F=ma/Toolkit: Before You Start

Problem-Solving Strategy

How to set a problem up before touching an equation: choosing axes, isolating the system, and checking limiting cases.

Frequent on exams20 min read30 problems

Two students get the same problem: a block on a ramp, a string over a pulley, a hanging weight. One of them fills three pages with sines and cosines, makes a sign error on page two, and gets a number that is wrong by a factor of two. The other writes four lines and is done. They know the same physics. The difference happened before either of them wrote an equation.

The last lesson gave you a way to make hard expressions small and manageable once you have them. This one is about the step before that: how to set a problem up so that the expressions are never hard in the first place. Three decisions do most of the work: what you draw a boundary around, which directions you call $x$ and $y$ , and which law you reach for. Get those three right and most olympiad mechanics becomes bookkeeping.

Draw the boundary first

Every equation in mechanics is about a system: a specific collection of stuff that you have mentally cut out of the world. Everything inside the boundary is the system. Everything outside can only touch the system through external forces, and only those appear in $\sum F = ma$ .

That word only is the whole trick. Forces between two parts of the same system come in equal and opposite pairs, so they cancel inside the sum and vanish from your equation. Move the boundary and a force that was giving you trouble can disappear.

Take two blocks touching each other on a frictionless table, $m_A = 3.0$ kg in front and $m_B = 5.0$ kg behind it, with a horizontal force $F = 12$ N pushing on the front block so that both slide together. The blocks push on each other with some contact force $C$ that you do not know. If you draw the boundary around both blocks, $C$ is internal and never appears:

$F = (m_A + m_B)\,a \quad\Rightarrow\quad a = \frac{12\ \text{N}}{8.0\ \text{kg}} = 1.5\ \text{m/s}^2.$

One line. Now suppose the contact force is what you actually want. Move the boundary so it cuts between the blocks, keeping only the front block inside. The cut turns $C$ from internal into external, which is exactly why you make it. The front block feels $F$ forward and $C$ backward from the block behind it, and it has the acceleration you just found:

$F - C = m_A a \quad\Rightarrow\quad C = 12 - (3.0)(1.5) = 7.5\ \text{N}.$

Two systems, two lines, no simultaneous equations. The general rule: put the boundary around everything when you want the motion, cut through the thing you want to know when you want a force.

Figure 1. The same pair of blocks, drawn twice. On the left the dashed boundary encloses both, so the contact force is internal and never enters the equation. On the right the boundary cuts between them and that same force becomes external. Notice that nothing physical changed, only where you drew the line.

Worth remembering

A force belongs on your diagram only if something outside the boundary is touching the system, or it is gravity. If you cannot name the object doing the pushing, the arrow is imaginary and it is costing you the problem.

Quick check

A rope connects two carts that are pulled along a track. You want the tension in the rope. Which system should you analyse?

Pick one option first.

Axes are free, so choose selfishly

Nothing in physics says $x$ must be horizontal. Axes are a bookkeeping choice you make, and the only rule is that they be perpendicular. So choose the pair that makes the most vectors lie flat along an axis.

A block of mass $m = 2.0$ kg on a frictionless incline of angle $\theta = 30^\circ$ makes the point. With horizontal and vertical axes, the normal force and the acceleration both have two components, and you end up solving a pair of coupled equations. Tilt the axes instead: let $x$ point down along the surface and $y$ point out perpendicular to it. Now the normal force is entirely along $y$ , and the acceleration is entirely along $x$ , because the block cannot burrow into the ramp or lift off it. Only the weight needs splitting.

The weight $mg$ points straight down, at angle $\theta$ from your new $y$ axis. So its component along the surface is $mg\sin\theta$ and its component into the surface is $mg\cos\theta$ . Watch which one gets the sine: the piece that pushes the block along the ramp has to vanish when $\theta = 0$ , and $\sin 0 = 0$ , so it is the sine. That check takes a second and saves the most common error in mechanics.

Along $x$ : $\;mg\sin\theta = ma$ , so

$a = g\sin\theta = (9.8)(0.500) = 4.9\ \text{m/s}^2.$

The mass cancelled, which it had to: every object falls the same way on a frictionless slope. Along $y$ the block does not accelerate, so the components must balance:

$N = mg\cos\theta = (2.0)(9.8)(0.866) = 17.0\ \text{N}.$

Less than the $19.6$ N weight, as it should be: the ramp only has to support the part of the weight pressing into it.

