Exercises — Normal force — reaction force, not always = mg

Free-body diagram (arrows on the book):

• $\uparrow N$ (table pushes up)
• $\downarrow mg$ (Earth pulls down)

WHAT case: horizontal surface, no extra vertical force, no vertical acceleration. WHY: the only vertical forces are gravity $mg$ down and $N$ up, and the book is at rest so $a_y=0$. $\sum F_y=N-mg=0\Rightarrow N=mg=4\times 10=\boxed{40\text{N}}$ This is the one case where $N=mg$.

Free-body diagram (arrows on the person):

• $\uparrow N$ (floor pushes up)
• $\downarrow mg$ (Earth pulls down)

WHERE does $N=m(g+a)$ come from? Take up as positive and let $a$ be the person's upward acceleration (so downward acceleration makes $a$ negative). The only two vertical forces are $N$ up and $mg$ down, so Newton's 2nd law reads $\sum F_y=N-mg=ma.$ Add $mg$ to both sides: $N=mg+ma=m(g+a).$ That is the whole derivation — the elevator "formula" is just $\sum F=ma$ rearranged. Now read off each case:

• (a) at rest, $a=0$: $N=mg$ — equal.
• (b) accelerating up, $a>0$: $N=m(g+a)$ — greater (you feel heavy).
• (c) accelerating down, $a<0$: $N=m(g+a)<mg$ — less (you feel light).
• (d) free fall, $a=-g$: $N=m(g-g)=0$ — zero, weightless.

$N$ is always perpendicular to the surface, pointing away from it into the object. On a ramp that is not straight up — it tilts by the same $25^{\circ }$ as the ramp. Straight up is the direction of gravity, not $N$.

Free-body diagram (arrows on the crate):

• $\uparrow N$ (floor pushes up)
• $\downarrow mg=60\text{N}$ (Earth pulls down)
• $\downarrow F=25\text{N}$ (your hand presses down)

No vertical acceleration, so the up-arrow must cancel both down-arrows. $\sum F_y=N-mg-F=0\Rightarrow N=mg+F=60+25=\boxed{85\text{N}}$ The floor pushes back harder than $mg$ — it must cancel both gravity and your push.

Free-body diagram (arrows on the crate):

• $\uparrow N$ (floor pushes up)
• $\uparrow T=20\text{N}$ (rope pulls up)
• $\downarrow mg=60\text{N}$ (Earth pulls down)

Two up-arrows now share the job of balancing gravity. $\sum F_y=N+T-mg=0\Rightarrow N=mg-T=60-20=\boxed{40\text{N}}$ The rope shares the load, so the floor pushes less.

Free-body diagram (arrows on the person):

• $\uparrow N$ (floor pushes up)
• $\downarrow mg=500\text{N}$ (Earth pulls down)

Take up as positive. Acceleration is downward, so $a=-2{\text{m/s}}^2$. $\sum F_y=N-mg=ma\Rightarrow N=m(g+a)=50(10-2)=\boxed{400\text{N}}$ Less than $mg=500\text{N}$ — you feel lighter. See Apparent Weight & Elevators.

What the figure shows. The black square is the block on the ramp. The black downward arrow is gravity $mg$. The red arrow points straight out of the ramp surface — that is the normal force $N$, the key object of this figure. The short black dashed line drawn from the block into the ramp is the perpendicular component of gravity, labelled $mg\cos \theta$; $N$ is exactly what balances that dashed piece.

