Corrected exercises: friction, circular motion and gravitation (PHYS 101)

The normal force first, because both coefficients multiply it. The floor is level and the push is horizontal, so nothing has a vertical component except $N$ and the weight: $N = mg = 25 \times 9.8 = 245$ N. Write this line once, use it three times. The ceiling on static friction is $f_{s,\max} = \mu_{s} N = 0.45 \times 245 = 110.25$ N and the sliding value is $f_{k} = \mu_{k} N = 0.32 \times 245 = 78.4$ N. Both numbers exist before you know what the crate does.

a) Compare $F = 60$ N with the ceiling $110.25$ N. The push is below it, so the crate does not move, and if it does not move its acceleration is zero. Newton's first law along the floor then says the horizontal forces cancel: $f = F = 60$ N. Notice what was NOT done here: $\mu_{s} N$ was not used as the answer. Static friction is not a formula, it is whatever number balances the push, and $\mu_{s} N$ is only the largest number it is allowed to take.

b) Same reasoning, $F = 110$ N, and $110 < 110.25$: still stuck, so $f = 110$ N and $a = 0$. The crate is $0.25$ N away from moving and the friction force is $110$ N, not $110.25$ N. Writing $f = 110.25$ N here is the single most common error of the chapter and it costs the whole question, because it puts a net force of $-0.25$ N on a crate that is standing still.

c) Now $130 > 110.25$: the crate breaks loose and slides. The moment it slides, friction stops adjusting and locks at $f_{k} = 78.4$ N. Newton's second law along the direction of motion: $a = \frac{F - f_{k}}{m} = \frac{130 - 78.4}{25} = \frac{51.6}{25} = 2.06$ m/s$^2$. Sanity check: the answer must be positive and small, a few m/s$^2$ for a shove on a heavy crate, and it is.

d) The graph is a straight line of slope $1$ from the origin to the point $(110.25;\ 110.25)$, then a vertical drop to $78.4$ N, then a horizontal line. The drop is the physical content of the chapter: at the instant the crate starts to slide, friction falls from $110.25$ N to $78.4$ N, so the net force jumps from $0$ to $110.25 - 78.4 = 31.85$ N and the crate leaps forward at $a = 31.85/25 = 1.27$ m/s$^2$ with no extra push. That is why a heavy box you are straining against suddenly shoots away.

The trap, and what it costs. Two symbols, $\mu_{s} N$ and $f_{s}$, are not the same object: the first is a ceiling, the second is the actual force, and $f_{s} \le \mu_{s} N$ is an INEQUALITY. Every friction question begins with the comparison test, $F$ against $\mu_{s} N$, and only then chooses between $f = F$ and $f = \mu_{k} N$. Skipping the test is worth 3 to 4 points on a midterm because every later number inherits the wrong friction.

Axes along and across the plank, as in chapter 4, and only the weight needs splitting. Across the plank: $N = mg\cos\theta$. Along the plank, taking down-slope as positive: $mg\sin\theta - f$. Nothing here is new except that $f$ is no longer zero.

a) At $\theta = 25^{\circ}$ the block is at rest, so the along-plank forces cancel: $f = mg\sin 25^{\circ} = 4.0 \times 9.8 \times 0.4226 = 16.6$ N, directed UP the plank. Then check that this is allowed: the ceiling is $\mu_{s} mg\cos 25^{\circ} = 0.60 \times 4.0 \times 9.8 \times 0.9063 = 21.3$ N, and $16.6 < 21.3$, so the block can indeed hold. Answering $21.3$ N here is the same error as in exercise 1 in a new costume, and it is worth 2 points.

b) The block is on the point of sliding when the demand meets the ceiling: $mg\sin\theta_{s} = \mu_{s}\, mg\cos\theta_{s}$, hence $\tan\theta_{s} = \mu_{s} = 0.60$ and $\theta_{s} = \arctan(0.60) = 31.0^{\circ}$. Check it against part a: $25^{\circ} < 31.0^{\circ}$, so the block was indeed still holding, and the two answers agree.

