Three concurrent forces keep a particle in equilibrium. By Lami’s theorem, each force is proportional to:

the sine of the angle between the other two forces

the cosine of the angle between the other two forces

the sine of the angle between the other two forces

the sine of the angle it makes with the horizontal

the angle between the other two forces

At an unloaded joint of a truss, three members meet, two of which are collinear. The force in the third member is:

equal to the force in either collinear member

indeterminate without the loads elsewhere

Which of the following statements about dry friction are correct? (Select all that apply.)

The angle of repose equals the angle of friction

Limiting friction is independent of the apparent area of contact

The angle of repose equals the angle of friction

Limiting friction is independent of the apparent area of contact

The kinetic coefficient of friction is generally greater than the static coefficient

The friction force on a body at rest always equals μN

Which of the following statements about shear force and bending moment are correct? (Select all that apply.)

The bending moment is a maximum where the shear force passes through zero

Under a uniformly distributed load the bending moment varies parabolically

A point of contraflexure is where the bending moment changes sign

The bending moment is a maximum where the shear force passes through zero

Under a uniformly distributed load the bending moment varies parabolically

For a cantilever with a point load at its free end, the bending moment is greatest at the free end

A point of contraflexure is where the bending moment changes sign

Engineering Mechanics for Mining: Force Systems, Equilibrium, Trusses and Frames, Friction, Particle Dynamics and Beams

The first of two chapters for Section 3 of the GATE Mining Engineering (MN) paper, Geomechanics and Ground Control. The section opens with a sub-heading that is a subject of its own — Engineering Mechanics — and it is the toolkit the rest of the section, and much of the machinery section, stands on, so it has this chapter; the rock mechanics and ground control that follow have the second. The syllabus names six items and this chapter takes them in order: equivalent force systems, the equations of equilibrium, two-dimensional frames and trusses, friction forces, particle kinematics and dynamics, and beam analysis. The examples are mining ones where they can be — a cage decelerating in a shaft, a rope on a drum, a mine car on an incline, a roof beam under its own weight.

1. Equivalent force systems and the equations of equilibrium

Any system of forces on a rigid body can be replaced by a single resultant force acting at a chosen point plus a couple, and two systems are equivalent if they have the same resultant and the same moment about every point. The moment of a force about a point is force × perpendicular distance, and Varignon’s theorem says the moment of the resultant equals the sum of the moments of its components. A couple — two equal, opposite, parallel forces — has zero resultant and the same moment Fd about every point, so it can be balanced only by another couple.

A body in the plane is in equilibrium when ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point — three equations, so a free body can yield at most three unknown reactions. For three concurrent forces in equilibrium, Lami’s theorem gives each force proportional to the sine of the angle between the other two: P/sin α = Q/sin β = R/sin γ. The first step of every problem is the free-body diagram: isolate the body, and draw every force the surroundings exert on it, including the reactions at supports — a roller gives one force normal to its surface, a pin two components, and a fixed support two components and a moment.

🧠 Locating a resultant of parallel forces

Parallel forces of 30 N at x = 0 and 50 N at x = 4 m have a resultant of 80 N, and Varignon puts it where 80x̄ = 30 × 0 + 50 × 4, i.e. x̄ = 2.5 m — nearer the larger force, as it must be.

2. Two-dimensional trusses and frames

A truss is an assembly of straight two-force members pinned at their ends and loaded only at the joints, so every member is in pure tension or compression. A plane truss with j joints and m members is statically determinate and just rigid when m = 2j − 3; with fewer members it is a mechanism, with more it is redundant (indeterminate). Member forces are found by the method of joints — two equilibrium equations at each joint, starting where only two forces are unknown — or by the method of sections, cutting through at most three members and taking moments about the intersection of two of them to get the third directly.