Figure 2. The weight split along tilted axes. Drag the angle to zero and watch mg sin θ collapse while mg cos θ grows to the full weight; push it to 60 degrees and the normal force falls to half the weight. The readout gives both numbers as you drag.

Where this goes wrong

When you tilt the axes, tilt them for everything. A very common wreck is resolving the weight along the slope, then writing the normal force as $mg$ anyway because that is what it was on flat ground. Every vector in the problem gets resolved in the same frame, or none do.

Let symmetry answer the question for you

If the whole setup is unchanged by some operation (a rotation, a reflection, a swap of two labels) then the answer must be unchanged by it too. That sounds abstract until you see how much arithmetic it deletes.

A point charge $q = 6.0\ \mu\text{C}$ sits at one corner of a cube. What is the electric flux through one of the three faces that do not touch that corner?

The honest way is a surface integral over a square with a charge at an awkward angle, which is unpleasant. Instead, build symmetry into the picture. Surround the charge with eight identical cubes, stacked to make one big cube with the charge at its centre. By Gauss's law the total flux out of that big cube is $q/\varepsilon_0$ . The eight little cubes are interchangeable by reflection, so each takes an eighth: $q/8\varepsilon_0$ through your cube.

Now look at one little cube. Three of its faces touch the charge, and the flux through those is zero because the field skims along them. The other three faces are interchangeable with each other by rotating the cube about its long diagonal, so they must split the $q/8\varepsilon_0$ equally, a third each:

$\Phi = \frac{q}{24\varepsilon_0} = \frac{6.0\times10^{-6}}{24\,(8.854\times10^{-12})} = \frac{6.0\times10^{-6}}{2.125\times10^{-10}} = 2.82\times10^{4}\ \text{N}\cdot\text{m}^2/\text{C}.$

No integral. The same reflex answers a whole family of questions in one step. Two identical charges, field at the midpoint between them: reflect the picture through that midpoint and it is unchanged, but every field vector flips, so the field must equal minus itself, so it is zero. A Wheatstone bridge whose left and right branches have the same ratio: the two midpoints sit at the same potential, so the current through the bridge resistor is zero and you can delete it from the circuit.

Quick check

A bridge has 10 ohm and 20 ohm in series on the left, the same 10 ohm and 20 ohm in series on the right, and a 50 ohm resistor joining the two midpoints. What current flows in the 50 ohm resistor?

Pick one option first.

When the ground moves under you

Some problems hide their difficulty in the choice of frame. A block rests on the frictionless $25^\circ$ face of a wedge, and the wedge is pushed horizontally so that the block rides along without sliding. What acceleration does that take?

The block has only two forces on it: gravity $mg$ down and the normal force $N$ perpendicular to the face. Here tilted axes are the wrong choice, because the acceleration is horizontal, not along the slope. Use horizontal and vertical axes and let $N$ be the one that splits. Vertically the block does not accelerate, so the vertical part of $N$ carries the weight: $N\cos\theta = mg$ . Horizontally, the leftover part of $N$ is the only force, so it supplies the acceleration: $N\sin\theta = ma$ . Divide the second by the first and $N$ cancels:

$a = g\tan\theta = (9.8)\tan 25^\circ = (9.8)(0.4663) = 4.57\ \text{m/s}^2.$

And if the block has mass $m = 1.5$ kg, the normal force follows from the vertical equation:

$N = \frac{mg}{\cos\theta} = \frac{(1.5)(9.8)}{0.9063} = 16.2\ \text{N},$

larger than the $14.7$ N weight, because the surface is doing two jobs at once.

Now the harder cousin, and this one takes practice. A block of $m = 1.0$ kg slides down the frictionless $30^\circ$ face of a wedge of $M = 4.0$ kg that is itself free to slide on a frictionless floor. Find the wedge's acceleration.

Choose the system generously first: block plus wedge together. The floor is frictionless and gravity is vertical, so there is no external horizontal force at all, and the horizontal momentum of the pair is conserved. Starting from rest, that gives $m\,v_x + M\,V = 0$ at every instant, and differentiating, $m\,a_x = -M\,A$ where $A$ is the wedge's acceleration. One unknown expressed in terms of the other, for free.

