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learning goal: to understand the role of friction in a state of equilibrium. a chair weighing 70.0 n rests on a level floor that is not frictionless. a man pushes on the chair with a force f

Sagot :

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  1. The magnitude of the normal force that the floor exerts on the chair is 93.79 N
  2. coefficient of static friction μ is 0.32

What is coefficient of static friction?

  • The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value. The formula fr = Fr/N serves as a representation of it.
  • "The highest ratio of applied force to normal force with no motion is called the coefficient of static friction." We all understand that a force called friction resists motion. If we carefully notice, there is a point at which the body opposes movement after force is applied to move it from rest.
  • The things that are creating friction determine the coefficient of static friction's value. Its value typically ranges from 0 to 1, but it can also be higher than 1.

Given details :

  • Weight of chair W = 70 N
  • Force P = 39 N
  • Angle = 38°

1.Force equilibrium in x direction

∑ [tex]F{x}[/tex] = 0

F - 39 cos Ф = 0

As Ф = 38°

F = 39 cos38°

F = 30.73 N

Force equilibrium in y direction

∑ [tex]F{y}[/tex] = 0

N = W + P sin38°

   =  70 + 39 sin38°

N = 93.79 N

Thus the normal force on chair = 93.79 N

From the free body diagram

F = μN

30.73 = μ x 93.79

μ = 30.73 / 93.79 = 0.32

Learn more about coefficient of static friction refer to :

https://brainly.com/question/25534663

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