c) Both sides of $mg\sin\theta_{s} = \mu_{s} mg\cos\theta_{s}$ carry the factor $mg$, which cancels. The tipping angle depends on the two surfaces and on nothing else: a $4$ kg block and a $400$ kg block let go at the same $31.0^{\circ}$. This is the standard laboratory measurement of $\mu_{s}$, and it is why you never need a balance to perform it.

d) At $\theta = 35^{\circ} > \theta_{s}$ the block slides, so friction locks at $f_{k} = \mu_{k} mg\cos\theta$ and points UP the plank, against the motion. Newton's second law along the plank: $ma = mg\sin\theta - \mu_{k} mg\cos\theta$, so $a = g(\sin\theta - \mu_{k}\cos\theta) = 9.8 \times (0.5736 - 0.45 \times 0.8192) = 9.8 \times 0.2050 = 2.01$ m/s$^2$ down the plank. The mass has cancelled again, which is the check to run before reaching for the calculator.

What the numbers must obey. The answer to d has to sit between $0$ and $g\sin 35^{\circ} = 5.62$ m/s$^2$, the frictionless value of chapter 4: friction can only slow a slide, never reverse it or speed it up. Getting $a$ larger than $5.62$ m/s$^2$ means a sign error on $f$; getting $a$ negative at an angle above $\theta_{s}$ means $\mu_{k}$ was used where $\mu_{s}$ belonged. And the block never accelerates UP the plank: at rest above $\theta_{s}$ it starts down, at rest below it stays put.

a) $45$ revolutions per minute is $45/60 = 0.75$ revolution per second, so one revolution takes $T = 1/0.75 = 1.33$ s. In one period the coin travels one circumference: $v = \frac{2\pi r}{T} = \frac{2\pi \times 0.12}{1.3333} = 0.565$ m/s. Slow, which is what you expect from a record player.

b) Two routes, and they must agree. $a_{c} = \frac{v^{2}}{r} = \frac{0.565^{2}}{0.12} = \frac{0.3198}{0.12} = 2.66$ m/s$^2$. Or, skipping the speed entirely, $a_{c} = \frac{4\pi^{2} r}{T^{2}} = \frac{4\pi^{2} \times 0.12}{1.3333^{2}} = \frac{4.738}{1.778} = 2.66$ m/s$^2$. The second form is the one to use whenever the data are a period or a rotation rate, because it never asks you to compute $v$ first. Both say the same thing: the speed of the coin never changes, and it is accelerating all the same, because the DIRECTION of its velocity turns through a full circle every $1.33$ s.

c) There is no string, no hand, nothing pushing the coin sideways except the surface it rests on: the centripetal force is supplied by STATIC friction, pointing from the coin toward the axis. Note that static friction is what holds the coin, not kinetic: the coin does not slide on the turntable, it rides with it. Newton's second law along the radius, positive toward the centre: $f_{s} = m a_{c}$, and the ceiling is $f_{s} \le \mu_{s} N = \mu_{s} mg$. So $m a_{c} \le \mu_{s} mg$, the mass cancels, and $\mu_{s} \ge \frac{a_{c}}{g} = \frac{2.6648}{9.8} = 0.272$. A copper coin on felt clears this easily, which is why the coin stays.

d) The trap is to read $a_{c} = v^{2}/r$ and conclude that a bigger radius means a smaller acceleration. It would, at fixed SPEED. Here the turntable rate is fixed, so $T$ is fixed and $v$ doubles along with $r$: use $a_{c} = 4\pi^{2}r/T^{2}$, which is proportional to $r$. The acceleration doubles to $5.33$ m/s$^2$ and the required coefficient doubles to $0.544$. Check by the other route: $v = 2\pi(0.24)/1.3333 = 1.131$ m/s and $v^{2}/r = 1.279/0.24 = 5.33$ m/s$^2$. Same number.

