Study Companion: S03 — Force and Its Rectangular Components
A force is the first object statics works with, and it carries more information than an arrow with a number on it. Where it acts matters to the inside of a body. For the body as a whole, only its line of action matters. Several forces through one point combine into one, and a force splits into parts along any two directions. This lesson builds those facts in order. It ends with the one tool used in every later lesson: write each force as x and y components, and add the columns.
Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, articles 2/1 ("Introduction"), 2/2 ("Force") and 2/3 ("Rectangular Components"), with Sample Problems 2/1–2/4. Marghitu, Dupac and Madsen, Statics with MATLAB, sections 1.4 ("Resolution of Vectors") and 1.5 ("Angle Between Two Vectors"), with Example 1.1. Harper, Solving Statics Problems in MATLAB, Problem 2/21 ("2D Rectangular Components"), which is Meriam's spike problem made parametric. Notation follows Meriam: unit vectors \mathbf i, \mathbf j (Marghitu prints them dotless). Harper's and Meriam's US-unit data are converted to SI where they first appear.
How to use this companion
Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson. Diagrams use x to the right and y up, with angles measured counter-clockwise from +x unless a figure says otherwise.
The puzzle
A spike is stuck in a timber beam, and you want to pull it straight out along its own axis. A block A sits in front of it, so no cable can run straight along the axis. Two cables are hooked to the spike's eye E instead. One, Q, passes 150\,\mathrm{mm} below the axis and pulls with 1.8\,\mathrm{kN}. The other, P, passes 120\,\mathrm{mm} above it. Both cross the front of the block 200\,\mathrm{mm} along the axis.
How hard must P pull so that the spike feels a pull along its axis only, and how large is that pull T? The tempting answer is "pull equally, P = 1.8\,\mathrm{kN}, and the pull is 3.6\,\mathrm{kN}." Both halves are wrong. The answer, P = 2.10\,\mathrm{kN} and T = 3.24\,\mathrm{kN}, comes in Nugget 06, once the earlier nuggets have built the tools.
Nugget 01: A Force Slides Along Its Line Without Changing Its External Effect
Claim. A force is fixed by its magnitude, its direction and its point of application. For the effects outside a rigid body, the point can slide anywhere along the force's line of action.
The picture. A steel rod 2\,\mathrm m long is bolted to a wall at its left end. A 5\,\mathrm{kN} pull acts along its axis, first at the free end, then at the midpoint M through a collar.
In both rows the wall must hold the rod back with 5\,\mathrm{kN} to the left. That is the external effect: what the supports feel. Inside the rod the two cases differ. With the pull at the end, every cross-section between the wall and the end carries 5\,\mathrm{kN} of tension. With the pull at M, only the left half is stretched and the right half carries nothing. That is the internal effect: the internal forces and the deformation of the material.
Check with numbers. Cut the rod at 0.5\,\mathrm m from the wall. In both rows the piece to the right of the cut is pulled 5\,\mathrm{kN} to the right, so the cut carries 5\,\mathrm{kN} of tension. Cut it at 1.5\,\mathrm m instead. With the pull at the end the cut still carries 5\,\mathrm{kN}. With the pull at M (1.0\,\mathrm m) nothing acts on the piece beyond the cut, so the cut carries 0. The wall reaction is 5\,\mathrm{kN} in both rows.
The idea. Meriam's starting point (2/2): a force is a vector, and its complete specification is magnitude, direction and point of application. A force fixed in this way is a fixed vector. Change any of the three and something changes, such as the force in a bracket's bolts or the stress in its material. External forces split into applied forces (the pull) and reactive forces (the wall's response). How internal forces deform a body depends on its material and belongs to strength of materials.
Statics mostly treats bodies as rigid (no deformation at all). For a rigid body only the external effects matter. Experience shows that they do not change when the force is moved along its line of action. That is the principle of transmissibility. A force treated this way is a sliding vector, fixed by its magnitude, direction and line of action; the point of application drops out.
The equation (physical law for a rigid body, from experience; an approximation for a real, slightly deformable one):
Assumptions: rigid body; the move is along the line, never across it.
Worked number. Meriam's Fig. 2/2 is a plate on a pin O and a roller C, with a horizontal force P that can be applied at A or at B on the same horizontal line. Take P = 800\,\mathrm N. Whichever point it is applied at, the pin and roller feel the same reactions, because the line of action is the same. The rod above makes the same point with numbers: 5\,\mathrm{kN} at the wall in both rows, and 5\,\mathrm{kN} or 0 in the outer half.
Real example. A crane hook pulls a load through a sling. For the crane's supports it makes no difference where along the sling line the force is "applied". For the sling's own stitching it makes all the difference.
Plausible wrong answer. "The force can be moved anywhere on the body." It can slide only along its own line. Moving it sideways changes its turning effect (S04, S05). Even sliding along the line changes the internal forces, as the rod shows.
Check. A 3\,\mathrm{kN} push acts along the axis of a column, at its top. Is the force at the base the same if the push is applied halfway down instead? Is the upper half of the column loaded in both cases? (Base: yes, 3\,\mathrm{kN}. Upper half: only when the push is at the top.)
Narration spine. Draw the rod and wall. Apply the pull at the end; the wall arrow appears, and the whole rod glows with tension. Slide the pull along the axis to M. The wall arrow does not change, but the outer half stops glowing. Finally try to slide the pull sideways off the axis; a red cross shows this move is not allowed.
Bridge. A force can now be drawn properly. Before adding several forces we need to know what kinds of force a body meets, and which force pairs with which, so that none is counted twice or left out.
