You must push a crate across a floor to a docking bay. The crate weighs 165 N. The coefficient of static friction between crate and floor is 0.510, and the coefficient of kinetic friction is 0.32. Your force on the crate is directed horizontally. (a) What magnitude of your push puts the crate on the verge of sliding? (b) With what magnitude must you then push to keep the crate moving at a constant velocity? (c) If, instead, you then push with the same magnitude as the answer to (a), what is the magnitude of the crate’s acceleration?

Weight of the crate:$W=mg=165\mathrm{N}$

Coefficient of static friction: ${\mu }_s=0.51$

Coefficient of kinetic friction:${\mu }_k=0.32$

The problem deals with the Newton’s laws of motion which describe the relations between the forces acting on a body and the motion of the body. Also it involves friction force. Draw a free body diagram for the given situation and then apply Newton’s first and second law.

Formula:

Frictional force is given by,

${\mathbf{f}}_{\mathbf{k}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{k}}{\mathbf{F}}_{\mathbf{N}}$

(a)

The push or the $F_{applied}$to get the crate just moving the applied force must be as big as the maximum force of static friction = $f_{s.max}={\mu }_s{F}_N$and here $F_N=W=165\mathrm{N}$

Hence, required push,

$$\begin{gathered} F_{applied}=f_{s.max} \\ ={\mu }_s{F}_N \\ \Rightarrow F_{applied}=0.51\times 165\mathrm{N} \\ \Rightarrow {\mathrm{F}}_{\mathrm{applied}}=84.15\mathrm{N} \end{gathered}$$

(b)

While in motion, constant velocity (zero acceleration) is maintained if the push is equal to the kinetic friction force

$$\begin{gathered} F\left(applied\right)=f_k \\ ={\mu }_s{F}_N \\ \Rightarrow F_{\mathit{a}\mathit{p}\mathit{p}\mathit{l}\mathit{i}\mathit{e}\mathit{d}}=0.32\times 165\mathrm{N} \\ \Rightarrow F_{\mathit{a}\mathit{p}\mathit{p}\mathit{l}\mathit{i}\mathit{e}\mathit{d}}\mathit{=}52.8\mathrm{N} \end{gathered}$$

(c)

Mass of the crate,

$$\begin{gathered} m=\frac{165\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =16.84\mathrm{kg} \end{gathered}$$

The acceleration, using the push from part (a), is:

$a=\frac{84.15\mathrm{N}-52.8\mathrm{N}}{16.84\mathrm{kg}}$

$\Rightarrow a=1.87m/s^2$