For the second equation you need the block alone, and the key is that the block's acceleration is measured in the ground frame while the constraint (that it stays on the face) lives in the wedge's frame. Write the block's acceleration as the wedge's acceleration plus a sliding acceleration $a_r$ along the face. Doing the two component equations for the block and eliminating $N$ gives the standard result

$A = \frac{m g \sin\theta\cos\theta}{M + m\sin^2\theta}.$

Before plugging in, test it. If $M \to \infty$ the denominator blows up and $A \to 0$ : a wedge nailed to the earth does not move. If $\theta \to 0$ the face is flat, nothing slides, and $\sin\theta\cos\theta \to 0$ kills $A$ . If $\theta \to 90^\circ$ the block falls vertically past a vertical face, pushing on nothing, and $\cos\theta \to 0$ kills $A$ again. Three checks passed, so plug in:

$A = \frac{(1.0)(9.8)(0.500)(0.866)}{4.0 + (1.0)(0.500)^2} = \frac{4.244}{4.25} = 1.0\ \text{m/s}^2.$

Check the formula before you trust the number

Those three limit checks were not decoration. A symbolic answer is a machine that makes predictions, and you can test it in ten seconds against cases whose answers you already know.

Run the checks in this order. Dimensions: both sides must carry the same units, and if they do not, stop. Limits: send each variable to zero, to infinity, or to equality with another, and ask whether what comes out is what physics demands. Symmetry: if swapping two labels leaves the problem unchanged, it must leave the answer unchanged. Sign: does a positive answer point where you said positive points? Magnitude: a $2$ kg block moving at $300$ m/s after a nudge, an efficiency of $1.4$ , a negative kinetic energy: each one means go back.

The Atwood machine is the cleanest place to practise. Two masses on a string over an ideal pulley give $a = (m_1 - m_2)g/(m_1 + m_2)$ and $T = 2m_1m_2g/(m_1 + m_2)$ . Set $m_1 = m_2$ : the acceleration vanishes, correct. Let $m_2 \to 0$ : $a \to g$ , free fall, correct. Swap the labels: $a$ changes sign, which is right because "down on the left" becomes "up on the left". For $m_1 = 5.0$ kg and $m_2 = 3.0$ kg,

$a = \frac{(5.0-3.0)(9.8)}{8.0} = 2.45\ \text{m/s}^2, \qquad T = \frac{2(5.0)(3.0)(9.8)}{8.0} = 36.8\ \text{N},$

and $T$ sits between the two weights, $29.4$ N and $49$ N, exactly as it must: the string has to hold back the heavy one and drag up the light one.

Figure 3. Make the two masses equal and watch the acceleration land on zero; shrink one mass and watch the curve flatten out against the a = g line it can never cross. The dot on the graph is your current setting.

Where this goes wrong

Plugging numbers in at the start throws away every one of these checks. Once $9.8$ has been multiplied into three other numbers you cannot send the mass to infinity and see what survives. Carry symbols to the last line, then substitute once.

Recognising which move a problem wants

The problem statement usually tells you, if you read it for structure rather than for numbers.

If two objects are joined and forced to move together (touching blocks, a string, a pulley) do the pair first, then cut. If the question asks for a tension, a contact force, or a force in a hinge, that is the signal to cut exactly there.

If the motion is constrained to a surface, tilt the axes to that surface. If instead the acceleration is along some other fixed direction, as with the driven wedge, keep the axes along the acceleration and resolve the normal force instead. The axis you refuse to tilt is the one the acceleration lies on.

If the setup has a repeated or interchangeable piece (identical charges, identical resistors on the edges of a cube, an infinite lattice that looks the same from every node) look for a symmetry or superposition argument before computing anything. Those problems are usually unfeasible by brute force and one line by symmetry, which is a strong hint about what was intended.

If the floor or the pivot is frictionless and nothing external pushes sideways, horizontal momentum is conserved and that is a free equation. If the question names a "before" and an "after" and asks nothing about the time in between, reach for a conservation law rather than for $F = ma$ .

And when the answer is a formula, spend the ten seconds on limits. That habit catches more mistakes than any amount of care in the algebra, and it is the one thing that separates the four-line solution from the three-page one.

Next you will turn measurements into answers, which needs its own discipline: graphs and uncertainty.

Problems

Work these on paper, then check your answer here.

Tilted Axes on a Frictionless Incline

EasyWritten for this guideaxesincline

A $2.0\,\text{kg}$ block slides on a frictionless incline of angle $\theta = 30^\circ$ . Find the magnitude of its acceleration along the incline, in $\text{m/s}^2$ .

m/s^2

Normal Force with Tilted Axes

EasyWritten for this guideaxesincline

The same $2.0\,\text{kg}$ block sits on a frictionless $30^\circ$ incline. Using axes along and perpendicular to the surface, find the normal force in $\text{N}$ .

