\theta,

the coefficient of friction can be expressed as


μ=fmg cosθ=F−mg sinθmg cosθ. μ=20−2⋅10 sin30°2⋅10 cosθ=0.58.\mu=\frac{f}{mg\text{ cos}\theta}=\frac{F-mg\text{ sin}\theta}{mg\text{ cos}\theta}.\\\space\\ \mu=\frac{20-2\cdot10\text{ sin}30°}{2\cdot10\text{ cos}\theta}=0.58.
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