Statics: Equivalent Forces, Equilibrium, Trusses and Frames, Friction and Stability

This chapter answers the first half of Section XE3.1, Mechanics of Rigid Bodies, which belongs to XE3 Solid Mechanics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam. The syllabus text is: “Equivalent forces and moments; equilibrium equations; analysis of determinate trusses and frames; sliding and sticking friction; the principle of minimum potential energy and its relation to stable equilibrium.” It teaches force-couple reduction, support reactions, trusses and frames, Coulomb friction with tipping, belts and wedges, and stability from potential energy. The numericals are couples and resultants, beam reactions, member forces, the friction a ladder needs, belt tensions and the critical spring stiffness of an inverted rod. Take g = 9.81 m/s² wherever weight appears.

1. Moments, couples and equivalent force systems

The moment of a force F about a point O is M_O = r × F. In a plane its size is F·d, where d is the perpendicular distance from O to the line of action, and anticlockwise is taken as positive. On a rigid body a force may slide along its line of action without changing its effect (transmissibility), but moving it off that line changes its moment. A couple is a pair of equal, opposite, parallel forces on different lines: the net force is zero and the moment F·d (d is the separation of the lines) is the same about every point, so a couple is a free vector that can be placed anywhere in its plane.

Force-couple replacement. A force F acting at A is equivalent to the same force F acting at O together with a couple of moment F·d, where d is the perpendicular distance from O to the original line, in the sense of the moment of F about O. Any coplanar system therefore reduces to a resultant R = ΣF at O and a resultant moment M_O = ΣM_O (moments of the forces plus any couples). If R ≠ 0 the system is one single force R whose line is displaced by d = M_O/R from O, to the side that reproduces the sense of M_O. If R = 0 and M_O ≠ 0 it is a pure couple. A distributed load is replaced by its resultant, which equals the area of the load diagram and acts through its centroid: a uniform load w over length L gives wL at mid-length, and a triangular load rising from zero to w₀ over length L gives w₀L/2 at L/3 from the large end.

Worked example. A 50 N downward force acts at x = 0.6 m on a beam, and a 20 N·m anticlockwise couple also acts on it. About x = 0: R = 50 N down and M_O = −50(0.6) + 20 = −10 N·m, i.e. 10 N·m clockwise. A single 50 N downward force at d = 10/50 = 0.20 m to the right of O gives that same clockwise moment (a downward force to the right of O turns the beam clockwise), so this is the equivalent single force.

⚠️ Couples are free, forces are not
A couple has the same moment about every point, but the moment of a force depends on the point chosen. Sliding a force along its line is valid for a rigid body only; in a deformable body the internal stresses change. The commonest error is to move a force off its line and forget the couple F·d.

2. Free-body diagrams, equilibrium equations and support reactions

A body is in equilibrium when ΣF = 0 and ΣM_O = 0 about any point O. In 2-D this gives three independent equations (ΣFx = 0, ΣFy = 0, ΣM = 0); two moment equations plus one force equation, or three moment equations about non-collinear points, serve equally well (for the first set the force direction must not be perpendicular to the line joining the two moment points). In 3-D there are six: ΣFx, ΣFy, ΣFz, ΣMx, ΣMy, ΣMz = 0. The method starts with the free-body diagram (FBD): isolate the body, draw its weight at the centre of gravity, show every applied load, and replace each support or contact by the reaction it can exert, in an assumed sense (a negative answer simply reverses it).

Support reactions
Support or contactUnknownsDirection
Roller, rocker, smooth surface1 force (2-D)Normal to the surface
Link or cable1 forceAlong the member; a cable pulls only
Pin (hinge)2 force componentsRx and Ry; no moment
Fixed (built-in) supportRx, Ry and a moment M (3)Prevents rotation too
Ball-and-socket (3-D)3 forcesNo moment
Fixed support (3-D)3 forces and 3 momentsSix unknowns
  • Two-force member: loaded at only two points and carrying no couple, it exerts forces that are equal, opposite and along the line joining the two points.
  • Three-force member: the three forces are concurrent, or all parallel.
  • Determinacy: as many unknown reactions as independent equations (three in 2-D) means statically determinate; more unknowns means indeterminate; and an equal count with reactions that are all concurrent or all parallel is an improper constraint, an unstable body.