Two Blocks as One System

EasyWritten for this guidesystemnewton

Blocks of mass $3.0\,\text{kg}$ and $5.0\,\text{kg}$ sit touching each other on a frictionless horizontal table. A horizontal force of $12\,\text{N}$ pushes on the $3.0\,\text{kg}$ block, so both slide together. Find the acceleration in $\text{m/s}^2$ .

m/s^2

Isolating the Second Block★

EasyWritten for this guidesystemnewton

For the same two blocks ( $3.0\,\text{kg}$ in front of a $5.0\,\text{kg}$ block, $12\,\text{N}$ applied to the $3.0\,\text{kg}$ one, frictionless table), find the magnitude of the contact force between the blocks, in $\text{N}$ .

Atwood Machine Acceleration

EasyWritten for this guideatwoodsystem

An ideal pulley carries a light string with $3.0\,\text{kg}$ on one side and $5.0\,\text{kg}$ on the other. Find the magnitude of the acceleration of either block, in $\text{m/s}^2$ .

m/s^2

Atwood Tension by Isolation

EasyWritten for this guideatwoodsystem

For the $3.0\,\text{kg}$ / $5.0\,\text{kg}$ Atwood machine above, find the string tension in $\text{N}$ .

Field at the Midpoint of Two Equal Charges

EasyWritten for this guidesymmetryelectrostatics

Two identical point charges $q = +4.0\,\text{nC}$ are held $6.0\,\text{cm}$ apart. Find the magnitude of the electric field at the midpoint of the line joining them, in $\text{N/C}$ .

N/C

Scale Reading in an Accelerating Elevator

EasyWritten for this guideframesnewton

A $60\,\text{kg}$ person stands on a bathroom scale in an elevator accelerating upward at $2.0\,\text{m/s}^2$ . Find the scale reading in $\text{N}$ .

Balanced Wheatstone Bridge

EasyWritten for this guidesymmetrycircuits

A bridge has $10\,\Omega$ and $20\,\Omega$ in series on the left branch, $10\,\Omega$ and $20\,\Omega$ in series on the right branch, and a $50\,\Omega$ resistor connecting the two midpoints. A $12\,\text{V}$ battery is across the outer terminals. Find the current in the $50\,\Omega$ bridge resistor, in $\text{A}$ .

10.

Choosing the Pivot on a Loaded Plank

EasyWritten for this guidetorquepivot

A uniform $20\,\text{kg}$ plank of length $4.0\,\text{m}$ rests on supports at its two ends. A $50\,\text{kg}$ person stands $1.0\,\text{m}$ from the left end. Find the upward force from the right support, in $\text{N}$ .

11.

Ring of Charge on Its Axis

NormalWritten for this guidesymmetryelectrostatics

A thin ring of radius $R = 0.10\,\text{m}$ carries a uniformly distributed charge $Q = 5.0\,\text{nC}$ . Find the magnitude of the electric field on the axis at distance $z = 0.10\,\text{m}$ from the center, in $\text{N/C}$ . Use $k = 8.99\times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ .

N/C

12.

Wedge Accelerated So the Block Does Not Slide

NormalWritten for this guideframeswedge

A block rests on the frictionless $25^\circ$ face of a wedge. The wedge is pushed horizontally so that the block stays at rest relative to the wedge. Find the required acceleration in $\text{m/s}^2$ .

m/s^2

13.

Normal Force on the Accelerating Wedge

NormalWritten for this guideframeswedge

For the same $25^\circ$ frictionless wedge accelerating so that a $1.5\,\text{kg}$ block rides along without sliding, find the normal force on the block in $\text{N}$ .

14.

Pulley with Friction: System First

NormalWritten for this guidesystemfriction

A $4.0\,\text{kg}$ block on a horizontal table is connected over an ideal pulley at the table edge to a $3.0\,\text{kg}$ hanging block. The coefficient of kinetic friction between block and table is $0.25$ . Find the magnitude of the acceleration in $\text{m/s}^2$ .

m/s^2

15.

Tension in the Friction Pulley Problem

NormalWritten for this guidesystemfriction

For the setup above ( $4.0\,\text{kg}$ on the table with $\mu_k = 0.25$ , $3.0\,\text{kg}$ hanging), find the string tension in $\text{N}$ .

16.

