Question

A man attempts to push a 19.8 kg crate across a warehouse floor. He slowly increases the force until the crate starts to move at a force of 34.8 N. He then maintains this 34.8 N force while the box accelerates at 0.357 m/s^2. What is the coefficient of static friction between the crate and the floor?

Answer 1

Answer:

 μ = 0.18

Explanation:

Let's use Newton's second Law, the coordinate system is horizontal and vertical

Before starting to move the box

Y axis

     N-W = 0

     N = W = mg

X axis

     F -fr = 0

     F = fr

The friction force has the formula

     fr = μ N

     fr =  μ m g

At the limit point just before starting the movement

     F = μ m g

     μ = F / m g

calculate

      μ = 34.8 / (19.8 9.8)

    μ = 0.18

Answer 2

Answer:

The coefficient of friction between the crate and the floor = 0.143

Explanation:

Frictional Force: This is the force that act between two surface in contact and tend to oppose their motion. it is measured in Newton (N)

F - F₁ = ma.................... Equation 1

Where F = Force of the crate, F₁ = frictional force, m = mass of the crate, a = acceleration of the crate

making F₁ the subject the equation 1

F₁ = F - ma .................... Equation 2

Given: F = 34. 8 N,  m = 19.8 kg. a = 0.357 m/s².

Substituting these values into equation 2

F₁ = 34.8 - (19.8×0.357)

F₁ = 34.8 - 7.07

F₁ = 27.73 N.

F₁ = μR ............... equation 3

making μ the subject of formula in equation 3

μ = F₁/R.............. Equation 4

Where F₁ = Frictional Force, μ = coefficient of static friction, R = Normal reaction.

R = mg,  

where g = 9.8 m/s², m = 19.8 kg

R = 19.8( 9.8) = 194.04 N,

R = 194.04 N, F₁ = 27.73 N

Substituting these values into equation 4

 μ  = 27.73/194.04

μ  = 0.143.

therefore the coefficient of friction between the crate and the floor = 0.143