WHAT we do: split gravity into a part perpendicular to the ramp ($mg\cos \theta$) and a part along it ($mg\sin \theta$). The red arrow ($N$) has to balance only the perpendicular piece. WHY $\cos \theta$: the angle between the vertical gravity arrow and the ramp's inward normal is exactly $\theta$, so the adjacent (perpendicular) component is $mg\cos \theta$. Perpendicular to the ramp there is no acceleration (the block slides along, not into, the surface): $\sum F_{\perp }=N-mg\cos \theta =0\Rightarrow N=mg\cos \theta =8\times 10\times \cos 40^{\circ }=80\times 0.766=\boxed{61.28\text{N}}$

Free-body diagram (arrows on the box — see figure):

• $\uparrow N$ (floor pushes up, red arrow in figure)
• $\downarrow mg=100\text{N}$ (Earth pulls down)
• the push $P$ at $30^{\circ }$ below horizontal splits into $\to P\cos 30^{\circ }$ and $\downarrow P\sin 30^{\circ }$

WHAT we do: the push has a horizontal part $P\cos 30^{\circ }$ and a vertical (downward) part $P\sin 30^{\circ }$. Only the vertical part changes $N$. WHY: $N$ is set by the vertical balance; horizontal components go into horizontal motion, not into the surface push. Vertical downward component: $P\sin 30^{\circ }=50\times 0.5=25\text{N}$. $\sum F_y=N-mg-P\sin 30^{\circ }=0\Rightarrow N=mg+P\sin 30^{\circ }=100+25=\boxed{125\text{N}}$ Pushing into the floor at an angle increases $N$ — which also increases friction, since $f=\mu N$ (see Friction).

Free-body diagram (arrows on the box):

• $\uparrow N$ (floor pushes up)
• $\downarrow mg=100\text{N}$ (Earth pulls down)
• the pull $T$ at $30^{\circ }$ above horizontal splits into $\to T\cos 30^{\circ }$ and $\uparrow T\sin 30^{\circ }$

Now the vertical component of the rope is upward: $T\sin 30^{\circ }=25\text{N}$. $\sum F_y=N+T\sin 30^{\circ }-mg=0\Rightarrow N=mg-T\sin 30^{\circ }=100-25=\boxed{75\text{N}}$ Pulling up at an angle reduces $N$ — the box becomes "lighter" against the floor.

Free-body diagram at lift-off (arrows on the box):

• $N=0$ (floor no longer pushes)
• $\uparrow T\sin 30^{\circ }$ (rope's vertical pull)
• $\downarrow mg=100\text{N}$ (Earth pulls down)

Set $N=0$: $mg-T\sin 30^{\circ }=0\Rightarrow T=\frac{mg}{\sin 30^{\circ }}=\frac{100}{0.5}=\boxed{200\text{N}}$. For $T>200\text{N}$ the box lifts off and $N=0$ (a surface can only push, never pull — so $N$ cannot go negative).

Free-body diagram (arrows on the block):

• $\uparrow N$ (floor pushes up)
• $\downarrow mg=30\text{N}$ (Earth pulls down)
• $\to f$ (friction from floor, horizontal — this is what accelerates the block)

The acceleration is horizontal; there is still no vertical acceleration. So the vertical balance is unchanged: $\sum F_y=N-mg=0\Rightarrow N=mg=3\times 10=\boxed{30\text{N}}$ The horizontal push comes from friction, not from $N$. Lesson: only acceleration perpendicular to the surface changes $N$.

Free-body diagram (arrows on the block — see figure):

• $\downarrow mg$ (Earth pulls down)
• red arrow $N$ perpendicular to the ramp, which splits into $\to N\sin \theta$ (horizontal) and $\uparrow N\cos \theta$ (vertical)

WHAT we do: work in ground axes (horizontal $x$, vertical $y$). Only two forces act on the block: gravity $mg$ down and $N$ perpendicular to the incline surface. The block does not accelerate vertically (it moves horizontally with the wedge), but it does accelerate horizontally at $a$. Vertical (no vertical acceleration): $N\cos \theta -mg=0\Rightarrow N=\frac{mg}{\cos \theta }(1)$ Horizontal (accelerates at $a$): $N\sin \theta =ma(2)$ Check the given $a$ by dividing (2) by (1): $\frac{N\sin \theta }{N\cos \theta }=\frac{ma}{mg}\Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{a}{g}\Rightarrow \tan \theta =\frac{a}{g}\Rightarrow a=g\tan \theta .$ The $N$ and $m$ cancel cleanly, and we recover exactly the acceleration promised in the problem — so equation (1) is the right expression for $N$. Numerically: $N=\frac{mg}{\cos \theta }=\frac{2\times 10}{\cos 30^{\circ }}=\frac{20}{0.866}=\boxed{23.09\text{N}}$ Notice $N>mg=20\text{N}$ now — bigger than on a static incline (where $N=mg\cos \theta =17.3\text{N}$). The horizontal acceleration forces the surface to press harder.