The lesson worth carrying. The two formulas for $a_{c}$ are the same formula, but they hold different things fixed in your head, and choosing the wrong one produces an answer that is wrong by a factor of four. Ask what the problem keeps constant, the speed or the rotation rate, before choosing. On the turntable, the outer coins fly off first, exactly the opposite of what $v^{2}/r$ suggests to a careless reader.

a) An object of mass $m$ on the surface feels $F = G\frac{Mm}{R^{2}}$, and by definition that same force is $mg_{M}$. The $m$ cancels, which is the whole reason a surface gravity exists at all: $g_{M} = \frac{GM}{R^{2}} = \frac{6.67\times 10^{-11} \times 6.42\times 10^{23}}{(3.39\times 10^{6})^{2}} = \frac{4.282\times 10^{13}}{1.149\times 10^{13}} = 3.73$ m/s$^2$. Order of magnitude check: Mars is about a tenth of Earth's mass and about half its radius, so $g$ should be roughly $0.1/0.25 = 0.4$ of Earth's, and $3.73/9.8 = 0.38$.

b) Mass is the amount of matter and it travels unchanged: $70$ kg on Mars, $70$ kg on Earth, $70$ kg in orbit. Weight is a FORCE, $w = mg$, and it changes with the planet: $w_{M} = 70 \times 3.726 = 261$ N, against $w_{E} = 70 \times 9.8 = 686$ N. Writing the Martian weight in kilograms is worth a point on any exam, and writing that her mass is smaller on Mars is worth the whole part.

c) $g(r) = \frac{GM}{r^{2}}$, and we want $g(r) = \frac{1}{2}g_{M}$, so $\frac{GM}{r^{2}} = \frac{1}{2}\frac{GM}{R^{2}}$. Everything cancels except the radii: $r^{2} = 2R^{2}$, so $r = R\sqrt{2} = 4.79\times 10^{6}$ m. That is the distance from the CENTRE. The altitude is $h = r - R = 4.794\times 10^{6} - 3.39\times 10^{6} = 1.40\times 10^{6}$ m, about $1400$ km. Forgetting the last subtraction turns an altitude into a radius and is the most frequent loss of the question.

d) At an altitude of one radius, the distance from the centre is $r = 2R$, and an inverse SQUARE divides by $2^{2} = 4$, not by $2$: $g(2R) = \frac{3.726}{4} = 0.932$ m/s$^2$. To halve $g$ you go out by a factor $\sqrt{2} \approx 1.41$, which is part c. The mental rule worth keeping: distance times $2$ means force divided by $4$, distance times $3$ means force divided by $9$.

Where the $r$ is measured, and why it matters. In $F = GMm/r^{2}$, $r$ is the separation of the CENTRES, never the height above the ground. For anything on or near a planet that distinction is the whole problem: at $400$ km above Earth, $r$ is $6770$ km and not $400$ km, so $g$ is still about $89$ per cent of its surface value. A solution that feeds an altitude into $r$ produces gravitational fields thousands of times too strong and should be caught by the order-of-magnitude check before it is handed in.

a) The satellite travels a circle at constant speed, so its acceleration is $v^{2}/r$ toward the centre. The only force on it is gravity, also toward the centre. Newton's second law along the radius therefore reads $G\frac{Mm}{r^{2}} = m\frac{v^{2}}{r}$. The mass of the SATELLITE cancels, which is the headline result: the orbit does not care whether you send up a bolt or a space station. Simplify one power of $r$: $v^{2} = \frac{GM}{r}$, so $v = \sqrt{GM/r}$. For the period, one lap is one circumference at that speed, $T = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{r}{GM}}$, and squaring gives $T^{2} = \frac{4\pi^{2}}{GM}r^{3}$, which is Kepler's third law with the constant made explicit.

b) The radius is measured from the CENTRE of the Earth: $r = 6.37\times 10^{6} + 4.00\times 10^{5} = 6.77\times 10^{6}$ m. Then $v = \sqrt{\frac{3.982\times 10^{14}}{6.77\times 10^{6}}} = \sqrt{5.882\times 10^{7}} = 7.67\times 10^{3}$ m/s, about $7.67$ km/s or $27\,600$ km/h. Period: $T = \frac{2\pi \times 6.77\times 10^{6}}{7669} = 5546$ s $= 92.4$ min. The astronauts see a sunrise roughly every hour and a half, which is the check everybody can remember.