Zero-force members: at an unloaded joint with two non-collinear members, both carry zero force; at an unloaded joint with three members of which two are collinear, the third carries zero force. Spotting these first shortens the work.
A frame has at least one multi-force member — one loaded along its length or with more than two pins — which carries bending as well as axial force. It is solved by dismembering it into free bodies and using Newton’s third law at each pin. Headframes over a shaft and loading-bin structures are frames.

3. Friction forces

Dry (Coulomb) friction resists impending or actual sliding. Up to the point of slipping the friction force F is whatever equilibrium needs, F ≤ μs N; at impending motion F = μs N, and once sliding F = μk N, with the kinetic coefficient usually a little below the static one. Friction is independent of the apparent area of contact. The angle of friction φ = tan⁻¹ μ equals the angle of repose — the steepest incline on which a block rests unaided. To push a block of weight W up an incline θ with a force parallel to it, P = W(sin θ + μ cos θ); down it, W(μ cos θ − sin θ) if that is positive.

A rope or belt wrapped through an angle θ (in radians) round a drum or pulley can hold a tension ratio of up to T₁/T₂ = e^μθ before it slips — the capstan (belt friction) equation. It governs belt conveyor drives, the friction (Koepe) winder and a rope snubbed round a post. With μ = 0.3 and half a turn (θ = π), e^0.3π = 2.566, so a 10 kN slack side holds up to 25.66 kN on the tight side. Wedges and screws are friction problems of the same kind, solved with the friction angle.

⚠️ F = μN only at the limit

A 1000 N block resting on a level floor with μ = 0.4 and a 100 N horizontal push feels a friction force of 100 N, not 400 N — friction matches the push until it reaches μN. Writing F = μN before checking whether the body is at the point of slipping is the commonest error in friction problems.

4. Particle kinematics and dynamics

For uniform acceleration a, v = u + at, s = ut + ½at² and v² = u² + 2as; a cage moving at 12 m/s brought to rest at 1.5 m/s² travels 12²/(2 × 1.5) = 48 m. For curvilinear motion the acceleration has a tangential part dv/dt and a normal part v²/r towards the centre. Newton’s second law, ΣF = ma, is the whole of particle dynamics; written as D’Alembert’s principle — add the inertia force −ma and treat the problem as statics — it is how hoisting loads are found. A cage of mass m accelerated upward at a hangs from a rope tension T = m(g + a), and the rope is least loaded during deceleration, m(g − a).

Two integrated forms save the time a free-body solution would spend on the path. Work–energy: the work of all forces equals the change in kinetic energy, ΣW = ½mv₂² − ½mv₁², so a 1000 kg car rolling at 5 m/s against a constant 500 N resistance stops in (½ × 1000 × 25)/500 = 25 m. Impulse–momentum: ∫F dt = mv₂ − mv₁, the tool for collisions and short, large forces, with the coefficient of restitution e = (separation speed)/(approach speed) completing a collision problem. Power is force × velocity, the link to the machinery chapter.

5. Beam analysis

A beam carries load by shear force V and bending moment M at each section. They are linked to the load intensity w by dV/dx = −w and dM/dx = V, so the bending moment is greatest where the shear force passes through zero, a uniformly distributed load makes M parabolic and V linear, and a point load makes a jump in V and a kink in M. A point where M changes sign is a point of contraflexure.

Maximum bending moments worth remembering
Beam and load	Maximum moment	Where
Simply supported, UDL w over span L	wL²/8	mid-span
Simply supported, central point load P	PL/4	mid-span
Cantilever, point load P at the free end	PL	fixed end
Cantilever, UDL w	wL²/2	fixed end
Fixed at both ends, UDL w	wL²/12 at the supports, wL²/24 at mid-span	supports

The bending stress follows the flexure formula σ = M y/I, greatest at the outer fibre, σmax = M/Z, where the section modulus of a rectangle b wide and h deep is Z = bh²/6. A 6 m simply supported roof beam under 10 kN/m has M = 10 × 36/8 = 45 kN·m; if it is 0.2 m wide and 0.3 m deep, Z = 0.2 × 0.09/6 = 0.003 m³ and σmax = 45 000/0.003 = 15 MPa. The same flexure formula, applied to a bed of roof rock spanning a gallery, is the starting point of the beam theory of roof design in the next chapter.

