Chapter 6, Problem 19P

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 component that washer must exert when she pulls the squeegee down the window at constant velocity.

Answer to Problem 19P

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

Given information:

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 $4.00\text{N}$.

Draw the diagram for the given condition.

Figure I

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

$F_{\text{H}}=F\sin \theta$

• $F$ is the force exerted by washer against the window.
• $F_{\text{H}}$ is the horizontal component of the force exerted by washer against the window.

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

$F_{\text{V}}=F\cos \theta$

• $F_{\text{V}}$ is the vertical component of the force exerted by washer against the window.

The net vertical force for the given system is given as,

$\text{Vertical}\text{force}=F_{\text{V}}+mg-F_{\text{s}}$ (I)

• $m$ is the mass of the squeegee.
• $g$ is the acceleration due to gravity.
• $F_{\text{s}}$ is the frictional force.

The expression for the frictional force is given as,

$F_{\text{s}}=\mu F\sin \theta$

• $\mu$ is the coefficient of kinetic friction.

At constant velocity the net vertical force is zero.

Substitute $\mu F\sin \theta$ for $F_{\text{s}}$ in equation (I).

$0=F_{\text{V}}+mg-\mu F\sin \theta$

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 $F\sin \theta$ 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 of the squeegee if the washer increases the downward force component by $25\%$.

Answer to Problem 19P

The acceleration of the squeegee when the washer increases the downward force component by $25\%$ is $3.18\text{m}/{\text{s}}^2$.

Explanation of Solution

Section 1:

To determine: The net vertical force when the downward force component is increased by $25\%$.

Answer: The net vertical force when the downward force component is increased by $25\%$ is $\text{0}\text{.5055}\text{N}$.

Given information:

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 $4.00\text{N}$.

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

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

The net vertical force is given as,

$F_{{\text{V}}_{\text{n}}}={F^{\prime }}_{\text{V}}+mg-\mu F_{\text{H}}$

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

$F_{{\text{V}}_{\text{n}}}=1.25\left(F_{\text{V}}\right)+mg-\mu \left(F_{\text{H}}\right)$ (III)

Substitute $4.00\text{N}$ for $F_{\text{H}}$, $160\text{g}$ 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).

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

Section 2:

To determine: The acceleration of the squeegee.

Answer: The acceleration of the squeegee is $3.18\text{m}/{\text{s}}^2$.

Given information:

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 $4.00\text{N}$.

Formula to calculate the force is,

$F_{{\text{V}}_{\text{n}}}=ma$

• $a$ is the acceleration.

Rearrange the above expression for $a$.

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

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

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

Conclusion:

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

To determine

The terminal velocity that the squeegee approaches.

Answer to Problem 19P

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

Explanation of Solution

Given information:

The mass of the squeegee is $160\text{g}$, the coefficient of kinetic friction between the squeegee and dry glass is $0.900$, the horizontal component of the force against the window is $4.00\text{N}$ and the fluid drag force is $-20.0v$.

According to the given condition,

$R=-\left({F^{\prime }}_{\text{V}}+mg\right)$

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 $0.205\text{m}/\text{s}$.