c) Here the period is the datum, so invert the law: $r^{3} = \frac{GM\,T^{2}}{4\pi^{2}}$ with $T = 24 \times 3600 = 86\,400$ s. Then $r^{3} = \frac{3.982\times 10^{14} \times 7.465\times 10^{9}}{39.478} = 7.530\times 10^{22}$ m$^3$, and the cube root is $r = 4.22\times 10^{7}$ m. Altitude: $4.22\times 10^{7} - 6.37\times 10^{6} = 3.59\times 10^{7}$ m, about $35\,900$ km, the number quoted for every television satellite. Taking the square root instead of the cube root is the classic slip, and it gives $2.7\times 10^{11}$ m, well past the Moon: the order of magnitude catches it instantly.

d) Kepler's third law says $T^{2}/r^{3}$ is the same for every satellite of the same planet, so $\frac{T_{2}}{T_{1}} = \left(\frac{r_{2}}{r_{1}}\right)^{3/2}$. Here $\frac{r_{2}}{r_{1}} = \frac{4.222\times 10^{7}}{6.77\times 10^{6}} = 6.236$ and $6.236^{3/2} = 15.58$. So $T_{2} = 92.4 \times 15.58 = 1440$ min $= 24$ h. The two independent calculations agree, which is what the ratio form is for: it needs no constants and no calculator memory, only the two radii.

Two things to keep. Orbital speed goes DOWN as you climb, $v \propto 1/\sqrt{r}$, which surprises everyone: the geostationary satellite moves at $3.07$ km/s, less than half the ISS. And the mass of the satellite is absent from every result on this page, so any answer of yours in which it survives contains an algebra error, not a physical effect.

Set the axes horizontal and vertical, not along the slope. This is the single decision the question turns on, and it is the opposite of exercise 2. On an incline a block accelerates ALONG the surface, so axes follow the surface. On a banked curve the car has no vertical acceleration at all and its acceleration is horizontal, toward the centre. Use horizontal and vertical axes and split the NORMAL force instead of the weight.

a) Two real forces on a frictionless banked car, $N$ perpendicular to the road and $mg$ down. Vertical, no acceleration: $N\cos\theta = mg$. Horizontal, toward the centre: $N\sin\theta = \frac{mv^{2}}{r}$. Divide the second by the first and both $N$ and $m$ vanish: $\tan\theta = \frac{v^{2}}{rg}$, so $v = \sqrt{rg\tan\theta} = \sqrt{85 \times 9.8 \times 0.2126} = \sqrt{177.1} = 13.3$ m/s, that is $47.9$ km/h. A gentle $12^{\circ}$ bank is designed for a slow ramp, which is what an exit is.

b) From the vertical equation, $N = \frac{mg}{\cos\theta} = \frac{1200 \times 9.8}{\cos 12^{\circ}} = \frac{11\,760}{0.9781} = 1.20\times 10^{4}$ N. That is LARGER than the weight $11\,760$ N, by about $2$ per cent, and it must be: the road has to hold the car up and push it round the bend at the same time. Writing $N = mg\cos\theta$ here, the formula from the incline, gives $11\,500$ N, smaller than the weight, and the sign of the error is visible without any calculation. That confusion is the most expensive single line of the chapter.

c) The ball moves in a HORIZONTAL circle whose radius is not the length of the string: $r = L\sin 25^{\circ} = 1.20 \times 0.4226 = 0.507$ m. Two real forces, tension $X$ along the string and weight $mg$ down. Vertical: $X\cos 25^{\circ} = mg$. Horizontal: $X\sin 25^{\circ} = mv^{2}/r$. Dividing again, $\tan 25^{\circ} = \frac{v^{2}}{rg}$, so $v = \sqrt{0.5071 \times 9.8 \times 0.4663} = \sqrt{2.318} = 1.52$ m/s. One revolution: $T = \frac{2\pi r}{v} = \frac{3.186}{1.522} = 2.09$ s. As a check, the tension is $X = \frac{mg}{\cos 25^{\circ}} = \frac{4.9}{0.9063} = 5.41$ N, again larger than the weight $4.9$ N.