Nugget 02: Every Force Is a Push or Pull From Something Else
Claim. Every force on a body comes from another body, either by contact or at a distance. Each one is paired with an equal and opposite force on that other body, never on the same body.
The picture. A 12\,\mathrm{kg} toolbox rests on a shelf. On the right it is isolated: drawn alone, with every force that acts on it.
Two bodies act on the box. The Earth pulls every particle of it downward; the total is the weight W, drawn at the centre of gravity G. The shelf pushes up on the whole base by contact. The free-body diagram's sign convention is +y upward.
Check with numbers. W = mg = 12\,\mathrm{kg} \times 9.81\,\mathrm{m/s^2} = 117.7\,\mathrm N downward. The box is at rest, so the shelf's push totals 117.7\,\mathrm N upward. Spread over a base of 0.40\,\mathrm m \times 0.25\,\mathrm m = 0.10\,\mathrm{m^2}, that is an average pressure of 117.7/0.10 = 1177\,\mathrm{Pa}.
The idea. Meriam (2/2) sorts forces two ways.
- Contact or body. A contact force comes from touching: the shelf on the box. A body force comes from where a body sits in a force field (gravitational, electric, magnetic): the weight.
- Concentrated or distributed. Every contact force really acts over an area. When that area is small compared with the body, the force is treated as concentrated at a point, with negligible loss of accuracy. A force can also be distributed over a volume (weight) or along a line (the weight of a hanging cable).
The weight is distributed over the volume, but it may be taken as one concentrated force through the centre of gravity. For a symmetric body that point is obvious; otherwise it is found by the methods of S17.
A force is measured by comparison: against known forces on a balance, or by the calibrated stretch of an elastic element (a spring balance, a load cell). The standard unit is the newton. (Meriam also uses the US pound; this book works in newtons throughout.)
Action and reaction. Newton's third law pairs every force with an equal, opposite, collinear force on the other body. To keep the pairs apart, isolate the body first, then draw only the forces exerted on it.
The equation (physical law, Newton's third law):
Worked number. The four forces in the toolbox story, each 117.7\,\mathrm N: the Earth pulls the box down; the box pulls the Earth up; the shelf pushes the box up; the box pushes the shelf down. The first two are one action–reaction pair, the last two another.
Real example. A load cell under a weighing platform measures the contact force of the platform on the cell. It reads the weight only because the load is at rest, so that the contact force and the weight happen to be equal.
Plausible wrong answer. "The shelf's push is the reaction to the weight." Both act on the box, so they cannot be an action–reaction pair. They are equal here only because the box is at rest. In a lift accelerating upward, the shelf's push exceeds the weight, while the third-law pairs stay exactly equal.
Check. A 4\,\mathrm{kg} lamp hangs from a cord. Name the reaction to the cord's pull on the lamp. (The lamp's pull on the cord, 39.24\,\mathrm N downward, acting on the cord.)
Narration spine. The box sits on the shelf. Lift it away, leaving a dashed outline: the box is isolated. Draw the weight arrow at G, then a row of small upward arrows along the base, merging into one 117.7\,\mathrm N arrow. Then split the screen: four arrows on four bodies, paired by colour.
Bridge. The toolbox had forces along one line. A real body has forces in several directions, applied at different points. We need to know when and how such forces combine into one.
Nugget 03: Concurrent Forces Add Where Their Lines Meet
Claim. Forces whose lines of action meet at one point add by the parallelogram law at that point. The resultant acts on a definite line, and a careless triangle can put it on the wrong one.
The picture. A plate carries 3\,\mathrm{kN} pointing along +x at A and 4\,\mathrm{kN} pointing along +y at B. The coordinates are A(0, 1)\,\mathrm m and B(2, 0)\,\mathrm m.
The two forces act at different points, but their lines meet at C(2, 1)\,\mathrm m. By transmissibility, slide each force along its own line to C. Now they act at one point, and the parallelogram there gives the resultant \mathbf R on a line through C. Replacing the two forces by \mathbf R at C changes nothing outside the plate.
Check with numbers. At C, R = \sqrt{3^2 + 4^2} = 5\,\mathrm{kN}, at \tan^{-1}(4/3) = 53.13^\circ above +x. If instead you slide the 4\,\mathrm{kN} force to the tip of the 3\,\mathrm{kN} arrow drawn at A (the triangle law), the closing side is again 5\,\mathrm{kN} at 53.13^\circ. But it now starts at A, on a parallel line 1.6\,\mathrm m away from the true one. Magnitude and direction are right, but the line of action is wrong.
The idea. Forces are concurrent at a point when their lines of action meet there. Their sum (the resultant) lies in their plane and passes through that point. The triangle law is a quick way to get the magnitude and direction. It moves one force off its own line, so it says nothing about where \mathbf R acts (Meriam's Fig. 2/3d). Draw the triangle for size, then put \mathbf R back through the point of concurrency.
Parallel forces never meet, so Meriam (2/2) uses a trick. Add two equal, opposite, collinear forces \mathbf F and -\mathbf F, one at each force. They cancel, so they change nothing. Now combine \mathbf F_1 + \mathbf F = \mathbf R_1 and \mathbf F_2 - \mathbf F = \mathbf R_2. These two are not parallel, so their lines meet, and the resultant passes through that meeting point. The same trick helps with two forces that are nearly parallel and meet off the page.
