Rough Surface with: Ms = 0.8 HK = 0.4 (a) Find the magnitude of the

force F needed to prevent the

book from sliding down the

rough wall.

F

M = 1.5 kg

600

(b) Find the minimum force F

needed to set the book in

motion up the rough wall with

constant velocity

a)    F = 18.375N, b) F = 24.5 N

Explanation:

This exercise can be solved using the translational equilibrium equations.

Let's start by fixing a reference system with the horizontal x axis and the vertical y axis, from the statement of the exercise I understand that the wall is vertical and the book is supported on it, therefore the applied force is in the direction towards the wall

a) In this part the force that does not allow the movement of the book is requested, therefore the static friction coefficient must be used (μ_s = 0.8)

X axis  

       F - N = 0

       N = F

Y axis

       fr - W = 0

       W = fr

where W is the weight of the book.

The friction force has the formula

       fr = μ_s N

we substitute

       mg = μ_s F

       F =

let's calculate

       F = 1.5 9.8 / 0.8

       F = 18.375N

b) In this case the book is moving so the friction coefficient to use is kinetic (   μ_K = 0.6)

       F =

we calculate

        F = 1.5 9.8 / 0.6

        F = 24.5 N

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.