Free-body diagram (arrows on the box):

• $\uparrow N$ (floor pushes up)
• $\downarrow mg=50\text{N}$ (Earth pulls down)
• push $P$ at $20^{\circ }$ below horizontal $\to$ $\to P\cos 20^{\circ }$ and $\downarrow P\sin 20^{\circ }$
• $\leftarrow f=\mu N$ (kinetic friction opposes motion)

Vertical: $N=mg+P\sin 20^{\circ }$ (the downward push adds to $N$). Horizontal, constant velocity ($a=0$): $P\cos 20^{\circ }=f=\mu N$. Substitute $N$: $P\cos 20^{\circ }=\mu (mg+P\sin 20^{\circ })$ $P\cos 20^{\circ }-\mu P\sin 20^{\circ }=\mu mg$ $P=\frac{\mu mg}{\cos 20^{\circ }-\mu \sin 20^{\circ }}=\frac{0.4\times 50}{0.9397-0.4\times 0.3420}=\frac{20}{0.8029}=\boxed{24.91\text{N}}$ Then $N=mg+P\sin 20^{\circ }=50+24.91\times 0.342=\boxed{58.52\text{N}}$. Pushing downward at an angle raises $N$, which raises friction — one reason it is harder to shove a box by pressing down-and-forward than by pulling up-and-forward.

Free-body diagram for $A$ (arrows on $A$):

• $\uparrow N_{AB}$ (block $B$ pushes up on $A$)
• $\downarrow m_{A}g$ (Earth pulls $A$ down)

Free-body diagram for $B$ (arrows on $B$):

• $\uparrow N_{Bfloor}$ (elevator floor pushes up on $B$)
• $\downarrow m_{B}g$ (Earth pulls $B$ down)
• $\downarrow N_{AB}$ (block $A$ presses down on $B$ — Newton's 3rd-law partner)

(a) Block $A$ alone. Accelerates up at $a$: $N_{AB}-m_{A}g=m_{A}a\Rightarrow N_{AB}=m_A(g+a)=2(10+5)=\boxed{30\text{N}}$ (b) Block $B$ alone. With $A$ pressing down on $B$ with $N_{AB}=30\text{N}$ (see Newton's Third Law): $N_{Bfloor}-m_{B}g-N_{AB}=m_{B}a$ $N_{Bfloor}=m_B(g+a)+N_{AB}=3(15)+30=45+30=\boxed{75\text{N}}$ Cross-check: treat $A+B$ as one $5\text{kg}$ system — floor push $=(m_A+m_B)(g+a)=5\times 15=75\text{N}$. ✔ Matches.

Free-body diagram for $A$ (arrows on $A$):

• $\uparrow N_{AB}$ (block $B$ pushes up on $A$)
• $\downarrow m_{A}g$ (Earth pulls $A$ down)

Setting the sign convention. Take up as positive. The block sits inside the elevator and moves with it, so the block's acceleration is the same as the elevator's. The elevator accelerates downward with magnitude $a$, so its acceleration points in the negative direction: the block's vertical acceleration is $a_y=-a$. This is why a downward $a$ enters the equation as a minus sign.

Newton's 2nd law for $A$: $N_{AB}-m_{A}g=m_A{a}_y=m_A(-a)\Rightarrow N_{AB}=m_A(g-a)$ Set $N_{AB}=0$: $g-a=0\Rightarrow a=g=\boxed{10{\text{m/s}}^2}$. This is exactly free fall — at that instant nothing pushes on $A$ from below, and it (and $B$) are weightless. For any $a>g$ the blocks would separate and float apart inside the cabin.