A 29 kg chair initially at rest on a horizontal floor requires a 171 N horizontal force to set it in motion. Once the chair is in motion, a 135 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? 017 (part 2 of 2) 10.0 points b) What is the coefficient of kinetic friction between the chair and the floor?

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moya1316 moya1316 · Answer 1 · 2019-11-01T13:40:20+01:00

Answer:

a) μ = 0.602 and b) μ = 0.475

Explanation:

a) Let's use Newton's second law for this exercise. The reference system the x axis is parallel to the floor and the axis and perpendicular.

Y Axis

N-W = 0

N = W

X axis

F-fr = ma

The friction force equation is

fr = μN

At the point where the movement begins the acceleration is zero

Let's replace

F = μ (mg) = 0

F = μ mg

μ = F / mg

μ = 171/29 9.8

μ = 0.602

b) in this case we have the same formula, but the friction is kinetic

μ = F2 / mg

μ = 135/29 9.8

μ = 0.475