d) Both problems are the same two lines: $X\cos\theta = mg$ vertically and $X\sin\theta = \frac{mv^{2}}{r}$ horizontally, where $X$ is the one tilted force, the tension for the pendulum and the normal force for the car. Dividing kills $X$ and $m$ together and leaves $\tan\theta = v^{2}/(rg)$ in both cases. In neither problem is there a force pushing outward. What an outward arrow would represent is the seat pressing on your back, an interaction between you and the car, and it has no place in the diagram of the CAR.

How to read $\tan\theta = v^{2}/(rg)$. It contains no mass, so a loaded truck and a motorcycle have the same design speed on the same ramp. It rises with $v^{2}$, so doubling the design speed demands $\tan\theta$ four times larger. And it is a DESIGN speed, not a limit: below it the car tends to slide down the bank and friction must point up the slope, above it friction must point down. Chapter 6 will not change any of this, because none of it is about energy.

a) At the top of the circle the centre is BELOW the bucket, so the inward direction is downward. Two real forces act: the rope pulls the bucket toward $O$, that is downward, and the weight also acts downward. Both arrows point the same way, which is the feature of this position and the reason it is examined. Taking downward as positive because that is toward the centre: $T + mg = \frac{mv^{2}}{L}$. There is no third arrow. Nothing holds the bucket up at the top, and the water does not need holding up.

b) The rope can pull but never push, so $T \ge 0$. The limiting case is $T = 0$, where gravity alone bends the path: $mg = \frac{mv_{\min}^{2}}{L}$, the mass cancels, and $v_{\min} = \sqrt{gL} = \sqrt{9.8 \times 1.10} = \sqrt{10.78} = 3.28$ m/s. Below that speed the circle of radius $1.10$ m would demand less centripetal force than gravity already supplies, so the bucket falls inside the circle and the rope goes slack. Note what $v_{\min}$ does not contain: the mass. A bucket of water and an empty bucket let go at the same speed.

c) $T = m\left(\frac{v^{2}}{L} - g\right) = 3.0 \times \left(\frac{4.5^{2}}{1.10} - 9.8\right) = 3.0 \times (18.41 - 9.80) = 25.8$ N. The bracket is positive because $4.5 > 3.28$, as it must be. A negative tension is not a physical answer, it is the signal that the speed given is below $v_{\min}$ and that the object has already left the circular path.

d) At the bottom the centre is ABOVE, so inward is upward, and the two forces now oppose each other: $T - mg = \frac{mv^{2}}{L}$, giving $T = m\left(\frac{v^{2}}{L} + g\right) = 3.0 \times \left(\frac{36}{1.10} + 9.8\right) = 3.0 \times 42.53 = 128$ N. Compare with the weight, $3.0 \times 9.8 = 29.4$ N: the rope carries about $4.3$ times the weight at the bottom and only $25.8$ N at the top. The rope breaks at the bottom of the swing, never at the top, and that is what the two signs are telling you.

Why the water does not fall out, said properly. Not because something throws it outward. At the top the bucket's bottom is above the water and can only PUSH the water downward, toward the centre. That push plus the water's own weight is exactly the centripetal force the circle demands. At $v_{\min}$ the bucket pushes with zero force and gravity does the whole job; below $v_{\min}$ gravity is more than the circle needs, so the water falls inside the circle, which from the ground looks like it landing on your head. An answer built on a centrifugal force gets the right feeling and the wrong physics, and examiners mark the reasoning.