Worked example. A simply supported beam of span 6 m (pin at A, roller at B) carries 5 kN/m over the whole span and a 10 kN point load 2 m from A. The uniform load has resultant 5 × 6 = 30 kN at 3 m. ΣM_A = 0: R_B(6) = 10(2) + 30(3) = 110, so R_B = 18.33 kN. ΣFy = 0: R_A = 40 − 18.33 = 21.67 kN. Check with moments about B: R_A(6) = 10(4) + 30(3) = 130, which agrees.

⚠️ Counting reactions is not enough
A plate held by a pin and a roller whose line of action passes through the pin has three unknowns and three equations, yet it can still rotate about the pin: the three reactions are concurrent and cannot balance a moment about that point. Choose moment centres where two unknown lines cross, so one equation yields one unknown.

3. Determinate trusses and frames

A plane truss is idealised with pin joints, loads and reactions applied only at joints, and weightless straight members. Every member is then a two-force member carrying an axial force only: tension (pulling on the joints at its ends) or compression (pushing). With m members, j joints and r reaction components there are m + r unknowns and 2j joint equations. m + r = 2j means statically determinate (if the members are properly arranged), m + r < 2j is a mechanism, and m + r > 2j is statically indeterminate. A simple truss grows from one triangle by adding two members per new joint, so m = 2j − 3, and with r = 3 it is determinate.

  • Method of joints: take a joint with at most two unknown member forces, apply ΣFx = 0 and ΣFy = 0 (tension positive, drawn away from the joint) and move on joint by joint. It yields every member.
  • Method of sections: pass an imaginary cut through at most three members whose lines are not all concurrent or all parallel, treat either side as a rigid body and write three equilibrium equations. Taking moments about the point where two cut members meet gives the third at once, so one member far from the supports is found without working along the truss.
  • Zero-force members (no calculation needed):
    (i) two non-collinear members meeting at a joint with no load and no support are both zero;
    (ii) at an unloaded joint where three members meet and two are collinear, the third is zero. They are kept for stability and other load cases.

Worked example (joints). Pin at A (0, 0), roller at B (4 m, 0), apex C (2 m, 3 m), load 12 kN down at C. Check: m + r = 3 + 3 = 2(3). By symmetry R_A = R_B = 6 kN. At joint A the member AC has tan θ = 3/2 to the horizontal (sin θ = 0.832, cos θ = 0.555). ΣFy: 6 = F_AC sin θ, so F_AC = 7.21 kN compression. ΣFx: F_AB = F_AC cos θ = 4.00 kN tension.

A frame contains at least one multi-force member, loaded at three or more points or by a non-axial load, so it is not a truss. Draw an FBD of the whole frame for the external reactions (it often gives only some of them), then dismember it: draw each member separately, show the force components at every pin, and apply Newton's third law (the force on one member at a pin equals and opposes that on its partner, drawn consistently). Each rigid member supplies three equations in 2-D, and a two-force member inside a frame has one unknown, its axial force, so spot those first.

⚠️ A loaded member is not a two-force member
A member that carries a load along its length, such as a beam with a uniform load, bends and is not a two-force member; treating it as one is the standard frame error.

4. Sliding and sticking friction, tipping, belts and wedges

Dry (Coulomb) friction acts along the contact tangent, opposing relative sliding or its tendency. While the surfaces stick, F is whatever equilibrium demands, up to the limit F ≤ μ_s N; at impending slip F = μ_s N exactly; once sliding, kinetic friction is F = μ_k N, normally with μ_k < μ_s, and nearly independent of contact area and speed. N must come from equilibrium, not be assumed equal to W. The friction angle φ = tan⁻¹ μ is the greatest angle the total contact reaction can make with the normal. On an incline of angle θ a block sticks when tan θ ≤ μ_s whatever its weight, and with a force P parallel to the plane it is about to move up when P = W(sin θ + μ_s cos θ).

Tipping versus sliding. A crate of weight W and base width b is pushed by a horizontal force P at height a above the floor. It slides when P = μ_s W. It tips about its leading bottom edge when the normal force has moved to that edge, so P·a = W·b/2 and P = Wb/(2a). Whichever force is smaller occurs first: the crate slides first if μ_s < b/(2a) and tips first if μ_s > b/(2a).