The equation (physical law: forces add as vectors, the parallelogram law; plus transmissibility for the line):
Worked number. A beam carries 300\,\mathrm N and 200\,\mathrm N, both downward, 1.0\,\mathrm m apart (right half of the figure). Add F = 100\,\mathrm N pointing along +x at the left force and -F at the right one. Then \mathbf R_1 = 100\mathbf i - 300\mathbf j (316.2\,\mathrm N) through the left end, and \mathbf R_2 = -100\mathbf i - 200\mathbf j (223.6\,\mathrm N) through the right end. Along \mathbf R_1's line, x = 100s and y = -300s. Along \mathbf R_2's, x = 1 - 100t and y = -200t. Equal y gives t = 1.5s. Equal x gives 100s = 1 - 150s, so s = 0.004. The lines meet at (0.4, -1.2)\,\mathrm m, and \mathbf R = \mathbf R_1 + \mathbf R_2 = -500\mathbf j\ \mathrm N acts on the vertical line x = 0.4\,\mathrm m. That is 0.4\,\mathrm m from the larger force and 0.6\,\mathrm m from the smaller, and indeed 300 \times 0.4 = 200 \times 0.6. (S06 finds this line faster with moments.)
Real example. A lifting beam carries a load on two slings. The crane hook must sit on the resultant's line, nearer the heavier sling, or the beam tilts as it is lifted.
Plausible wrong answer. "The triangle law gives the resultant." It gives the right size and direction and the wrong line, 1.6\,\mathrm m away in the example. On a rigid body the line matters: it decides how the body turns (S04).
Check. Two forces of 6\,\mathrm{kN} along +x at (0, 2)\,\mathrm m and 8\,\mathrm{kN} along -y at (3, 0)\,\mathrm m. Through which point does their resultant act? (Through (3, 2)\,\mathrm m, where the lines meet: 10\,\mathrm{kN} at 53.13^\circ below +x.)
Narration spine. Two arrows sit on the plate at A and B. Dashed lines grow from them and meet at C; both arrows slide along their lines to C. The parallelogram is drawn and the 5\,\mathrm{kN} diagonal appears. A ghost triangle at A gives the same arrow, offset; a red bracket marks the 1.6\,\mathrm m error. Then the beam: two downward arrows, two small green arrows added, two slanted resultants whose lines meet below the beam, and the 500\,\mathrm N arrow drops through the meeting point.
Bridge. Adding forces is half the job. Statics also needs to split a force into parts along chosen directions, for instance along the members of a bracket. A split can be done in two ways, and only one of them is correct.
Nugget 04: Components Are Not Projections
Claim. The vector components of a force along two axes are the sides of a parallelogram that rebuilds the force. Its perpendicular projections onto the same axes are different numbers, unless the axes are perpendicular.
The picture. A 1000\,\mathrm N force along x, with two oblique axes: a at 20^\circ above x and b at 40^\circ below.
The orange arrows come from completing the parallelogram: from the tip of \mathbf F, draw lines parallel to each axis back to the other axis. They are the vector components \mathbf F_a and \mathbf F_b, and they add back to \mathbf F. The green arrows come from dropping perpendiculars from the tip of \mathbf F onto each axis. They are the orthogonal projections \mathbf P_a and \mathbf P_b, and they do not add back to \mathbf F.
Check with numbers. The parallelogram's triangle has angles 20^\circ (at O, between \mathbf F and a), 40^\circ (between \mathbf F and b) and 180^\circ - 60^\circ = 120^\circ (opposite \mathbf F). The law of sines gives
The projections are P_a = 1000\cos20^\circ = 939.7\,\mathrm N and P_b = 1000\cos40^\circ = 766.0\,\mathrm N. Adding the projections as vectors gives 1470\mathbf i - 171\mathbf j\ \mathrm N, of size 1480\,\mathrm N, nowhere near the 1000\,\mathrm N we started with. Adding the components gives exactly 1000\mathbf i\ \mathrm N.
The idea. Resolution is addition run backwards (Marghitu 1.4). For any two non-parallel directions in a plane, a force has exactly one pair of vector components along them, and they must add back to the force. A projection answers a different question: how much of the force lies along one axis, ignoring the other. The two agree only when the axes are perpendicular, because then the parallelogram is a rectangle and its sides are the shadows. Marghitu adds that a vector equation \mathbf v = \mathbf 0 means every component is zero: three scalar equations in space, two in a plane.
The equation (mathematical identity: the law of sines in the parallelogram's triangle; \phi_a, \phi_b are the angles from \mathbf F to each axis):
Worked number. Meriam's Sample Problem 2/4 finds the projection of a resultant onto an axis b. Two forces of 100\,\mathrm N and 80\,\mathrm N, 50^\circ apart, have R = \sqrt{80^2 + 100^2 + 2(80)(100)\cos50^\circ} = 163.4\,\mathrm N. With b along the 80\,\mathrm N force, the projection is 80 + 100\cos50^\circ = 144.3\,\mathrm N: the projections of the two forces add, because each is just a shadow on the same line. Sample Problem 2/3 does the nonrectangular case. A 500\,\mathrm N force at 60^\circ below x, resolved along x and along y' (30^\circ above x, so at right angles to the force), has components 1000\,\mathrm N along x and -866\,\mathrm N along y' by the law of sines. Both are larger than the force itself.
Real example. A wall bracket holds a sign with a horizontal tie and an inclined strut. The sign's weight splits into a component along the tie and a component along the strut. Those components, not the projections, are the forces the two members carry. The same split, member by member, is how trusses are solved (S13).