a) TRUE. The model $f_{k} = \mu_{k} N$ says friction depends on how hard the surfaces are pressed together and on the pair of materials, and on nothing else. Doubling the contact area halves the pressure at every point and leaves the total force unchanged, which is why a brick slides the same way on its face and on its edge. It is an approximation, and it fails at very high speeds and for tyres near their limit, but on this course it is exact. Be careful not to over-correct a true statement: rewriting a true line costs marks as surely as missing a false one.

b) FALSE. $\mu_{s}N$ is a CEILING, not a value: $f_{s} \le \mu_{s}N$. The actual static friction takes whatever value is needed to keep the block still, and it equals $\mu_{s}N$ in exactly one situation, on the point of slipping. Correct version: the force of static friction on a block at rest is whatever is needed to balance the other forces along the surface, and it cannot exceed $\mu_{s}N$. The number from exercise 1: with $N = 245$ N and $\mu_{s} = 0.45$ the ceiling is $110.25$ N, but under a push of $60$ N the friction force is $60$ N.

c) FALSE. The forces on an object are the ones other objects exert on it: gravity, a normal force, tension, friction. Circular motion adds none. The phrase centripetal force names the RESULTANT of those real forces, taken along the direction of the centre, and its value is $mv^{2}/r$. Correct version: for an object in uniform circular motion, the resultant of the real forces points at the centre and has magnitude $mv^{2}/r$. Adding an extra arrow labelled $F_{c}$ counts the same force twice and makes the resultant zero, so the object should travel in a straight line, which it plainly does not.

d) FALSE, and doubly so. The figure shows $g$ against altitude: at $400$ km, $g = \frac{GM}{r^{2}} = \frac{3.982\times 10^{14}}{(6.77\times 10^{6})^{2}} = 8.69$ m/s$^2$, that is $89$ per cent of the $9.81$ m/s$^2$ at the ground. Gravity is what holds the station in orbit at all, as exercise 5 shows. Correct version: astronauts float because the station and everything in it are in free fall together, accelerating toward the Earth at $8.69$ m/s$^2$, so nothing presses on the floor. What is zero is the apparent weight, that is the normal force, never $g$.

e) FALSE. On a flat curve the friction force supplies the whole centripetal force: $f_{s} = \frac{mv^{2}}{r}$ with $f_{s} \le \mu_{s}mg$. The mass cancels and leaves $\mu_{s} \ge \frac{v^{2}}{rg}$. For $r = 85$ m at $13.3$ m/s that is $\mu_{s} \ge 0.213$, whatever the vehicle weighs. Correct version: the required coefficient of static friction depends only on the speed and the radius, so both vehicles need the same value. The truck does need a larger friction FORCE, since $f_{s} = mv^{2}/r$ grows with its mass, but its normal force grows in the same proportion, and the two effects cancel exactly.

a) $12\,000$ revolutions per minute is $200$ revolutions per second, so $T = \frac{1}{200} = 5.0\times 10^{-3}$ s. With a period in hand, use the period form and never compute the speed first: $a_{c} = \frac{4\pi^{2}r}{T^{2}} = \frac{4\pi^{2} \times 0.085}{(5.0\times 10^{-3})^{2}} = \frac{3.356}{2.5\times 10^{-5}} = 1.34\times 10^{5}$ m/s$^2$. Watch the two unit conversions that decide the answer: $8.5$ cm is $0.085$ m, and rpm is not a frequency in hertz until it is divided by $60$. A factor of $60$ in a squared quantity is a factor of $3600$ in the result.

b) Relative centrifugal field: $\frac{a_{c}}{g} = \frac{134\,227}{9.8} = 1.37\times 10^{4}$, quoted as about $13\,700$ times $g$, which is in the ordinary range of a benchtop machine. The speed at the far end is $v = \frac{2\pi r}{T} = \frac{2\pi \times 0.085}{5.0\times 10^{-3}} = 107$ m/s, roughly $385$ km/h, and this is why a rotor that comes loose destroys the instrument.

