Chapter 6, Problem 6.34P

Review. A window washer pulls a rubber squeegee down a very tall vertical window. The squeegee has mass 160 g and is mounted on the end of a light rod. The coefficient of kinetic friction between the squeegee and the dry glass is 0.900. The window washer presses it against the window with a force having a horizontal component of 4.00 N. (a) If she pulls the squeegee down the window at constant velocity, what vertical force component must she exert? (b) The window washer increases the downward force component by 25.0%, while all other forces remain the same. Find the squeegee’s acceleration in this situation. (c) The squeegee is moved into a wet portion of the window, where its motion is resisted by a fluid drag force R proportional to its velocity according to R = −20.0v, where R is in newtons and v is in meters per second. Find the terminal velocity that the squeegee approaches, assuming the window washer exerts the same force described in part (b).

To determine

The vertical force with which the washer pulls the squeegee down the window at constant velocity.

Answer to Problem 6.34P

The vertical component of the force that washer must exert when she pulls the squeegee down the window at constant velocity is $2.03\text{N}$.

Explanation of Solution

The mass of the squeegee is $160\text{g}$, the coefficient of kinetic friction between the squeegee and dry glass is $0.900$ and the horizontal component of the force against the window is $4.00\text{N}$ which is the normal force.

Write the expression for the net vertical force for the given system is given as,

    ${\displaystyle \sum F_{\text{y}}}=F_{\text{V}}+mg-F_{\text{s}}$                                                                (I)

Here, ${\displaystyle \sum F_{\text{y}}}$ is the net vertical force, $m$ is the mass of the squeegee, $g$ is the acceleration due to gravity and $F_{\text{s}}$ is the frictional force.

Write the expression for the frictional force

    $F_{\text{k}}=\mu N$

Here, $\mu$ is the coefficient of kinetic friction.

At constant velocity the net vertical force is zero.

Substitute $\mu N$ for $F_{\text{s}}$ and $0$ for ${\displaystyle \sum F_{\text{y}}}$ in equation (I).

    $0=F_{\text{V}}+mg-\mu N$

Rearrange the above expression for $F_{\text{V}}$.

    $F_{\text{V}}=\mu F\sin \theta -mg$                                                                                  (II)

Substitute $0.900$ for $\mu$, $160\text{g}$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$ and $4.00\text{N}$ for $N$ in equation (II).

    $\begin{array}{c}{F}_{\text{V}}=\left(0.900\right)\left(4.00\right)-\left(160\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =2.03\text{N}\end{array}$

Conclusion:

Therefore, the vertical component of the force that washer must exert when she pulls the squeegee down the window at constant velocity is $2.03\text{N}$.

To determine

The acceleration when downward pulling force increases by $25\%$.

Answer to Problem 6.34P

The acceleration when downward pulling force increases by $25\%$ is $3.18\text{m}/{\text{s}}^2$.

Explanation of Solution

The new vertical component of the force exerted by washer against the window is given as,

    ${F^{\prime }}_{\text{V}}=1.25F_{\text{V}}$

Write the expression for the new net vertical force for the given system

    ${\displaystyle \sum F_{\text{y}}{}^{\prime }}=F_{\text{V}}{}^{\prime }+mg-F_{\text{s}}$

Substitute $1.25F_{\text{V}}$ for ${F^{\prime }}_{\text{V}}$ in above equation.

    ${\displaystyle \sum F_{\text{y}}{}^{\prime }}=1.25\left(F_{\text{V}}{}^{\prime }\right)+mg-\mu N$                                                                  (III)

Substitute $4.00\text{N}$ for $N$, $160\text{g}$ for $m$, $0.900$ for $\mu$, $9.8\text{m}/{\text{s}}^2$ for $g$ and $2.03\text{N}$ for $F_{\text{V}}$ in equation (III) to find ${\displaystyle \sum F_{\text{y}}{}^{\prime }}$.

    $\begin{array}{c}{\displaystyle \sum F_{\text{y}}{}^{\prime }}=1.25\left(2.03\text{N}\right)+\left(160\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right)\left(9.8\text{m}/{\text{s}}^2\right)-\left(0.900\right)4.00\text{N}\\ =\text{0}\text{.5055}\text{N}\end{array}$

Write the formula to calculate the force

    $F_{{\text{V}}_{\text{n}}}=ma$

Here, $a$ is the acceleration.

Rearrange the above expression for $a$.

    $a=\frac{F_{{\text{V}}_{\text{n}}}}{m}$                                                                                               (IV)

Conclusion:

Substitute $160\text{g}$ for $m$ and $\text{0}\text{.5055}\text{N}$ for $F_{{\text{V}}_{\text{n}}}$ in equation (IV).

    $\begin{array}{c}a=\frac{0.5055\text{N}}{\left(160\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right)}\\ =3.18\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration when downward pulling force increases by $25\%$ is $3.18\text{m}/{\text{s}}^2$.

To determine

The terminal velocity that the squeegee approaches.

Answer to Problem 6.34P

The terminal velocity that squeegee approaches is $0.205\text{m}/\text{s}$ in the downward direction.

Explanation of Solution

After reaching the terminal velocity there is no acceleration.

Write the new force equation taking into account the drag force

    $R+{F^{\prime }}_{\text{V}}+mg=0$

Substitute $-20.0v$ for $R$ and $1.25\left(F_{\text{V}}\right)$ in above expression.

    $20.0v=\left(1.25\left(F_{\text{V}}\right)+mg\right)$

Rearrange the above expression for $v$.

    $v=\frac{\left(\left(1.25\left(F_{\text{V}}\right)+mg\right)\right)}{20.0}$                                                                                   (V)

Substitute $160\text{g}$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$ and $2.03\text{N}$ for $F_{\text{V}}$ in equation (V).

    $\begin{array}{c}v=\frac{\left(1.25\left(2.03\text{N}\right)+\left(160\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right)9.8\text{m}/{\text{s}}^2\right)}{20.0}\\ =0.205\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the terminal velocity that squeegee approaches is $0.205\text{m}/\text{s}$ in the downward direction.