In Fig. 6-61 a fastidious worker pushes directly along the handle of a mop with a force. The handle is at an angle$\mathit{\theta }$with the vertical, and${\mathit{\mu }}_{\mathbf{s}}$and${\mathit{\mu }}_{\mathbf{k}}$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 mis in its head. (a) If the mop head moves along the floor with a constant velocity, then what is F? (b) Show that if$\mathit{\theta }$.is less than a certain value${\mathit{\theta }}_{\mathbf{0}}$, then$\overrightarrow{\mathbf{f}}$(still directed along the handle) is unable to move the mop head. Find${\mathit{\theta }}_{\mathbf{0}}$.

The handle is at an angle with the vertical,${\mu }_s$ and${\mu }_k$ are the coefficients of static and kinetic friction between the head of the mop and the floor

Find the components of force in horizontal and vertical direction. And write the equation in X and Y direction and solve them to find the value of force.

Formula:

$\mathbf{F}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{k}}\mathbf{mg}}{\mathbf{sin\theta }\mathbf{-}{\mathbf{\mu }}_{\mathbf{k}}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{\theta }}$

(a)

The x component of F contributes to the motion of the crate while its y component indirectly contributes to the inhibiting effects of friction Along the y direction, we have$F_N-F\cos \mathbb{\theta }\mathbb{-}mg\mathbb{=}0$ and along x direction we have$F\sin \theta -f_k=0$ (since it is not accelerating, according to the problem). Also,$f_k={\mu }_k{F}_N$ . Solving these equations for F yields

$F=\frac{{\mu }_{k}mg}{\sin \theta -{\mu }_k\cos \theta }$

(b)

When $\theta <{\theta }_0={\tan }^{-1}{\mu }_s$, F will not be able to move the mop head.