A $50-\mathrm{N}$ box is slid straight across the floor at constant speed by a force of $25 \mathrm{~N}$, as depicted in Fig. $4-3(a)$. How large a friction force impedes the motion of the box? (b) How large is the normal force? $(c)$ Find $\mu_{k}$ between the box and the floor.
The forces acting on the box are shown in Fig. $4-3(a)$. The friction force is $F_{f}$, and the normal force, the supporting force exerted by the floor, is $F_{N}$. The free-body diagram and components are drawn in Fig. $4-3(b)$. Because the box is moving with constant velocity, it is in equilibrium. The first condition for equilibrium, taking to the right as positive,
$$
\pm \sum F_{x}=0 \quad \text { or } \quad 25 \cos 40^{\circ}-F_{f}=0
$$
(a) We can solve for the friction force $F_{f}$ at once to find that $F_{f}=19.2 \mathrm{~N}$, or to two significant figures, $F_{\mathrm{f}}=19 \mathrm{~N}$
(b) To find $F_{N}$, use the fact that
$$
+\uparrow \sum F_{y}=0 \quad \text { or } \quad F_{N}+25 \sin 40^{\circ}-50=0
$$
Solving gives the normal force as $F_{N}=33.9 \mathrm{~N}$ or, to two significant figures, $F_{N}=34 \mathrm{~N}$.
(c) From the definition of $\mu_{k}$,
$$
\mu_{k}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{19.2 \mathrm{~N}}{33.9 \mathrm{~N}}=0.57
$$