Plausible wrong answer. "The component along a is F\cos\phi_a." On oblique axes that is the projection. Here it is 939.7\,\mathrm N instead of 742.2\,\mathrm N, an error of 27 %.
Check. If a and b were perpendicular, with \mathbf F at 30^\circ to a, what would F_a and P_a be for F = 1000\,\mathrm N? (Both 1000\cos30^\circ = 866.0\,\mathrm N, since \sin60^\circ/\sin90^\circ = \cos30^\circ.)
Narration spine. One arrow, two dashed oblique axes. Lines parallel to the axes sweep out from the tip and the orange components snap into place. Their head-to-tail chain lands on the tip. Then perpendiculars drop from the tip and the green projections appear, longer than the components. Chained head to tail they overshoot to 1480\,\mathrm N. Finally rotate b until it is perpendicular to a and watch the orange and green arrows merge.
Bridge. With oblique axes every force needs its own law-of-sines triangle. Perpendicular axes remove that work: components become projections, and adding forces becomes adding numbers.
Nugget 05: Rectangular Components Turn Addition Into Arithmetic
Claim. With perpendicular x and y axes, each force becomes two signed numbers, and the resultant is found by adding the x numbers and the y numbers separately.
The picture. Three forces act at point A of a bracket. \mathbf F_1 = 750\,\mathrm N at 40^\circ above x. \mathbf F_2 = 520\,\mathrm N rising 12 for every 5 to the left. \mathbf F_3 = 1300\,\mathrm N pulling towards a point B that is 0.25\,\mathrm m to the right of A and 0.60\,\mathrm m below it. Each force's direction is given a different way, as in Meriam's Sample Problem 2/1.
Each force is the diagonal of a rectangle whose sides lie along x and y. Those sides are its rectangular components. The signs come from the quadrant the force points into: \mathbf F_2 points left, so its x component is negative; \mathbf F_3 points down, so its y component is negative.
Check with numbers. Read each force's components the way its direction is given.
- \mathbf F_1, by its angle: F_{1x} = 750\cos40^\circ = 574.5\,\mathrm N, F_{1y} = 750\sin40^\circ = 482.1\,\mathrm N.
- \mathbf F_2, by its slope triangle with sides 5, 12 and hypotenuse 13: F_{2x} = -520 \times \frac{5}{13} = -200\,\mathrm N, F_{2y} = 520 \times \frac{12}{13} = 480\,\mathrm N. No angle is needed.
- \mathbf F_3, by a unit vector from A to B: \overrightarrow{AB} = 0.25\mathbf i - 0.60\mathbf j\ \mathrm m, of length \sqrt{0.25^2 + 0.60^2} = 0.65\,\mathrm m, so \mathbf n_{AB} = 0.3846\mathbf i - 0.9231\mathbf j and \mathbf F_3 = 1300\,\mathbf n_{AB} = 500\mathbf i - 1200\mathbf j\ \mathrm N.
Add the columns:
| Force | F_x (N) | F_y (N) |
|---|---|---|
| \mathbf F_1 | 574.5 | 482.1 |
| \mathbf F_2 | −200.0 | 480.0 |
| \mathbf F_3 | 500.0 | −1200.0 |
| \mathbf R | 874.5 | −237.9 |
R = \sqrt{874.5^2 + 237.9^2} = 906.3\,\mathrm N, at \tan^{-1}(-237.9/874.5) = -15.22^\circ, that is 15.22^\circ below +x. Since R_x > 0 and R_y < 0, the fourth quadrant is correct.
The idea. Meriam (2/3) writes \mathbf F = \mathbf F_x + \mathbf F_y: the parallelogram with perpendicular sides. Each vector component is a scalar times a unit vector, \mathbf F_x = F_x\mathbf i and \mathbf F_y = F_y\mathbf j. The scalar components F_x and F_y carry signs. The magnitude F never does. When a force and its components appear in one diagram, draw one set dashed so they are not read as three separate forces.
The formulas F_x = F\cos\theta, F_y = F\sin\theta hold only when \theta is measured from +x. Meriam's Fig. 2/6 shows the other common cases. If the angle \beta is measured from the y-axis, then F_x = F\sin\beta and F_y = F\cos\beta. If the force leans back past an axis, then F_x = -F\cos\beta. If an angle is given relative to an inclined member, then F_x = F\cos(\beta - \alpha). Sketch the force on its axes and read off which side is adjacent; do not apply a formula blindly.
Marghitu (1.4) gives the same results: in a Cartesian basis the components are unique, \lvert\mathbf v\rvert = \sqrt{v_x^2 + v_y^2}, and sums and differences go component by component. In 1.5 he defines the angle between two vectors (put them tail to tail, 0^\circ to 180^\circ) and the direction cosines \cos\theta_x = F_x/F and \cos\theta_y = F_y/F, which are the same numbers as the components of the unit vector.
The equation (definition of the components, and the mathematical identity that vector addition becomes column sums):
Here \theta is measured counter-clockwise from +x, and its quadrant is set by the signs of F_x and F_y.
Worked number. Meriam's Sample Problem 2/1 has forces of 600, 500 and 800\,\mathrm N at one point. The first is given by an angle: 600\cos35^\circ = 491\,\mathrm N and 600\sin35^\circ = 344\,\mathrm N. The second is given by a 3-4-5 slope: (-400, 300)\,\mathrm N. The third is given by two points: \overrightarrow{AB} = 0.2\mathbf i - 0.4\mathbf j\ \mathrm m, which gives (358, -716)\,\mathrm N. These are the same three ways used above.
