contestada

In the figure a fastidious worker pushes directly along the handle of a mop with a force Upper F Overscript right-arrow EndScripts. The handle is at an angle θ = 44° with the vertical, and μs = 0.68 and μk = 0.32 are the coefficients of static and kinetic friction between the head of the mop and the floor. Ignore the mass of the handle and assume that all the mop's mass m = 0.64 kg is in its head.
(a) If the mop head moves along the floor with a constant velocity, then what is F?
(b) If θ is less than a certain value θ0, then Upper F Overscript right-arrow EndScripts (still directed along the handle) is unable to move the mop head. Find θ0.

Respuesta :

Answer:

Part a)

[tex]F = 4.32 N[/tex]

Part b)

[tex]\theta = 34.2 degree[/tex]

Explanation:

Part a)

let force is applied at constant value F

Now by vertical direction force balance we know that

[tex]Fcos44 + mg = F_n[/tex]

now the friction force on the mop while it moves with uniform speed is given as

[tex]F_k = \mu_k F_n[/tex]

[tex]F_k = 0.32(Fcos44 + mg)[/tex]

Now for uniform speed of mop the horizontal force must be balanced on it

so we have

[tex]Fsin44 = 0.32(Fcos44 + mg)[/tex]

[tex]Fsin44 - 0.32 Fcos44 = 0.32 mg[/tex]

[tex]F = \frac{0.32mg}{sin44 - 0.32cos44}[/tex]

[tex]F = \frac{0.32(0.64)(9.81)}{sin44 - 0.32cos44}[/tex]

[tex]F = 4.32 N[/tex]

Part b)

for a certain value of angle if Mop is not moved then we will have

[tex]F_x = F_f[/tex]

so we can say that in that case the friction must be static friction

so we will have

[tex]Fsin\theta = \mu_s(mg + F cos\theta)[/tex]

[tex]F sin\theta = 0.68(0.64\times 9.81 + F cos\theta)[/tex]

[tex]F(sin\theta - 0.68cos\theta) = 4.27[/tex]

[tex]F = \frac{4.27}{sin\theta - 0.68 cos\theta}[/tex]

If it will not move than

[tex]sin\theta < 0.68 cos\theta[/tex]

[tex]\theta = 34.2 degree[/tex]

Otras preguntas