🎯 Why rock beams fail in tension

The outer fibres of a bent beam carry the largest stress, and on the underside of a roof bed that stress is tensile. Rock is typically ten or more times weaker in tension than in compression, so a roof bed cracks at mid-span underneath, or at the abutments on top, long before it crushes.

Key takeaways

Any force system reduces to a resultant plus a couple; equilibrium in the plane is three equations, so a free body yields at most three unknowns.
A plane truss is determinate and just rigid when m = 2j − 3; find zero-force members first, then joints or sections.
Friction is at most μN and equals μN only at impending slip; ropes on drums obey T₁/T₂ = e^μθ with θ in radians.
A hoisted mass pulls T = m(g + a) while accelerating upward; v² = u² + 2as gives stopping distances, and work–energy avoids the path.
Mmax = wL²/8 and PL/4 for simple spans; σ = M/Z with Z = bh²/6, and the maximum moment sits where the shear force is zero.

Practice questions (13)

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.

Two parallel forces of 30 N and 50 N act in the same direction at x = 0 and x = 4 m on a bar. The distance of their resultant from x = 0, in m, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.5
R = 30 + 50 = 80 N, and by Varignon 80x̄ = 30 × 0 + 50 × 4 = 200, so x̄ = 2.5 m, nearer the larger force. Taking the midpoint (2 m) ignores the unequal magnitudes, and 50 × 4/30 = 6.67 m divides by the wrong force.
Three concurrent forces keep a particle in equilibrium. By Lami’s theorem, each force is proportional to:
1. the cosine of the angle between the other two forces
2. the sine of the angle between the other two forces
3. the sine of the angle it makes with the horizontal
4. the angle between the other two forces
Show answer
Answer: B — the sine of the angle between the other two forces
Lami: P/sin α = Q/sin β = R/sin γ, where each angle is the one between the other two forces — it is the sine rule applied to the closed force triangle. The cosine appears in the resolution of forces, not in Lami, and the angle with the horizontal plays no role.
A plane pin-jointed truss has 6 joints. For it to be statically determinate and just rigid, the number of members must be:
1. 8
2. 9
3. 10
4. 12
Show answer
Answer: B — 9
m = 2j − 3 = 2 × 6 − 3 = 9. With 8 members it would be a mechanism and collapse; with 10 or more it would be redundant, needing compatibility equations beyond statics. The count 3j − 6 = 12 is the rule for a space truss.
At an unloaded joint of a truss, three members meet, two of which are collinear. The force in the third member is:
1. equal to the force in either collinear member
2. zero
3. half the sum of the other two
4. indeterminate without the loads elsewhere
Show answer
Answer: B — zero
Resolving perpendicular to the collinear pair leaves the third member’s force alone in the equation, with nothing to balance it, so it must be zero. The two collinear members then carry equal forces; the result does not depend on the loads elsewhere in the truss.
Which of the following statements about dry friction are correct? (Select all that apply.)
1. The angle of repose equals the angle of friction
2. Limiting friction is independent of the apparent area of contact
3. The kinetic coefficient of friction is generally greater than the static coefficient
4. The friction force on a body at rest always equals μN
Show answer
Answer: A — The angle of repose equals the angle of friction; B — Limiting friction is independent of the apparent area of contact
On an incline at the angle of repose, tan θ = μ, which is the definition of the friction angle; and Coulomb friction depends on the normal force, not the contact area. Kinetic friction is usually slightly less than static, and a body at rest feels only as much friction as equilibrium needs, up to μN.