Real example. Every cable, strut and contact on a crane boom's tip is added this way. Sample Problem 2/2 combines a 600\,\mathrm{lb} (2669\,\mathrm N) cable pull with an 800\,\mathrm{lb} (3559\,\mathrm N) horizontal pull at the top of a structure. Three methods give the same answer: drawing the parallelogram to scale, the law of cosines, and components. The result is R = 524\,\mathrm{lb} (2330\,\mathrm N) at 48.6^\circ below the horizontal. Components are the method that scales to many forces.
Plausible wrong answer. "\theta = \tan^{-1}(F_y/F_x) from the calculator."
A calculator returns angles between -90^\circ and 90^\circ only. Here that
happens to be right, because R_x > 0. Marghitu's Example 1.1 shows the failure
case. Three plane vectors sum to (-1.565, 0.550), and his atan gives
-19.36^\circ, but the vector lies in the second quadrant at 160.64^\circ. Use
the signs of the components, or atan2.
Check. A force of 750\,\mathrm N points 35^\circ below +x. Write it in \mathbf i, \mathbf j form. (614.4\mathbf i - 430.2\mathbf j\ \mathrm N.)
Second view (Marghitu, MATLAB). A force is a row of numbers, and adding forces is adding rows:
F1 = 750*[cosd(40) sind(40)]; % N
F2 = 520*[-5 12]/13;
nAB = [0.25 -0.60]/norm([0.25 -0.60]);
F3 = 1300*nAB;
R = F1 + F2 + F3 % [874.5 -237.9]
norm(R) % 906.3
atan2d(R(2), R(1)) % -15.22 deg
Narration spine. Three arrows leave A. For each, a dashed rectangle blooms with its corner at the tip, and two signed numbers drop into a table column: an angle for \mathbf F_1, a 5-12-13 triangle for \mathbf F_2, and the segment AB shrinking to a unit arrow for \mathbf F_3. The columns sum, and the resultant arrow is rebuilt from its two totals. A calculator readout flashes -15.22^\circ and is checked against the quadrant.
Bridge. The axes have so far been horizontal and vertical. Nothing requires that. The spike in the puzzle has its own natural direction, and choosing axes along it makes the question almost answer itself.
Nugget 06: Choose the Axes the Problem Hands You
Claim. Axes are a choice of convenience. Put one along the direction the question is about, and the condition you need becomes a single component equation.
The picture. Return to the spike. Put x along the spike and y across it. The requirement "the resultant points along the spike" is now one statement: the y components of the two cable forces cancel.
Placed head to tail, \mathbf P climbs 1.08\,\mathrm{kN} and \mathbf Q drops the same 1.08\,\mathrm{kN}. The triangle closes on the axis, and the closing side is \mathbf T. Its length is the sum of the two horizontal runs, 1.80 + 1.44.
Check with numbers. The cable P runs 200\,\mathrm{mm} along and 120\,\mathrm{mm} up, a length of \sqrt{200^2 + 120^2} = 233.2\,\mathrm{mm}. The cable Q runs 200\,\mathrm{mm} along and 150\,\mathrm{mm} down, a length of 250\,\mathrm{mm}. So
For no sideways pull, P_y = +1.08\,\mathrm{kN}:
That answers the puzzle: P = 2.10\,\mathrm{kN} and T = 3.24\,\mathrm{kN}. The equal-pull guess, P = 1.8\,\mathrm{kN}, gives P_y = 0.926\,\mathrm{kN} against Q_y = 1.08\,\mathrm{kN}. That leaves 0.154\,\mathrm{kN} pulling the spike sideways and only 2.98\,\mathrm{kN} along it. Adding magnitudes (3.9 or 3.6\,\mathrm{kN}) ignores the angles entirely.
The idea. Meriam (2/3) says reference axes are "a matter of arbitrary convenience". Pick them from the geometry: along the dimensions given, along a member, or along the direction a condition is stated in. Sample Problem 2/3 shows how simple a rotated axis can make things. A 500\,\mathrm N force is 250\mathbf i - 433\mathbf j\ \mathrm N in the x-y axes, but with x' chosen along the force it is just 500\,\mathrm N along x' and 0 along y'. The n-t axes of Meriam's problems (normal and tangential to a bar or a path) are the same idea.
With the spike axes, the general case gives two equations in two unknowns. For angles \theta (cable P) and \beta (cable Q) measured from the axis:
so that
The equation (mathematical identity: the component condition, applied to the forces of the problem):
Worked number (second view, Harper's parametric study). Harper asks how the answer changes with the distance d from the eye to the block. Keep the 120\,\mathrm{mm} and 150\,\mathrm{mm} offsets and Q = 1.8\,\mathrm{kN}. Then \tan\theta = 120/d and \tan\beta = 150/d. Substituting into the formula for T above gives T = 2.25\,Q\,d/\sqrt{d^2 + 150^2} (with d in millimetres).
| d (mm) | P (kN) | T (kN) |
|---|---|---|
| 50 | 1.850 | 1.281 |
| 100 | 1.950 | 2.247 |
| 150 | 2.037 | 2.864 |
| 200 | 2.099 | 3.240 |
| 300 | 2.167 | 3.622 |
| very large | 2.25 | 4.05 |
The closer the block, the more of each cable's pull is wasted sideways. At d = 50\,\mathrm{mm} the two cables together pull the spike out with only 1.28\,\mathrm{kN}. Harper's own numbers are in US units: a 400\,\mathrm{lb} (1779\,\mathrm N) cable with offsets of 4 and 6\,\mathrm{in} (101.6 and 152.4\,\mathrm{mm}). At d = 8\,\mathrm{in} (203.2\,\mathrm{mm}) it gives P = 537\,\mathrm{lb} (2389\,\mathrm N) and T = 800\,\mathrm{lb} (3559\,\mathrm N), Meriam's answer to his Problem 2/23.