Cube of Resistors: Body Diagonal★

NormalWritten for this guidesymmetrycircuits

Twelve identical $6.0\,\Omega$ resistors form the edges of a cube. Find the resistance between two opposite corners of the cube (a body diagonal), in $\Omega$ .

17.

Infinite Ladder Network

NormalWritten for this guidesymmetrycircuits

An infinite ladder is built from $1.0\,\Omega$ resistors: a series $1.0\,\Omega$ into a node, a $1.0\,\Omega$ shunt to the bottom rail, then the pattern repeats forever. Find the input resistance in $\Omega$ .

18.

Elastic Collision via the CM Frame

NormalWritten for this guidecm-framecollision

A $2.0\,\text{kg}$ ball moving at $+5.0\,\text{m/s}$ collides head-on and elastically with a stationary $3.0\,\text{kg}$ ball. Find the velocity of the $2.0\,\text{kg}$ ball afterward, in $\text{m/s}$ (sign included, with $+$ along the initial motion).

m/s

19.

Energy Method on a Rough Incline

NormalWritten for this guideenergyfriction

A block is released from rest and slides $3.0\,\text{m}$ down a $30^\circ$ incline with coefficient of kinetic friction $0.20$ . Find its speed at the bottom, in $\text{m/s}$ .

m/s

20.

Fixing an Unknown Coefficient by a Limiting Case

NormalWritten for this guidelimiting-caseatwood

A student derives the Atwood acceleration as $a = \dfrac{(m_1-m_2)g}{m_1 + n\,m_2}$ but cannot remember the number $n$ . Determine $n$ by taking the limit $m_1 \to 0$ , where the remaining block must fall freely, then evaluate $a$ for $m_1 = 6.0\,\text{kg}$ and $m_2 = 4.0\,\text{kg}$ . Give the magnitude of $a$ in $\text{m/s}^2$ .

m/s^2

21.

Speed of a Point on a Rolling Wheel

NormalWritten for this guideframesrolling

A wheel of radius $0.35\,\text{m}$ rolls without slipping with center speed $3.0\,\text{m/s}$ . Find the ground-frame speed of the point on the rim level with the axle (a quarter turn from the contact point), in $\text{m/s}$ .

m/s

22.

Pendulum in an Accelerating Train

NormalWritten for this guideframespseudo-force

A small bob hangs from the ceiling of a train accelerating horizontally at $2.5\,\text{m/s}^2$ . In the steady state, find the angle the string makes with the vertical, in degrees.

degrees

23.

Cube of Resistors: Face Diagonal

HardWritten for this guidesymmetrycircuits

Twelve identical $8.0\,\Omega$ resistors form the edges of a cube. Find the resistance between two corners on the same face that are diagonally opposite (a face diagonal), in $\Omega$ .

24.

Uniform Sphere with an Off-Center Cavity★

HardWritten for this guidesuperpositionelectrostatics

A sphere of radius $R = 0.10\,\text{m}$ carries uniform charge density $\rho = 2.0\times10^{-6}\,\text{C/m}^3$ , except for an empty spherical cavity of radius $R/2$ whose center is $d = 0.050\,\text{m}$ from the sphere's center. Find the magnitude of the electric field inside the cavity, in $\text{N/C}$ . Use $\varepsilon_0 = 8.854\times10^{-12}\,\text{F/m}$ .

N/C

25.

Ladder Against a Frictionless Wall

HardWritten for this guidetorquepivot

A uniform ladder leans at $60^\circ$ above the horizontal against a frictionless vertical wall, with friction only at the floor. Find the minimum coefficient of static friction at the floor for the ladder not to slip. Answer is dimensionless.

26.

Block Sliding on a Free Wedge

HardWritten for this guidesystemwedge

A $1.0\,\text{kg}$ block slides down the frictionless $30^\circ$ face of a $4.0\,\text{kg}$ wedge that is itself free to slide on a frictionless floor. Find the magnitude of the wedge's acceleration, in $\text{m/s}^2$ .

m/s^2

27.

Energy Lost in a Perfectly Inelastic Collision

HardWritten for this guidecm-frameenergy

A $2.0\,\text{kg}$ cart moving at $6.0\,\text{m/s}$ strikes and sticks to a stationary $4.0\,\text{kg}$ cart on a frictionless track. Find the kinetic energy converted to other forms, in $\text{J}$ .

28.