Worked example (ladder). A 5 m uniform ladder of mass 20 kg rests with its foot 3 m from a smooth vertical wall, so its top is 4 m up. W = 20 × 9.81 = 196.2 N acts 1.5 m horizontally from the foot. Moments about the foot: N_w(4) = 196.2(1.5), so N_w = 73.6 N (a smooth wall gives a horizontal force only). ΣFy: N = 196.2 N. ΣFx: F = N_w = 73.6 N. No slip needs F ≤ μ_s N, so μ_s ≥ 73.6/196.2 = 0.375.

Belt (or rope) over a fixed drum at impending slip: T₁/T₂ = e^(μβ), where β is the angle of contact in radians (1.5 turns is 3π) and T₁ is the larger tension, on the side towards which the belt tends to slip relative to the drum. It does not depend on the drum radius. Wedge rule: draw the friction at each face opposing the relative slip of the two touching faces, put N perpendicular to each face, and write equilibrium for each body. A symmetric wedge of half-angle α driven into a slot stays in place (self-locks) when tan α ≤ μ_s, because the outward push 2N sin α is then no larger than the grip 2μ_s N cos α.

⚠️ Friction is not always μN
In a static problem F = μ_s N holds only when slip is about to occur; otherwise F follows from ΣF = 0, and a push below the limit is met by an equal friction force. Friction at a rolling contact without slip is static and does no work.

5. Minimum potential energy and stability of equilibrium

When only conservative forces do work (weights and springs), the system has a potential energy V(q) in terms of a coordinate q that fixes its configuration: V = m g y_G (height of the centre of gravity above a fixed datum, y up) plus spring energy ½kδ² (δ is the extension from the unstretched length; ½k_tθ² for a torsion spring). Equilibrium needs dV/dq = 0, which is the same as zero virtual work for every admissible displacement. The stability of that position follows from the curvature of V.

Stability from V(q) at an equilibrium position
TypeTestAfter a small disturbancePicture
Stabled²V/dq² > 0 (V is a minimum)Returns to the positionBall in a bowl
Unstabled²V/dq² < 0 (V is a maximum)Moves further awayBall on a hilltop
NeutralV constant, all derivatives zeroStays at the new positionBall on a level floor

If d²V/dq² = 0, examine higher derivatives: the first non-zero one being even and positive means stable, even and negative means unstable, and odd means unstable on one side (semi-stable). Worked example (spring-loaded inverted rod). A uniform rod of mass m and length L is hinged at its lower end and held vertical (θ = 0) by a torsion spring that is unstressed at θ = 0. V = ½k_tθ² + mg(L/2) cos θ, so dV/dθ = k_tθ − (mgL/2) sin θ, which vanishes at θ = 0 for every k_t. At that position d²V/dθ² = k_t − mgL/2. The vertical position is stable if k_t > mgL/2, unstable if k_t < mgL/2, and the critical stiffness is mgL/2. For m = 2 kg and L = 1 m, k_cr = 2(9.81)(1)/2 = 9.81 N·m/rad. For a point mass m on a light rod the critical value is mgL, so do not mix the two cases.

Rocking bodies. A body with a curved base rocking on a horizontal plane is stable if its centre of gravity lies below the centre of curvature of the contact. A solid hemisphere resting on its curved face has its centre of gravity 3R/8 below the centre of the sphere, so V = mg[R − (3R/8) cos θ] and d²V/dθ² = 3mgR/8 > 0: stable.

⚠️ A stationary V gives equilibrium only
dV/dq = 0 locates equilibrium positions; the sign of d²V/dq² decides stability. Measure heights from a fixed datum with y upward, so that the gravity term is +mgy: a sign slip turns a stable equilibrium into an unstable one.

Key takeaways

  • A force equals the same force at another point plus a couple F·d, and a couple has the same moment about every point. A coplanar system reduces to R = ΣF and M_O = ΣM_O; a distributed load acts as its area at its centroid.
  • Equilibrium gives three equations in 2-D and six in 3-D from a free-body diagram. Two-force members are axial and three-force members are concurrent; equal counts do not prove determinacy if the reactions are concurrent or parallel.
  • For a plane truss m + r = 2j means determinate. Joints give two equations each and a section three; frames are dismembered using Newton's third law, and a loaded member is a beam, not a two-force member.
  • Friction is F ≤ μ_s N while sticking, F = μ_s N only at impending slip, and μ_k N while sliding. A crate tips before sliding if μ_s > b/(2a). A belt gives T₁/T₂ = e^(μβ) with β in radians.
  • Equilibrium of a conservative system is dV/dq = 0, stable if d²V/dq² > 0, unstable if negative, neutral if V is constant. A uniform inverted rod on a torsion spring is stable when k_t > mgL/2.
  • Exam traps: moving a force off its line without its couple, taking N = W or F = μN while sticking, mgL instead of mgL/2 for a uniform rod, and degrees in the belt exponent.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A force of 40 N acts at point A of a rigid body, and its line of action is 0.5 m (perpendicular distance) from a point O. The force is to be replaced by an equivalent system acting at O. What is needed?