Q = 1.8; % kN
d = 50:5:300; % mm
theta = atan(120./d); beta = atan(150./d);
P = Q*sin(beta)./sin(theta);
T = Q*(sin(beta).*cot(theta) + cos(beta));
plot(d, P, d, T), xlabel('d (mm)'), ylabel('force (kN)')
Harper's point is that the computer substitutes \theta(d) and \beta(d) for you. There is no need to simplify by hand before plotting.
Real example (one you can see). Pulling a nail with a claw hammer, or a tent peg with a loop of rope, at a steep angle wastes most of the pull sideways. The rope's pull along the peg is the component along the peg, not the pull on the rope.
Plausible wrong answer. "Symmetric cables, so P = Q." The cables are not symmetric: Q leaves at 36.87^\circ and P at 30.96^\circ. The shallower cable must pull harder to match the steeper one's sideways part.
Check. If the block were moved so that both cables passed 150\,\mathrm{mm} from the axis, what would P be? (P = Q = 1.8\,\mathrm{kN}: now the geometry really is symmetric.)
Narration spine. Rotate the whole picture until the spike is horizontal and call that x. Each cable's arrow drops a dashed vertical shadow. Slide \mathbf Q to the tip of \mathbf P and grow P until the two vertical shadows match and the triangle closes on the axis. The closing arrow is T = 3.24\,\mathrm{kN}. Then slide the block towards the spike and let the P and T curves trace out beside it.
Synthesis
The puzzle needed every nugget. The cables are forces with lines of action (Nugget 01). They pull on the spike, which is the body we isolate (Nugget 02). Their lines meet at the eye, so they add there (Nugget 03). We resolve along perpendicular axes (Nugget 05), not oblique ones (Nugget 04). We chose those axes along the spike (Nugget 06), and the answer was one line, \Sigma F_y = 0. The broader insight is that a force has two lives. As a vector it adds by geometry, and that geometry tells us where the resultant acts. As a pair of signed numbers it adds by arithmetic, and that tells us how big it is. Statics uses both: the geometry for lines of action and moments, the numbers for sums.
- Central idea: a force is a sliding vector for rigid-body external effects, and it is added and split by the parallelogram law, which rectangular components reduce to column sums.
- Key relationship: F_x = F\cos\theta, F_y = F\sin\theta and R_x = \Sigma F_x, R_y = \Sigma F_y, with signs set by the quadrant.
- Most important assumption: transmissibility holds only for the external effects on a rigid body. Internal forces change when a force slides along its line.
- Open question: a force slid sideways off its line changes how the body turns. How large is that turning effect, and how does it depend on where the line of action lies? (S04.)
Computed views
The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Marghitu's and Harper's programs; the Python is a translation, and both print the same numbers.
Python (numpy). S03_force_components.py solves the
spike for P and T over Harper's range of d, adds three forces by components, resolves
Sample Problem 2/3's force along x and the oblique y' by solving the 2\times2 system
[\mathbf e_x\ \mathbf e_{y'}]\,\mathbf c = \mathbf F, and finds where \mathbf R_1 and
\mathbf R_2 meet.
Part 1: the spike, Q = 1.8 kN, offsets 120 and 150 mm
d (mm) P (kN) T (kN)
50 1.850 1.281
100 1.950 2.247
150 2.037 2.864
200 2.099 3.240
300 2.167 3.622
1000000 2.250 4.050
equal pulls of 1.8 kN leave 0.154 kN sideways
Part 2: three forces by components
R = [ 874.5 -237.9] N, |R| = 906.3 N, at -15.22 deg
SP 2/1: [[491.0, 344.0], [-400.0, 300.0], [358.0, -716.0]] N
Part 3: oblique axes and concurrent lines
SP 2/3: F = [ 250. -433.] N, along x and y' [1000. -866.] N
SP 2/4: R = 163.4 N, projection on b 144.3 N
lines of R1 and R2 meet at [ 0.4 -1.2] m, R = [ 0. -500.] N
MATLAB (Octave-compatible). S03_force_components.m
does the sweep with element-wise operators and the two oblique and intersection problems with
the backslash operator.
d (mm) P (kN) T (kN)
50 1.850 1.281
100 1.950 2.247
150 2.037 2.864
200 2.099 3.240
300 2.167 3.622
1000000 2.250 4.050
R = [874.5 -237.9] N, |R| = 906.3 N at -15.22 deg
SP 2/3: along x 1000.0 N, along y-prime -866.0 N
lines meet at (0.400, -1.200) m, R = [0 -500] N
Traps
- Sliding a force to change the inside. Transmissibility keeps the wall reaction at 5\,\mathrm{kN}, but the rod's outer half goes from 5\,\mathrm{kN} of tension to none. Use it for external effects only. (From Nugget 01.)
- Sliding a force sideways. Only motion along the line of action is allowed. Any sideways shift changes the turning effect. (From Nugget 01.)
- Weight and normal force as an action–reaction pair. Both act on the same body. The reaction to the Earth's pull on a box is the box's pull on the Earth. (From Nugget 02.)
- Triangle law, wrong line. The triangle drawn at A gives the right 5\,\mathrm{kN} at 53.13^\circ, but on a line 1.6\,\mathrm m from the true one through C. (From Nugget 03.)