Flux Through One Face from a Corner Charge

HardWritten for this guidesymmetrygauss

A point charge $q = 6.0\,\mu\text{C}$ sits at one corner of a cube. Find the electric flux through one of the three faces that do not touch that corner, in $\text{N}\cdot\text{m}^2/\text{C}$ . Use $\varepsilon_0 = 8.854\times10^{-12}\,\text{F/m}$ .

N·m^2/C

29.

Infinite Grid of Resistors★

Very HardWritten for this guidesymmetrysuperpositioncircuits

An infinite square lattice of nodes is wired with a $3.0\,\Omega$ resistor along every edge. Find the resistance between two adjacent nodes, in $\Omega$ .

30.

Cheapest Angle to Drag a Block Uphill

Very HardWritten for this guidelimiting-casefrictionoptimization

A $5.0\,\text{kg}$ block is dragged at constant speed up a $25^\circ$ incline with $\mu_k = 0.35$ . The applied force may point at any angle above the incline surface. Find the smallest force magnitude that works, in $\text{N}$ .

Solutions

1. Tilted Axes on a Frictionless Incline Answer: 4.9 m/s^2

Tilt the axes so $x$ points down the slope. Then the normal force has no $x$ component and the motion is one-dimensional in $x$ . Newton's law along the slope gives $ma = mg\sin\theta$ , so $a = g\sin 30^\circ = 4.9\,\text{m/s}^2$ . The mass cancels, which is the usual sign that the axis choice was right.

2. Normal Force with Tilted Axes Answer: 17 N

Perpendicular to the incline there is no acceleration, because the block moves along the surface. So $N - mg\cos\theta = 0$ , giving $N = (2.0)(9.8)\cos 30^\circ = 17\,\text{N}$ . With horizontal-vertical axes you would need both components of $\vec a$ and get the same answer with more algebra.

3. Two Blocks as One System Answer: 1.5 m/s^2

Treat the two blocks as a single system. The contact force between them is internal and drops out, so $F = (m_1+m_2)a$ and $a = 12/8.0 = 1.5\,\text{m/s}^2$ .

4. Isolating the Second Block Answer: 7.5 N

To expose an internal force you must isolate one body. The $5.0\,\text{kg}$ block feels only the contact force $F_c$ , so $F_c = (5.0)(1.5) = 7.5\,\text{N}$ . Isolating the $3.0\,\text{kg}$ block instead gives $12 - F_c = (3.0)(1.5)$ , the same result.

5. Atwood Machine Acceleration Answer: 2.45 m/s^2

Because the string is inextensible, unroll the system into one line of mass $m_1+m_2$ driven by the net weight $(m_2-m_1)g$ . Then $a = (5.0-3.0)(9.8)/8.0 = 2.45\,\text{m/s}^2$ , and the tension never appears.

6. Atwood Tension by Isolation Answer: 36.8 N

Tension is internal to the whole system, so isolate one block. For the rising $3.0\,\text{kg}$ block, $T - m_1 g = m_1 a$ , giving $T = 3.0(9.8+2.45) = 36.8\,\text{N}$ . Check: $T$ lies between $m_1g = 29.4\,\text{N}$ and $m_2g = 49\,\text{N}$ , as it must.

7. Field at the Midpoint of Two Equal Charges Answer: 0 N/C

The setup is unchanged if you reflect it through the midpoint, but that reflection reverses any field vector there. A vector equal to its own negative is zero, so $E = 0$ . No calculation is needed.

8. Scale Reading in an Accelerating Elevator Answer: 708 N

Work in the ground frame with $+y$ up and isolate the person. Then $N - mg = ma$ , so $N = 60(9.8+2.0) = 708\,\text{N}$ . Limiting check: with $a=0$ this returns $588\,\text{N}$ , the person's weight.

9. Balanced Wheatstone Bridge Answer: 0 A

The two branches are identical, so the two midpoints divide the same voltage in the same ratio and sit at equal potential. With no potential difference across it, the bridge resistor carries zero current.

10. Choosing the Pivot on a Loaded Plank Answer: 220 N

Take torques about the left support, which kills the unknown left reaction. Then $N_R(4.0) = (20)(9.8)(2.0) + (50)(9.8)(1.0) = 882\,\text{N}\cdot\text{m}$ , so $N_R = 220\,\text{N}$ . The person is nearer the left support, so the right support should carry less than half the total weight of $686\,\text{N}$ ; it does.