    1. The same 40 N force at O alone, because a force on a rigid body may be moved to any point without changing its effect
    2. The same 40 N force at O together with a couple of 20 N·m, in the same sense as the moment of the original force about O
    3. The same 40 N force at O together with a couple of 20 N·m, in the sense opposite to the moment of the original force about O
    4. A force of 20 N at O together with a couple of 40 N·m, because the force and the distance simply exchange roles
    Show answer

    Answer: B — The same 40 N force at O together with a couple of 20 N·m, in the same sense as the moment of the original force about O

    Shifting a force from its line of action to O keeps the same force F and adds a couple equal to the moment of F about O, here 40 × 0.5 = 20 N·m, in the same sense as that moment, so the turning effect on the body is unchanged. Moving the force without the couple is the tempting answer, but a force may be moved freely only along its own line of action. The opposite sense would change the turning effect, and the force magnitude does not change when it is moved.

  2. Two parallel and opposite forces of 30 N each act along lines 0.4 m apart on a plate, forming a couple. A point P in the plane of the plate lies outside the pair, 1.0 m from the nearer line of action. The moment of the couple about P is:

    1. 42 N·m, since the farther force is 1.4 m from P
    2. 30 N·m, since the nearer force is 1.0 m from P
    3. Zero, since the two forces are equal and opposite and cancel
    4. 12 N·m, the same as about any other point of the plane
    Show answer

    Answer: D — 12 N·m, the same as about any other point of the plane

    About P the two forces give 30(1.4) − 30(1.0) = 12 N·m, which is F·d = 30 × 0.4, independent of where P lies: a couple has the same moment about every point. Taking the moment of only one force gives 42 or 30 N·m, which is incorrect because the other force also has a moment about P. The forces cancel as a resultant force, but not as a moment, so the net moment is not zero.

  3. A rigid rectangular plate in a vertical plane is held by a pin at corner A and by a straight link attached at corner B, and the line of the link passes through the pin at A. A vertical load acts at the centre of the plate. Which statement is correct?

    1. The three reaction components are concurrent at A, so none can balance the load's moment about A: the plate is improperly constrained despite three unknowns
    2. The plate is statically determinate, because there are three unknown reactions and three independent equilibrium equations available in the plane
    3. The plate is statically indeterminate to the first degree, because the link and the pin both resist force along the same line through A
    4. The plate is a mechanism with two degrees of freedom, because a link whose line passes through the pin adds no restraint
    Show answer

    Answer: A — The three reaction components are concurrent at A, so none can balance the load's moment about A: the plate is improperly constrained despite three unknowns

    The pin supplies two force components at A and the link force passes through A, so none of the three reactions has a moment about A. The load's moment about A is then unbalanced and the plate turns about A; its one degree of freedom (not two) is the rotation. Equal counts (3 = 3) are the tempting argument, but determinacy also needs reactions that are neither all concurrent nor all parallel. It is not indeterminate: fewer, not more, constraints than needed act in the rotation sense.

  4. A cantilever beam of length 4 m is fixed at its left end and carries a downward distributed load that decreases linearly from 6 kN/m at the fixed end to zero at the free end. The magnitude of the fixed-end moment is ____ kN·m.

    Numerical answer — type the value.

    Show answer

    Answer: 16

    The resultant is the area of the load triangle: ½ × 6 × 4 = 12 kN. It acts at the centroid, one third of the length from the large end: 4/3 m from the fixed end. Moment = 12 × 4/3 = 16 kN·m. Treating the load as uniform 6 kN/m would give 48 kN·m, and placing the resultant at mid-length would give 24 kN·m.