- Projection for component. On oblique axes, F\cos\phi_a = 939.7\,\mathrm N is the projection. The component is 742.2\,\mathrm N. (From Nugget 04.)
- F\cos\theta with the angle from the wrong axis. If \beta is measured from y, then F_x = F\sin\beta. Sketch the rectangle first. (From Nugget 05.)
- A negative magnitude. F is never negative; the signs belong to F_x and F_y. Writing F = -200\,\mathrm N for a force pointing left mixes the two. (From Nugget 05.)
- The calculator's quadrant. \tan^{-1}(0.550/{-1.565}) returns -19.36^\circ
for a vector at 160.64^\circ. Use the signs, or
atan2. (From Nugget 05.) - Adding magnitudes. The spike's cables give 3.24\,\mathrm{kN} along the axis, not 2.10 + 1.8 = 3.9\,\mathrm{kN}. (From Nugget 06.)
Self-check
- A 750\,\mathrm N force points down and to the right at 35^\circ below the positive x-axis. Write it in terms of \mathbf i and \mathbf j, and name its scalar and vector components.
- A 2.6\,\mathrm{kN} force points down and to the left along a slope that drops 5 for every 12 horizontally. Write it in terms of \mathbf i and \mathbf j.
- The line of action of a 5\,\mathrm{kN} force runs from A(-4, -1)\,\mathrm m towards B(8, 4)\,\mathrm m. Find its x and y components.
- A bar rises at 60^\circ to the horizontal. A 1.2\,\mathrm{kN} force acts on it horizontally, pointing to the left. Take n along the bar (up the bar) and t perpendicular to it, 90^\circ counter-clockwise from n. Find F_n and F_t.
- Two forces act on a bracket. F_1 = 3.5\,\mathrm{kN} points downward at 20^\circ to the left of the vertical. F_2 = 2\,\mathrm{kN} points downward at an angle \theta to the right of the vertical. Find \theta so that the resultant is vertical, and find the resultant.
- Two people move a sofa. One pulls with 420\,\mathrm N along +x, the other with 300\,\mathrm N at 60^\circ above +x. Find the resultant as a vector, its magnitude, and its angle with +x.
- A 6\,\mathrm{kN} force acts at 40^\circ above +x. Axis a is at 15^\circ and axis b at 75^\circ. Find the components F_a and F_b, and the projections P_a and P_b.
- A tower top A is 40\,\mathrm m above ground. Cable AB goes to a ground anchor 30\,\mathrm m to the left of the tower's base and carries 10\,\mathrm{kN}. Cable AC goes to an anchor on a hillside 45\,\mathrm m to the right and 15\,\mathrm m above the base. Find the tension T in AC that makes the two cables' combined pull on A vertical, and the size of that pull.
- A 6\,\mathrm{kN} force along +x acts at (0, 2)\,\mathrm m, and an 8\,\mathrm{kN} force along -y acts at (3, 0)\,\mathrm m. Find the resultant and the point where its line of action crosses the x-axis.
- A beam carries two upward forces, 400\,\mathrm N and 600\,\mathrm N, 1.5\,\mathrm m apart. Find their resultant and its line by Meriam's trick of adding 200\,\mathrm N and -200\,\mathrm N along the beam.
- Two forces lie in a plane: 40\,\mathrm N at 160^\circ and 25\,\mathrm N at 250^\circ, both measured counter-clockwise from +x. Find the resultant's magnitude and direction. What does a calculator's \tan^{-1} return?
- A 4\,\mathrm{kg} lamp hangs at rest from a cord. List every force on the lamp with its size, classify each (contact or body), and give the reaction to each.
Fully worked solutions
1. F_x = 750\cos35^\circ = 614.4\,\mathrm N and F_y = -750\sin35^\circ = -430.2\,\mathrm N, negative because the force points down. \mathbf F = 614.4\mathbf i - 430.2\mathbf j\ \mathrm N. The scalar components are 614.4\,\mathrm N and -430.2\,\mathrm N. The vector components are \mathbf F_x = 614.4\mathbf i\ \mathrm N and \mathbf F_y = -430.2\mathbf j\ \mathrm N.
2. The slope triangle has legs 12 (horizontal) and 5 (vertical), so its hypotenuse is 13. The force points left and down, so both components are negative: F_x = -2.6 \times \frac{12}{13} = -2.4\,\mathrm{kN} and F_y = -2.6 \times \frac{5}{13} = -1.0\,\mathrm{kN}. So \mathbf F = -2.4\mathbf i - 1.0\mathbf j\ \mathrm{kN}.
3. \overrightarrow{AB} = (8 - (-4))\mathbf i + (4 - (-1))\mathbf j = 12\mathbf i + 5\mathbf j\ \mathrm m, of length \sqrt{144 + 25} = 13\,\mathrm m. Then F_x = 5 \times \frac{12}{13} = 4.615\,\mathrm{kN} and F_y = 5 \times \frac{5}{13} = 1.923\,\mathrm{kN}.
4. \mathbf n = \cos60^\circ\,\mathbf i + \sin60^\circ\,\mathbf j = 0.5\mathbf i + 0.866\mathbf j and \mathbf t = -0.866\mathbf i + 0.5\mathbf j. With \mathbf F = -1.2\mathbf i\ \mathrm{kN}: F_n = \mathbf F\cdot\mathbf n = -1.2 \times 0.5 = -0.600\,\mathrm{kN}, which points down the bar. F_t = \mathbf F\cdot\mathbf t = -1.2 \times (-0.866) = 1.039\,\mathrm{kN}. Check: -0.600\,\mathbf n + 1.039\,\mathbf t = (-0.300 - 0.900)\mathbf i + (-0.520 + 0.520)\mathbf j = -1.2\mathbf i. Because n and t are perpendicular, the components equal the projections.