11. Ring of Charge on Its Axis Answer: 1590 N/C

Rotational symmetry about the axis means the components perpendicular to the axis from opposite bits of the ring cancel, so only the axial component survives. Each element contributes $k\,dq/(z^2+R^2)$ times the factor $z/\sqrt{z^2+R^2}$ , and every element has the same distance, so $E = \frac{kQz}{(z^2+R^2)^{3/2}} = 1.59\times 10^3\,\text{N/C}.$

12. Wedge Accelerated So the Block Does Not Slide Answer: 4.57 m/s^2

Use horizontal and vertical axes here, not tilted ones, because the acceleration is horizontal. Vertically $N\cos\theta = mg$ ; horizontally $N\sin\theta = ma$ . Dividing gives $a = g\tan 25^\circ = 4.57\,\text{m/s}^2$ . Limiting check: $\theta \to 0$ gives $a \to 0$ .

13. Normal Force on the Accelerating Wedge Answer: 16.2 N

The vertical equation alone fixes $N$ , since the block has no vertical acceleration: $N\cos\theta = mg$ , so $N = (1.5)(9.8)/\cos 25^\circ = 16.2\,\text{N}$ . It exceeds $mg = 14.7\,\text{N}$ because the surface must also supply the horizontal push.

14. Pulley with Friction: System First Answer: 2.8 m/s^2

Take both blocks plus string as one system moving with common speed. The driving force is the hanging weight $m_2 g$ , the retarding force is friction $\mu m_1 g$ , and the tension is internal: $a = \frac{m_2 g - \mu m_1 g}{m_1+m_2} = \frac{29.4-9.8}{7.0} = 2.8\,\text{m/s}^2.$

15. Tension in the Friction Pulley Problem Answer: 21 N

Isolate the hanging block. Then $m_2 g - T = m_2 a$ , so $T = 3.0(9.8-2.8) = 21\,\text{N}$ . Isolating the table block gives $T - \mu m_1 g = m_1 a$ , i.e. $9.8 + 11.2 = 21\,\text{N}$ , a useful cross-check.

16. Cube of Resistors: Body Diagonal Answer: 5 Ω

Symmetry about the body diagonal makes the three corners next to the input equipotential, and likewise the three next to the output. Merging equipotential nodes turns the cube into three resistors in parallel, then six, then three: $R = \frac{R_0}{3}+\frac{R_0}{6}+\frac{R_0}{3} = \frac{5R_0}{6} = 5.0\,\Omega.$

17. Infinite Ladder Network Answer: 1.62 Ω

The ladder is self-similar, so removing one stage leaves the same network of resistance $R$ . Then $R = 1 + \dfrac{1\cdot R}{1+R}$ , i.e. $R^2 - R - 1 = 0$ , so $R = (1+\sqrt 5)/2 = 1.62\,\Omega$ . The negative root is discarded as unphysical.

18. Elastic Collision via the CM Frame Answer: -1 m/s

In the center-of-mass frame an elastic head-on collision simply reverses each velocity. Here $v_{cm} = (2)(5)/5 = 2.0\,\text{m/s}$ , so the light ball approaches the CM at $+3.0$ and leaves at $-3.0$ ; back in the lab frame that is $2.0-3.0 = -1.0\,\text{m/s}$ . It rebounds because the target is heavier.

19. Energy Method on a Rough Incline Answer: 4.38 m/s

Energy accounting avoids solving for the acceleration and then the kinematics. Gravity does $mgL\sin\theta$ and friction removes $\mu mgL\cos\theta$ , so $v = \sqrt{2gL(\sin\theta-\mu\cos\theta)} = 4.38\,\text{m/s}.$ The mass cancels, as it must for a problem with only gravity and dry friction.

20. Fixing an Unknown Coefficient by a Limiting Case Answer: 1.96 m/s^2

A limiting case can pin down a forgotten constant. Setting $m_1 = 0$ gives $a = -g/n$ , and free fall requires magnitude $g$ , so $n = 1$ . Then $a = (2.0)(9.8)/10.0 = 1.96\,\text{m/s}^2$ .

21. Speed of a Point on a Rolling Wheel Answer: 4.24 m/s

Rolling motion is translation of the center plus rotation about it, and the two contributions at this point are perpendicular: $v_{cm}$ forward and $\omega R = v_{cm}$ vertical. So $v = \sqrt 2\,(3.0) = 4.24\,\text{m/s}$ . Equivalently, the contact point is instantaneously at rest and this point is $\sqrt 2 R$ from it.