  5. Which of the following statements about the equilibrium of rigid bodies in a plane are correct?

    1. A rigid body in a plane has three independent equilibrium equations, and they may be replaced by three moment equations about non-collinear points
    2. A member in equilibrium under three non-parallel coplanar forces has the lines of action of those forces concurrent at one point
    3. The moment of a couple about a point increases as the point is taken farther from the couple
    4. In a plane problem a pin support gives two reaction components and a built-in support gives two forces and a moment
    Show answer

    Answer: A — A rigid body in a plane has three independent equilibrium equations, and they may be replaced by three moment equations about non-collinear points; B — A member in equilibrium under three non-parallel coplanar forces has the lines of action of those forces concurrent at one point; D — In a plane problem a pin support gives two reaction components and a built-in support gives two forces and a moment

    Three independent equations in 2-D can be chosen as three moment equations about non-collinear points. A three-force member needs concurrent forces, otherwise their moments about the intersection of two of them cannot sum to zero. A pin gives Rx and Ry, and a built-in support adds a reactive moment. The moment of a couple is F·d about every point, so it does not grow with distance from the couple.

  6. A plane pin-jointed truss has 13 members and 8 joints and is supported by one pin and one roller. It is:

    1. A mechanism, because a truss with 8 joints needs 2 × 8 = 16 members to be rigid
    2. Statically indeterminate to the first degree, because the unknowns exceed the number of joints
    3. Statically determinate, because m + r = 13 + 3 = 16 = 2j
    4. Statically indeterminate to the second degree, because the support has three reaction components
    Show answer

    Answer: C — Statically determinate, because m + r = 13 + 3 = 16 = 2j

    The unknowns are the 13 member forces plus the 3 reaction components, 16 in all, and the joint equations number 2 × 8 = 16, so m + r = 2j and the truss is determinate when properly arranged. A truss needs 2j − 3 = 13 members when the supports provide three reactions, not 2j; that is why a count of 16 members would be wrong. The count of unknowns is compared with 2j, not with j.

  7. A plane pin-jointed truss is built from equilateral triangles of side 2 m; take √3 = 1.732. Its bottom joints are at (0, 0), (2, 0), (4, 0), (6, 0) and (8, 0) m and its top joints at (1, 1.732), (3, 1.732), (5, 1.732) and (7, 1.732) m. Members join consecutive bottom joints, consecutive top joints, and each top joint to the two bottom joints nearest it. The truss has a pin at (0, 0) and a roller at (8, 0), and carries one 10 kN vertical load at (4, 0). The force in the top-chord member joining (3, 1.732) and (5, 1.732) is ____ kN in magnitude, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 11.55

    Check: m = 4 + 3 + 8 = 15, j = 9, r = 3, and 15 + 3 = 18 = 2j. By symmetry each reaction is 5 kN. A vertical cut at x = 3.5 m crosses only the top chord (3 to 5), the diagonal from (3, 1.732) to (4, 0) and the bottom chord (2 to 4). Take the left part and moments about (4, 0), where the diagonal and bottom chord meet: the 5 kN reaction acts 4 m to the left, giving 20 kN·m, which the top-chord force F balances at arm 1.732 m: F × 1.732 = 20, so F = 11.55 kN, compressive.

  8. At a joint of a plane truss there is no external load and no support. Two members meet at the joint at an angle other than 0° and 180°. The forces in the two members are:

    1. Equal and opposite, one tension and one compression, which keeps the joint in equilibrium at any angle
    2. Both zero, because no two forces along different lines can balance with nothing else on the pin
    3. Equal in magnitude, with a value that depends on the angle between the members
    4. Undetermined from the joint alone, because the two joint equations then give only one relation
    Show answer

    Answer: B — Both zero, because no two forces along different lines can balance with nothing else on the pin

    With only two forces along different lines, ΣF = 0 at the pin can hold only if both vanish: a force along one line has a component perpendicular to the other member that nothing balances. Equal and opposite forces would have to be collinear, so that arrangement is impossible for non-collinear members. This is the first zero-force rule, and it needs no angle.

