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 $R = 220.0 v$ , 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).

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Solution

(a) Since the window is vertical, the normal force is horizontal and is given by $n = 4.00 N$ . To find the vertical component of the force, we note that the force of kinetic friction is given by
$f_{k} = μ_{k} n = 0.900 (4.00 N) = 3.60 N$ upwards.
to oppose downward motion. Newton’s second law then becomes
$s u m F_{y} = m a_{y} : + 3.6 N - (0.16 k g) (9.8 m / s^{2}) + P_{y} = 0$
$P_{y} = - 2.03 N = 2.03 N$ down.
(b) Now, with the increased downward force, Newton’s second law gives
$\sum F_{y} = m a_{y}$
$+ 3.60 N - (0.160 k g) (9.80 m / s^{2}) - 1.25 (2.03 N) = 0.160 k g a_{y}$
then,
$a_{y} = - 0.508 N / 0.16 k g = - 3.18 m / s^{2} = 3.18 m / s^{2}$ down.
(c) At terminal velocity,
$\sum F_{y} = m a_{y} : + (20.0 N . s / m) v_{T} - (0.160 k g) (9.80 m / s^{2}) - 1.25 (2.03 N) = 0$
Solving for the terminal velocity gives
$v_{T} = 4.11 N / (20 N \cdot s / m) = 0.205 m / s$ down

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