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A 21 kg chair initially at rest on a horizontal floor requires a 175 N horizontal force to set it in motion. Once the chair is in motion, a 137 N horizontal force keeps it moving at a constant velocity. The acceleration of gravity is 9.81 m/s 2 . a) What is the coefficient of static friction between the chair and the floor?

Respuesta :

Answer:[tex]\mu =0.85 [/tex]

Explanation:

Given

mass of chair m=21 kg

Force required to set chair in motion is 175 N

Once the chair is in motion 137 N is require to move it with constant velocity

i.e. 175 N is the amount of force needed to just overcome static friction and 137 is the kinetic friction force

thus

[tex]f_s(static\ friction)=\mu \cdot N[/tex]

where [tex]\mu [/tex]is the coefficient of static friction and N is Normal reaction

[tex]N=mg[/tex]

[tex]f_s=\mu mg[/tex]

[tex]\mu mg=175[/tex]

[tex]\mu =\frac{175}{21\times 9.8}[/tex]

[tex]\mu =0.85 [/tex]

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