  9. Which of the following statements about trusses and frames are correct?

    1. In the method of sections, moments about the point where two cut members meet give the force in the third cut member directly
    2. At a pin joining two frame members, the force that one member exerts on the other is equal and opposite to the force the other exerts on it
    3. A beam carrying a uniformly distributed load along its length may be treated as a two-force member in a frame
    4. A truss with properly arranged members and m + r = 2j is statically determinate
    Show answer

    Answer: A — In the method of sections, moments about the point where two cut members meet give the force in the third cut member directly; B — At a pin joining two frame members, the force that one member exerts on the other is equal and opposite to the force the other exerts on it; D — A truss with properly arranged members and m + r = 2j is statically determinate

    Moments about the meeting point of two cut members remove those two unknowns. Newton's third law governs the pin forces between frame members. m + r = 2j with a proper arrangement means determinate. A member with a load along its length bends, since it carries forces at more than two points, so it is not a two-force member.

  10. A 10 kg block rests on a rough horizontal floor with μ_s = 0.4 and μ_k = 0.3. Take g = 9.81 m/s². A horizontal push of 30 N is applied to it. The friction force acting on the block is:

    1. 30 N, equal to the push, because the block sticks below the 39.2 N limit
    2. 39.2 N, equal to the limiting static friction μ_s N, which always acts while there is a push
    3. 29.4 N, equal to the kinetic friction μ_k N, because the block is just starting to move
    4. Zero, because the push of 30 N is smaller than the weight of the block, 98.1 N
    Show answer

    Answer: A — 30 N, equal to the push, because the block sticks below the 39.2 N limit

    Here N = 98.1 N, so the limiting static friction is 0.4 × 98.1 = 39.2 N. The push of 30 N is below that limit, so the block sticks and friction equals the push, 30 N, from ΣFx = 0. Writing F = μ_s N (39.2 N) is the tempting slip: that value holds only at impending slip. Kinetic friction 29.4 N applies only once sliding. Comparing the push with the weight has no meaning for horizontal friction.

  11. A uniform rectangular crate of weight W has base width 0.8 m and height 1.6 m and rests on a floor where μ_s = 0.45. A horizontal force P is applied to a vertical side face at 1.2 m above the floor, in the direction that tends to tip the crate about its lower leading edge, and is increased slowly from zero. What happens first?

    1. The crate slides when P reaches 0.45 W, because friction is overcome before the crate can rotate at all
    2. The crate slides and tips together, because at a force of 0.45 W both limiting conditions are reached at once
    3. The crate neither slides nor tips, because the applied force must exceed the weight W to move the crate
    4. The crate tips about the leading bottom edge at P = 0.333 W, which is below the sliding force 0.45 W
    Show answer

    Answer: D — The crate tips about the leading bottom edge at P = 0.333 W, which is below the sliding force 0.45 W

    Sliding needs P = μ_s W = 0.45 W. Tipping needs the normal force to have moved to the leading edge, so P × 1.2 = W × 0.4, giving P = 0.333 W. The smaller force occurs first, so the crate tips. Equivalently μ_s = 0.45 exceeds b/(2a) = 0.333. The crate that slides first would need μ_s below 0.333, so the answer that has it sliding first reverses the comparison.

  12. A uniform ladder of length 5 m and mass 20 kg leans against a smooth vertical wall, with its foot on a rough horizontal floor 3 m from the wall. A person of mass 60 kg stands on the ladder at a point 3 m along its length from the foot. Take g = 9.81 m/s². The minimum coefficient of static friction at the floor that prevents slipping is ____, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.43

    The ladder makes cos θ = 3/5 with the floor and its top is 4 m up. Floor normal force N = (20 + 60)(9.81) = 784.8 N. Moments about the foot, with a horizontal wall force N_w: N_w(4) = 196.2(2.5 × 0.6) + 588.6(3 × 0.6) = 294.3 + 1059.5 = 1353.8, so N_w = 338.4 N. Friction at the floor equals N_w, so μ_s ≥ 338.4/784.8 = 0.431, which is 0.43.

  13. A rope passes over a fixed rough drum with μ = 0.3 and is on the verge of slipping. When the angle of contact is 180° the ratio of the larger to the smaller tension is e^(0.3π) ≈ 2.57. If the rope is wound through 360° instead, all else unchanged, the ratio becomes about:

    1. 5.13, twice as large, because the angle of contact has doubled
    2. 2.57, unchanged, because the ratio depends only on μ and not on the angle
    3. 6.59, the square of 2.57, because the exponent doubles
    4. 3.57, one more than before, because each half-turn adds a fixed tension
    Show answer

    Answer: C — 6.59, the square of 2.57, because the exponent doubles

    T₁/T₂ = e^(μβ). Doubling β from π to 2π doubles the exponent, from 0.942 to 1.885, so the ratio is e^1.885 = 6.59, the square of 2.57. Doubling the ratio (5.13) treats the relation as linear, which is the tempting slip. The ratio does depend on the contact angle, and it is multiplicative, not additive, so adding one is also wrong.