c) At fixed radius, $a_{c} = 4\pi^{2}rf^{2}$ with $f$ the rotation rate, so the field goes as the SQUARE of the rate. Doubling the field needs the rate multiplied by $\sqrt{2} = 1.414$: $12\,000 \times 1.414 = 16\,970$ rpm, call it $17\,000$. Doubling the rpm, the instinctive answer, would multiply the field by four and can exceed the rated maximum of the rotor, which is a safety matter and not only an arithmetic one. The same square law is why a protocol is always quoted in times $g$ and not in rpm: the rpm figure means nothing without the radius of the rotor.

d) Everything inside the tube is being carried around a circle, so everything needs an inward resultant of $m a_{c}$. For the liquid this comes from the pressure of the liquid behind it, higher near the wall, lower near the axis, exactly as the water in a bucket. A cell denser than the plasma has MORE mass in the same volume, so it needs more inward force than the plasma pressure around it supplies. The shortfall means its inward acceleration is smaller than the circle requires, so it cannot keep to its circle and drifts outward, toward the bottom of the tube, until it reaches the wall. Nothing threw it outward: it simply was not pulled inward hard enough.

The same reversal as the whole chapter. Sedimentation is usually explained by a centrifugal force pushing the heavy particles out, which predicts the right destination for the wrong reason and fails as soon as you ask what object exerts that force. Written with real forces only, the account also explains why a cell LESS dense than the medium, a fat globule in milk, moves INWARD: it needs less inward force than the surrounding pressure supplies, so it is pushed toward the axis, and the cream rises to the middle.

a) On a circular arc at constant speed the acceleration is centripetal: $a_{c} = \frac{v^{2}}{r} = \frac{180^{2}}{650} = \frac{32\,400}{650} = 49.8$ m/s$^2$. Already about five times $g$, and no force has been named yet. Reading this number before touching the forces is the fastest way to see where the problem is going.

b) At the bottom of the arc the centre of the circle is ABOVE the pilot, so inward is upward. Two real forces on her: the seat pushing up with $N$, and her weight $mg$ down. Newton's second law along the radius, upward positive: $N - mg = \frac{mv^{2}}{r}$, so $N = m\left(g + \frac{v^{2}}{r}\right) = 65 \times (9.8 + 49.85) = 65 \times 59.65 = 3878$ N. Her true weight is $65 \times 9.8 = 637$ N, so the ratio is $\frac{3878}{637} = 6.09$. Above the greyout threshold: without a pressure suit this manoeuvre is on the edge of consciousness.

c) Notice that the ratio never needed the mass: $\frac{N}{mg} = 1 + \frac{v^{2}}{rg}$. Setting that to $5$ gives $\frac{v^{2}}{rg} = 4$, so $v = \sqrt{4gr} = \sqrt{4 \times 9.8 \times 650} = \sqrt{25\,480} = 160$ m/s, about $575$ km/h. Check against part b: $160 < 180$, and a slower pass through the same arc must be gentler, so the inequality points the right way. Alternatively the pilot could keep $180$ m/s and open the arc out to a larger radius, $r = \frac{v^{2}}{4g} = 827$ m, which is exactly what a pilot does in practice.

d) Over a crest the centre of the circle is BELOW, so inward is downward and the two forces swap roles: $mg - N = \frac{mv^{2}}{r}$, giving $N = m\left(g - \frac{v^{2}}{r}\right)$. Apparent weightlessness is $N = 0$, which needs $\frac{v^{2}}{r} = g$, that is $v = \sqrt{gr} = \sqrt{9.8 \times 650} = 79.8$ m/s, about $287$ km/h. This is the same condition as the bucket at the top of its circle in exercise 7, $v = \sqrt{gr}$, and for the same reason: gravity alone bends the path, so nothing else needs to touch the object.

One equation, two signs, three sensations. Write $N = m\left(g \pm \frac{v^{2}}{r}\right)$ and let the position choose the sign: plus at the bottom of a valley, minus at the top of a hill, and the answers to the whole chapter of apparent weight follow. The pilot is heavy at the bottom and light at the top, an astronaut in orbit is permanently at the $N = 0$ case, and in none of these does $g$ change by a measurable amount. What changes is the force of the seat, which is the only thing your body can actually feel.