22. Pendulum in an Accelerating Train Answer: 14.3 degrees

In the train frame add a pseudo-force $ma$ opposite the acceleration; the bob then hangs along the effective gravity $\vec g_{\text{eff}}$ . Its tilt is $\theta = \arctan(a/g) = \arctan(2.5/9.8) = 14.3^\circ$ . Limiting check: $a \to 0$ gives a vertical string.

23. Cube of Resistors: Face Diagonal Answer: 6 Ω

The mirror plane containing the two chosen corners forces symmetric pairs of nodes to sit at equal potential. Merging those pairs collapses the cube to a simple series-parallel network with $R = \tfrac{3}{4}R_0 = 6.0\,\Omega$ . Sanity check: this is less than the body-diagonal value $\tfrac{5}{6}R_0 = 6.7\,\Omega$ , as it should be for corners that are closer together.

24. Uniform Sphere with an Off-Center Cavity Answer: 3760 N/C

Superpose a full sphere of density $+\rho$ with a small sphere of density $-\rho$ at the cavity's center. Inside a uniform sphere $\vec E = \rho\vec r/(3\varepsilon_0)$ measured from its own center, so the two position-dependent terms cancel and leave the constant field $E = \frac{\rho d}{3\varepsilon_0} = 3.76\times10^3\,\text{N/C}.$ The field in the cavity is uniform, which no direct integration would reveal so cheaply.

25. Ladder Against a Frictionless Wall Answer: 0.289

Take torques about the base, which eliminates both floor forces at once and leaves only the wall force $N_w$ and the weight. That gives $N_w L\sin\theta = mg(L/2)\cos\theta$ , so $N_w = \tfrac12 mg\cot\theta$ . Horizontal balance requires friction $f = N_w$ with $N_{\text{floor}} = mg$ , hence $\mu_{\min} = \tfrac12\cot 60^\circ = 0.289$ .

26. Block Sliding on a Free Wedge Answer: 1 m/s^2

Horizontal momentum of the block-plus-wedge system is conserved, so work with the system horizontally and with the block in the wedge frame along the incline. The standard result is $A = \frac{mg\sin\theta\cos\theta}{M+m\sin^2\theta} = \frac{(1.0)(9.8)(0.5)(0.866)}{4.0+0.25} = 1.0\,\text{m/s}^2.$ Limiting check: $M\to\infty$ gives $A\to 0$ , the fixed-wedge case.

27. Energy Lost in a Perfectly Inelastic Collision Answer: 24 J

In the CM frame the final state is at rest, so exactly the CM-frame kinetic energy is lost. That energy is $\tfrac12 \mu v_{\text{rel}}^2$ with reduced mass $\mu = m_1m_2/(m_1+m_2) = 4/3\,\text{kg}$ : $\Delta E = \tfrac12(4/3)(6.0)^2 = 24\,\text{J}.$ Note that only the relative velocity matters, never the frame you happen to watch from.

28. Flux Through One Face from a Corner Charge Answer: 28200 N·m^2/C

Build a big cube of eight copies with the charge at its center. By symmetry each of the six big faces takes $q/(6\varepsilon_0)$ , and each big face is made of four of the small faces in question, so one small face gets $q/(24\varepsilon_0) = 2.82\times10^4\,\text{N}\cdot\text{m}^2/\text{C}$ . The three faces touching the charge get zero flux, since the field is parallel to them.

29. Infinite Grid of Resistors Answer: 1.5 Ω

Inject current $I$ at node $A$ alone; by fourfold symmetry $I/4$ leaves along each of its four edges. Separately extract $I$ at the neighbor $B$ alone; again $I/4$ flows into $B$ along the connecting edge. Superposing the two problems is exactly the physical situation, and the shared edge carries $I/2$ , so $V_{AB} = (I/2)R_0$ and $R_{AB} = R_0/2 = 1.5\,\Omega$ .

30. Cheapest Angle to Drag a Block Uphill Answer: 34.2 N

Tilt the axes to the incline and let the force make angle $\alpha$ with the surface. Then $F\cos\alpha = mg\sin\theta + \mu(mg\cos\theta - F\sin\alpha)$ , so $F = \frac{mg(\sin\theta+\mu\cos\theta)}{\cos\alpha+\mu\sin\alpha}.$ The denominator is maximal when $\tan\alpha = \mu$ (here $\alpha = 19.3^\circ$ ), where it equals $\sqrt{1+\mu^2}$ , giving $F_{\min} = 34.2\,\text{N}$ . Limiting check: with $\mu=0$ the best angle is along the incline and $F = mg\sin\theta$ .