  14. For a conservative system described by a single coordinate q with potential energy V(q), an equilibrium position q₀ is stable if:

    1. dV/dq = 0 and d²V/dq² > 0 at q₀, so that V has a local minimum there
    2. dV/dq = 0 and d²V/dq² < 0 at q₀, so that V has a local maximum there
    3. V(q₀) = 0 and dV/dq > 0 at q₀, so that the energy rises on both sides
    4. d²V/dq² = 0 and dV/dq ≠ 0 at q₀, so that the curvature vanishes
    Show answer

    Answer: A — dV/dq = 0 and d²V/dq² > 0 at q₀, so that V has a local minimum there

    Equilibrium needs dV/dq = 0, and a minimum of V (positive second derivative) is the stable case: a small displacement raises the energy and the restoring force returns the system. A maximum is unstable. A non-zero first derivative means it is not an equilibrium position at all, and the value of V itself, being only a matter of datum, says nothing about stability.

  15. The potential energy of a system is V(x) = 2x³ − 9x² + 12x joules, with x in metres, and only conservative forces act. Which statement about its equilibrium positions is correct?

    1. x = 1 m is stable and x = 2 m is unstable, since V is larger at x = 1 m
    2. Both x = 1 m and x = 2 m are stable, since V rises on moving away from each
    3. Both x = 1 m and x = 2 m are unstable, since the cubic term dominates V
    4. x = 1 m is unstable and x = 2 m is stable, since V″ is −6 and +6 there
    Show answer

    Answer: D — x = 1 m is unstable and x = 2 m is stable, since V″ is −6 and +6 there

    V′ = 6x² − 18x + 12 = 6(x − 1)(x − 2), so equilibrium is at x = 1 and x = 2. V″ = 12x − 18 gives −6 at x = 1 (a local maximum, unstable) and +6 at x = 2 (a local minimum, stable). Comparing the values of V is misleading: V(1) = 5 J is larger than V(2) = 4 J, but that is a local maximum next to a local minimum, and stability is decided by curvature at each point.

  16. Which of the following statements about equilibrium and stability are correct?

    1. A ball resting on the top of a fixed smooth sphere is in unstable equilibrium
    2. A uniform rod of mass m and length L, hinged at its lower end and held vertical by a torsion spring that is unstressed when vertical, is stable if the spring stiffness exceeds mgL/2
    3. In neutral equilibrium V is constant over a range of q, so dV/dq and d²V/dq² are both zero there
    4. A position where dV/dq = 0 is necessarily a stable equilibrium position
    Show answer

    Answer: A — A ball resting on the top of a fixed smooth sphere is in unstable equilibrium; B — A uniform rod of mass m and length L, hinged at its lower end and held vertical by a torsion spring that is unstressed when vertical, is stable if the spring stiffness exceeds mgL/2; C — In neutral equilibrium V is constant over a range of q, so dV/dq and d²V/dq² are both zero there

    A ball on top of a sphere sits at a maximum of its gravitational energy, so it is unstable. For the rod, V = ½k_tθ² + mg(L/2) cos θ gives V″(0) = k_t − mgL/2, positive for stiffness above mgL/2. In neutral equilibrium V does not change with q, so all its derivatives vanish. dV/dq = 0 alone includes maxima, which are unstable, so it does not guarantee stability.

  17. A uniform rod of mass 3 kg and length 0.8 m is hinged at its lower end and held vertical by a torsion spring that exerts no torque when the rod is vertical. Take g = 9.81 m/s². The smallest torsional stiffness for which the vertical position is stable is ____ N·m/rad, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 11.77

    V = ½k_tθ² + mg(L/2) cos θ, so V″(0) = k_t − mgL/2. Stability needs V″ > 0, so the critical stiffness is mgL/2 = 3 × 9.81 × 0.8/2 = 11.772 N·m/rad, i.e. 11.77. Using mgL for the rod, as for a point mass at the tip, would give 23.54.