Exercises — Free body diagrams — systematic drawing technique

Scan the boundary of the book.

• Long-range: gravity ==$mg$, straight down== (always present near Earth).
• Contact: the book touches only the table → ==normal force $N$, straight up== (⊥ to the horizontal surface).

Two arrows only. No friction (nothing is pushing it sideways, no slip tendency), no "applied" force. Since the book is at rest, $N=mg$.

• Gravity $mg$ down.
• Contact: only the string touches it → ==tension $T$ up==, pointing away from the ball along the string (strings only pull).

No normal force (nothing pushes on it), no friction. Because it hangs still, $T=mg$.

Forces on the box (vertical only): gravity $mg$ down, tension $T$ up, normal $N$ up. Box stays on floor ⇒ $a_y=0$, so $\sum F_y=0$: $N+T-mg=0\Rightarrow N=mg-T.$ $N=(4.0)(9.8)-25=39.2-25=14.2\text{N}.$ The upward pull shares the job of holding the box, so the floor pushes less than the full weight. Sanity check: if $T$ reached $39.2\text{N}$, then $N=0$ and the box is about to leave the floor. ✓

Horizontal axis: the only horizontal force is $F$. By Newton's Second Law, $F=ma\Rightarrow a=\frac{F}{m}=\frac{6.0}{2.0}=3.0{\text{m/s}}^2,$ directed along $F$. Vertically $N=mg=19.6\text{N}$, but it plays no role since $a_y=0$.

Tilt the axes: $x$ down-slope, $y$ ⊥ to slope (see Inclined Plane Problems). Gravity points straight down, but neither of our tilted axes is vertical, so gravity has a piece along each axis — we must resolve it. Why $\sin$ and $\cos$ appear (look at the figure): drop gravity's arrow onto the two axes and it forms a right triangle whose acute angle equals the slope angle $\theta$ (the incline angle re-appears between the vertical and the perpendicular-to-slope direction). In that triangle:

• the side along the slope is opposite $\theta$, and opposite/hypotenuse $=\sin \theta$, giving $mg\sin \theta$;
• the side into the surface is adjacent to $\theta$, and adjacent/hypotenuse $=\cos \theta$, giving $mg\cos \theta$.

Sanity check on which is which: as $\theta \to 90^{\circ }$ (a vertical wall) the slope becomes vertical, so all of gravity should act along it — and indeed $\sin 90^{\circ }=1$. ✓

(a) No motion off the surface ⇒ $a_y=0$: $N-mg\cos \theta =0\Rightarrow N=mg\cos \theta =(5.0)(9.8)\cos 30^{\circ }=42.44\text{N}.$ (b) Along slope: $mg\sin \theta =ma\Rightarrow a=g\sin \theta =9.8\sin 30^{\circ }=4.9{\text{m/s}}^2.$ Mass cancels — heavier blocks slide with the same $a$.

Constant velocity ⇒ $a=0$ ⇒ forces balance on both axes. Here $f_k$ is the kinetic friction force defined at the top of the page, tied to the surface by $f_k={\mu }_{k}N$.

• $y:N=mg\cos \theta =42.44\text{N}$ (same perpendicular balance as Ex 3.1).
• $x:$ the block slides down, so kinetic friction acts up-slope, opposing slip: $mg\sin \theta -f_k=0\Rightarrow f_k=mg\sin \theta =(5.0)(9.8)\sin 30^{\circ }=24.5\text{N}.$ Then, using $f_k={\mu }_{k}N$ (so ${\mu }_k=f_k/N$): ${\mu }_k=\frac{f_k}{N}=\frac{24.5}{42.44}=\mathrm{0.577.}$ Note ${\mu }_k=\frac{mg\sin \theta }{mg\cos \theta }=\tan \theta$ exactly at constant velocity — a classic result.

Two FBDs — one per body (Tension in Strings and Pulleys). Same $T$ (ideal string), same $|a|$ (inextensible).

• $A$ (horizontal): $T=m_{1}a$.
• $B$ (vertical, taking down as positive): $m_{2}g-T=m_{2}a$.

Add to cancel $T$: $m_{2}g=(m_1+m_2)a\Rightarrow a=\frac{m_2}{m_1+m_2}g=\frac{2.0}{5.0}(9.8)=3.92{\text{m/s}}^2.$ Then $T=m_{1}a=(3.0)(3.92)=11.76\text{N}$. Check: $T<m_{2}g=19.6\text{N}$ — the string must pull less than $B$'s weight, else $B$ couldn't fall. ✓

Now $A$ feels kinetic friction opposing its motion (it moves toward the pulley, so friction acts backward):

• $A$ vertical: $N=m_{1}g=(3.0)(9.8)=29.4\text{N}$, so $f_k={\mu }_{k}N=0.20(29.4)=5.88\text{N}$.
• $A$ horizontal: $T-f_k=m_{1}a$.
• $B$: $m_{2}g-T=m_{2}a$.

Add the two driving equations to cancel $T$: $m_{2}g-f_k=(m_1+m_2)a\Rightarrow a=\frac{m_{2}g-{\mu }_k{m}_{1}g}{m_1+m_2}.$ $a=\frac{19.6-5.88}{5.0}=\frac{13.72}{5.0}=2.744{\text{m/s}}^2.$ Friction stole part of the drive, so $a$ dropped from $3.92$ to $2.744{\text{m/s}}^2$. ✓

Sign convention: choose up-slope as positive $x$ (the direction the push drives the blocks); down-slope is negative. Gravity's along-slope piece $mg\sin \theta$ therefore enters with a minus sign, exactly as pictured in Ex 3.1.

Treat both blocks as one system to find $a$ (the $C$–$D$ contact force is internal, so it drops out): $F-(m_C+m_D)g\sin \theta =(m_C+m_D)a.$ $a=\frac{F}{m_C+m_D}-g\sin \theta =\frac{60}{6.0}-9.8\sin 30^{\circ }=10-4.9=5.1{\text{m/s}}^2(\text{up-slope, positive}).$ (b) Now isolate $C$ (the lower block). By Newton's Third Law, the upper block $D$ pushes $C$ up the slope (it is shoving $C$ ahead of it in the direction of motion), so the contact force $P$ points in the positive $x$ direction on $C$ — that is why $P$ is taken positive up-slope. Gravity still pulls $C$ down-slope ($-m_{C}g\sin \theta$). Newton's 2nd law on $C$ along $x$: $P-m_{C}g\sin \theta =m_{C}a\Rightarrow P=m_C(a+g\sin \theta )=4.0(5.1+4.9)=40.0\text{N}.$ Check by isolating $D$: $F-P-m_{D}g\sin \theta =m_{D}a\Rightarrow 60-40-2.0(4.9)=20-9.8=10.2=m_{D}a=2.0(5.1)$. ✓

"Verge, frictionless" ⇒ just balance along the slope, $a=0$ (up-slope positive): $F-mg\sin \theta =0\Rightarrow F=mg\sin \theta =(1.0)(9.8)\sin 30^{\circ }=4.9\text{N}.$ Perpendicular axis is untouched by $F$ (it's along the slope), so as in Ex 3.1: $N=mg\cos \theta =(1.0)(9.8)\cos 30^{\circ }=8.487\text{N}.$ Any $F<4.9\text{N}$ and the block accelerates down; any $F>4.9\text{N}$ and it accelerates up. This threshold is exactly the down-slope gravity component. ✓