A mine car of weight 1000 N is to be pulled up a 30° incline by a force parallel to the incline. The coefficient of friction is 0.2. The force needed to start it moving up, in N, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 673.21
P = W(sin θ + μ cos θ) = 1000 × (0.5 + 0.2 × 0.86603) = 500 + 173.21 = 673.21 N. Friction acts down the slope because motion is up it; subtracting it (326.79 N) is the force to lower the car at constant speed, and using μ sin θ swaps the components.
A rope passes half-way round a fixed drum (angle of wrap 180°) with a coefficient of friction of 0.3. If the slack-side tension is 10 kN, the largest tight-side tension before slipping, in kN, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 25.66
T₁ = T₂ e^μθ with θ = π rad: e^0.3π = e^0.9425 = 2.5663, so T₁ = 25.66 kN. Using θ = 180 in degrees gives an absurd e⁵⁴; using 1 + μθ in place of the exponential gives 10 × 1.94 = 19.4 kN.
A cage moving down a shaft at 12 m/s is brought to rest with a uniform deceleration of 1.5 m/s². The distance it travels while stopping, in m, is ____.
Numerical answer — type the value.
Show answer
Answer: 48
v² = u² + 2as with v = 0: s = u²/(2a) = 144/3 = 48 m. Forgetting the factor 2 gives 96 m, and s = u/a = 8 is the stopping time in seconds, not the distance.
A loaded cage of total mass 5000 kg is accelerated upward at 1.5 m/s². Taking g = 9.81 m/s², the rope tension during acceleration, in kN, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 56.55
T − mg = ma, so T = m(g + a) = 5000 × 11.31 = 56 550 N = 56.55 kN. The static load is 49.05 kN; m(g − a) = 41.55 kN is the tension while decelerating upward or accelerating downward.
A 1000 kg mine car rolling at 5 m/s on level track is brought to rest by a constant resistance of 500 N. The distance it rolls, in m, is ____.
Numerical answer — type the value.
Show answer
Answer: 25
Work–energy: F s = ½mv², so s = (0.5 × 1000 × 25)/500 = 12 500/500 = 25 m. Dropping the ½ gives 50 m; the deceleration route, a = 0.5 m/s² and s = v²/2a, gives the same 25 m.
A simply supported beam of span 6 m carries a uniformly distributed load of 10 kN/m over its whole span. The maximum bending moment, in kN·m, is ____.
Numerical answer — type the value.
Show answer
Answer: 45
Mmax = wL²/8 = 10 × 36/8 = 45 kN·m, at mid-span where the shear force is zero. wL²/2 = 180 is the cantilever value, and treating the whole 60 kN as a central point load gives PL/4 = 90.
The beam of span 6 m under 10 kN/m has a rectangular section 0.2 m wide and 0.3 m deep. The maximum bending stress, in MPa, is ____.
Numerical answer — type the value.
Show answer
Answer: 15
M = 45 kN·m and Z = bh²/6 = 0.2 × 0.09/6 = 0.003 m³, so σ = M/Z = 45 000/0.003 = 15 × 10⁶ Pa = 15 MPa. Swapping b and h (Z = 0.3 × 0.04/6 = 0.002 m³) gives 22.5 MPa — the beam turned on its side — and using I in place of Z omits the distance y.
Which of the following statements about shear force and bending moment are correct? (Select all that apply.)
1. The bending moment is a maximum where the shear force passes through zero
2. Under a uniformly distributed load the bending moment varies parabolically
3. For a cantilever with a point load at its free end, the bending moment is greatest at the free end
4. A point of contraflexure is where the bending moment changes sign
Show answer
Answer: A — The bending moment is a maximum where the shear force passes through zero; B — Under a uniformly distributed load the bending moment varies parabolically; D — A point of contraflexure is where the bending moment changes sign
dM/dx = V makes M stationary where V = 0, and integrating a constant w twice gives a parabola; contraflexure is defined by the change of sign of M. A cantilever’s moment is zero at the free end and greatest (PL) at the fixed end, so the third statement reverses the two ends.