A worker pushes horizontally on a $\mathbf{35}\mathbf{kg}$crate with a force of magnitude $\mathbf{110}\mathbf{}\mathbf{N}$. The coefficient of static friction between the crate and the floor is $\mathbf{0}\mathbf{.}\mathbf{37}$. (a) What is the value of ${\mathit{f}}_{\mathbf{a}\mathbf{,}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{x}}$under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker’s $\mathbf{110}\mathbf{}\mathbf{N}$push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

Mass,$\mathrm{m}=35\mathrm{kg}$

Coefficient of static friction,${\mathrm{\mu }}_{\mathrm{s}}=.037$

External force on the crate,$\mathrm{F}=110\mathrm{N}$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion along vertical and horizontal direction. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

${\mathrm{F}}_{\mathrm{net}}=\sum _{}\mathrm{ma}$

where, F is the net force, m is mass and a is an acceleration.

Free body Diagram of Crate:

By using Newton’s 2nd law along vertical direction (along y),

$\sum _y=m{a}_y$

Since crate is not moving upward, $a_y‐0$

$\mathrm{N}-{\mathrm{F}}_{\mathrm{g}}=0$

$\mathrm{N}={\mathrm{F}}_{\mathrm{g}}$

Relation between maximumstatic frictional force and normal force is,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{s},\max }={\mathrm{\mu }}_{\mathrm{s}}\mathrm{N} \\ ={\mathrm{\mu }}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{g}} \\ =\left(0.37\right)\left(35\right)\left(9.81\right) \\ =127\mathrm{N} \end{gathered}$$

Hence, the value of ${\mathrm{f}}_{\mathrm{s},\max }$under the circumstances is 127 N.

As.$\mathrm{F}=110\mathrm{N}<{\mathrm{f}}_{\mathrm{s},\max }$

Hence, the crate does not move.

By using Newton’s 2nd law along the horizontal direction,

$\sum _{}{\mathrm{F}}_x={\mathrm{ma}}_x$

Since crate is not moving, ${\mathrm{a}}_{\mathrm{x}}=0$

$$\begin{gathered} \mathrm{F}-{\mathrm{f}}_{\mathrm{s}}=0 \\ {\mathrm{f}}_{\mathrm{s}}=\mathrm{F} \\ =110\mathrm{N} \end{gathered}$$

Hence, the frictional force on the crate from the floor is $110\mathrm{N}$.

Let, the upward force apply by the worker isF’, then by using Newton’s 2nd law,

$\sum _{}{\mathrm{F}}_y={\mathrm{ma}}_y$

since crate is not moving upward, ${\mathrm{a}}_{\mathrm{y}}=0$

$$\begin{gathered} \mathrm{N}+{\mathrm{F}}^1-{\mathrm{F}}_{\mathrm{g}}=0 \\ \mathrm{N}={\mathrm{F}}_{\mathrm{g}}-{\mathrm{F}}^1 \end{gathered}$$

In order to move the crate F must satisfy the condition,

$$\begin{gathered} \mathrm{F}>{\mathrm{\mu }}_{\mathrm{s}}\left(\mathrm{mg}-{\mathrm{F}}^1\right) \\ 110>\left(0.37\right)\left[\left(35\right)\left(9.81\right)-{\mathrm{F}}^1\right] \\ {\mathrm{F}}^1>46.05\mathrm{N} \end{gathered}$$

Since,F’ is slightly greater than $46.05\mathrm{N}$.

Hence, the least vertical pull that will allow the first worker’s $110\mathrm{N}$push to move the crate is 47 N.

If the horizontal force applied by the worker is F’ and to move the box the total forward force should overcome the backward static frictional force $127\mathrm{N}$

Thus, ${\mathrm{F}}_1{\mathrm{F}}^1>127\mathrm{N}$

$110\mathrm{N}+{\mathrm{F}}^1>127\mathrm{N}$

Therefore, the least magnitude of F’ to move the box should be $17\mathrm{N}$.