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Kay [80]
3 years ago
6

Rough Surface with: Ms = 0.8 HK = 0.4

Physics
1 answer:
vlabodo [156]3 years ago
3 0

Answer:

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 = \frac{mg}{\mu_s }

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 = \frac{mg}{\mu_K }

       

we calculate

        F = 1.5 9.8 / 0.6

        F = 24.5 N

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For this exercise the force is given by Coulomb's law

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directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

             r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

             r₁₃ = 0.50 m

            F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

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The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

           tan θ = y / x

           θ = tan⁻¹ y / x

          θ = tan⁻¹ 0.3 / 0.4

           tea = 36.87º

    The angle from the positive side of the x-axis is

         θ ’= 180 - θ

        θ ’= 180 - 36.87

        θ ’= 143.13º

       sin143.13 = F_13y / F₁₃

           F_13y = F₁₃ sin 143.13

           F{13y} = 1.697 10⁻² sin 143.13

           F_13y = 1.0183 10⁻² N

            cos 143.13 = F_13x / F₁₃

           F₁₃ₓ = F₁₃ cos 143.13

           F₁₃ₓ = 1.697 10⁻² cos 143.13

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Now we can find the components of the resultant force

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          Fx = -1,357 10-2 +0

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          Fy = F13y + F12y

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We use the Pythagorean theorem to find the modulus

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         tan tea = Fy / Fx

          tea = tan-1 (0.110183 / -0.01357)

          tea = 1,448 rad

to find the angle about the positive side of the + x axis

           tea '= pi - 1,448

           Tea = 1.6936 rad

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