5. Components: \mathbf F_1 = (-3.5\sin20^\circ, -3.5\cos20^\circ) = (-1.197, -3.289)\,\mathrm{kN} and \mathbf F_2 = (2\sin\theta, -2\cos\theta). A vertical resultant needs R_x = 0, so 2\sin\theta = 1.197, \sin\theta = 0.5985 and \theta = 36.77^\circ. Then R_y = -(3.289 + 2\cos36.77^\circ) = -(3.289 + 1.602) = -4.891\,\mathrm{kN}, so the resultant is 4.891\,\mathrm{kN} straight down.
6. \mathbf R = (420 + 300\cos60^\circ)\mathbf i + 300\sin60^\circ\,\mathbf j = 570\mathbf i + 259.8\mathbf j\ \mathrm N. R = \sqrt{570^2 + 259.8^2} = 626.4\,\mathrm N, at \tan^{-1}(259.8/570) = 24.50^\circ. Both components are positive, so the first quadrant is correct.
7. \mathbf F is 25^\circ from a and 35^\circ from b, and the axes are 60^\circ apart, so the triangle's angle opposite \mathbf F is 120^\circ. By the law of sines, F_a = 6\sin35^\circ/\sin120^\circ = 6 \times 0.5736/0.8660 = 3.974\,\mathrm{kN} and F_b = 6\sin25^\circ/\sin120^\circ = 6 \times 0.4226/0.8660 = 2.928\,\mathrm{kN}. The projections are P_a = 6\cos25^\circ = 5.438\,\mathrm{kN} and P_b = 6\cos35^\circ = 4.915\,\mathrm{kN}. The components and projections differ because the axes are not perpendicular.
8. Take A as the origin. \overrightarrow{AB} = -30\mathbf i - 40\mathbf j\ \mathrm m, of length 50\,\mathrm m, so \mathbf T_{AB} = 10(-0.6\mathbf i - 0.8\mathbf j) = -6\mathbf i - 8\mathbf j\ \mathrm{kN}. \overrightarrow{AC} = 45\mathbf i - 25\mathbf j\ \mathrm m, of length \sqrt{2025 + 625} = 51.48\,\mathrm m. A vertical resultant needs R_x = 0: -6 + T \times 45/51.48 = 0, so T = 6 \times 51.48/45 = 6.864\,\mathrm{kN}. Its vertical part is 6.864 \times 25/51.48 = 3.333\,\mathrm{kN} downward, so R = 8 + 3.333 = 11.33\,\mathrm{kN} downward.
9. \mathbf R = 6\mathbf i - 8\mathbf j\ \mathrm{kN}, so R = 10\,\mathrm{kN} at \tan^{-1}(-8/6) = -53.13^\circ. The lines y = 2 and x = 3 meet at (3, 2), so \mathbf R acts through (3, 2)\,\mathrm m. Moving along \mathbf R from there, y falls by 8 for every 6 in x. Reaching y = 0 takes a fall of 2, so x = 3 + 2 \times \frac{6}{8} = 4.5\,\mathrm m. The line crosses the x-axis at (4.5, 0)\,\mathrm m.
10. Put the 400\,\mathrm N force at x = 0 and the 600\,\mathrm N force at x = 1.5\,\mathrm m. Add 200\mathbf i\ \mathrm N at the first and -200\mathbf i\ \mathrm N at the second. \mathbf R_1 = 200\mathbf i + 400\mathbf j through (0, 0): x = 200s, y = 400s. \mathbf R_2 = -200\mathbf i + 600\mathbf j through (1.5, 0): x = 1.5 - 200t, y = 600t. Equal y gives t = \frac{2}{3}s. Equal x gives 200s = 1.5 - 133.3s, so s = 0.0045. The lines meet at (0.9, 1.8)\,\mathrm m. \mathbf R = 1000\mathbf j\ \mathrm N, acting upward on the line x = 0.9\,\mathrm m, 0.9\,\mathrm m from the smaller force. Check: 400 \times 0.9 = 600 \times 0.6 = 360.
11. \mathbf F_1 = 40(\cos160^\circ, \sin160^\circ) = (-37.59, 13.68)\,\mathrm N and \mathbf F_2 = 25(\cos250^\circ, \sin250^\circ) = (-8.55, -23.49)\,\mathrm N. So \mathbf R = -46.14\mathbf i - 9.81\mathbf j\ \mathrm N and R = \sqrt{46.14^2 + 9.81^2} = 47.17\,\mathrm N. The calculator's \tan^{-1}(-9.81/{-46.14}) = \tan^{-1}(0.2127) returns 12.01^\circ. Both components are negative, so the vector is in the third quadrant: 180^\circ + 12.01^\circ = 192.01^\circ.
12. Two forces act on the lamp. Its weight, W = 4 \times 9.81 = 39.24\,\mathrm N downward, is a body force exerted by the Earth. The cord's pull, 39.24\,\mathrm N upward, is a contact force (equal to W because the lamp is at rest). The reaction to the weight is the lamp's pull on the Earth, 39.24\,\mathrm N upward, acting on the Earth. The reaction to the cord's pull is the lamp's pull on the cord, 39.24\,\mathrm N downward, acting on the cord.