A member pinned at both ends and loaded only at its pins, with its weight neglected, carries in equilibrium:

two equal, opposite and collinear forces along the line joining the pins

two equal forces perpendicular to the member

a force and a couple at each pin

At an unloaded joint of a plane truss, exactly two non-collinear members meet. The forces in these members are:

equal and opposite, but not zero

equal to half the support reaction

indeterminate without the method of sections

Which statements about rolling resistance are true?

The force needed is roughly proportional to the load and inversely proportional to the wheel radius

The coefficient of rolling resistance has the dimension of length

It is usually much smaller than sliding friction for the same load

It is independent of the wheel radius

The force needed is roughly proportional to the load and inversely proportional to the wheel radius

The coefficient of rolling resistance has the dimension of length

It is usually much smaller than sliding friction for the same load

Which statements about the instantaneous centre of zero velocity of a rigid body in plane motion are true?

For a wheel rolling without slipping it is the point of contact

The velocity of any point is ω times its distance from it, perpendicular to the line joining them

For a body in pure translation it lies at infinity

For a wheel rolling without slipping it is the point of contact

The velocity of any point is ω times its distance from it, perpendicular to the line joining them

Its acceleration is always zero

For a body in pure translation it lies at infinity

Engineering Mechanics: Equilibrium, Friction, Belts, Trusses and Rigid Bodies in Plane Motion

Section A.2 names free-body diagrams and equilibrium; friction and its applications — rolling friction and the belt–pulley; trusses and frames; and the kinematics and dynamics of rigid bodies in plane motion. A manipulator is a chain of rigid bodies carrying loads through friction contacts and belt drives, so this is the statics and dynamics a robot designer checks before anything moves. Each topic is worked in the form GATE sets it: a force to impend motion, a belt’s power, a member force, an acceleration.

1. Free-body diagrams and equilibrium

A free-body diagram isolates one body and draws every external force on it: applied loads, weight, and the reactions of whatever was cut away — a roller gives one force normal to its surface, a pin two components, a fixed support two components and a moment. In the plane, equilibrium is ΣF_x = 0, ΣF_y = 0, ΣM = 0: three equations, so a body with three unknown reactions is statically determinate. A two-force member in equilibrium carries equal, opposite, collinear forces along the line joining its two pins; a three-force body has its three forces concurrent (or parallel).

Worked: a 100 N block rests on a floor with μ_s = 0.3 and is pulled by a force P at 30° above the horizontal. The pull lifts part of the weight, so N = 100 − P sin 30° and motion impends when P cos 30° = 0.3(100 − 0.5P): P = 30/(0.866 + 0.15) ≈ 29.5 N, less than the 30 N a horizontal pull would need.

2. Friction: sliding, rolling, and the belt on a pulley

Coulomb friction resists relative sliding with a force up to μ_s N before slip and μ_k N (μ_k < μ_s) during it; the angle of friction φ has tan φ = μ, and a block on an incline starts to slide when tan θ = μ_s. The first question is always whether the body moves: compute the friction needed for equilibrium and compare it with μ_s N. Rolling resistance arises because a loaded wheel and its surface deform, moving the normal reaction ahead of the centre by a distance a, the coefficient of rolling resistance, which has units of length; the force to keep the wheel rolling is about F = W a/r — proportional to load and inversely to wheel radius, and usually far smaller than sliding friction.

A flat belt on the point of slipping over a pulley has T₁/T₂ = e^μθ, θ the angle of lap in radians, T₁ the tight-side tension; for a V-belt of groove angle 2β, e^{μθ/sin β}. The power transmitted at belt speed v is P = (T₁ − T₂)v. With μ = 0.3 and θ = π, T₁/T₂ = e^0.942 ≈ 2.566; a slack side of 200 N gives T₁ ≈ 513 N, and at 10 m/s the belt transmits about 3.13 kW. At high speed the centrifugal tension mv² adds to both sides and reduces the net grip.

⚠️ Degrees in the exponent

The lap angle in e^μθ must be in radians: 180° is π. Putting 180 into the exponent gives e⁵⁴, a tension ratio no belt has ever had.

3. Trusses and frames

A truss is made 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 — 9 members for 6 joints; fewer is a mechanism, more is redundant. The method of joints applies ΣF_x = ΣF_y = 0 at each pin, starting where only two members are unknown; the method of sections cuts through at most three members and takes moments about the intersection of two of them to find the third directly. At an unloaded joint where two non-collinear members meet, both are zero-force members; at an unloaded joint of three members, two collinear, the third is zero-force.

Worked: two equal members meet at an apex carrying 10 kN downward, each inclined at 60° to the horizontal and pinned at its base. By symmetry each carries F, and vertical equilibrium at the apex gives 2F sin 60° = 10, so F = 10/√3 ≈ 5.77 kN, compressive. Flatten the members to 30° and the force rises to 10 kN: shallow trusses carry large member forces. A frame differs in having at least one multi-force member, which is analysed as a rigid body with its own free-body diagram rather than joint by joint.

4. Kinematics of rigid bodies in plane motion

A rigid body in plane motion has one angular velocity ω and one angular acceleration α. Any two points are related by v_B = v_A + ω × r_B/A and a_B = a_A + α × r_B/A − ω² r_B/A. At any instant there is an instantaneous centre of zero velocity about which the body is momentarily rotating: every point moves perpendicular to its line to the IC with speed ω × (distance), and a body in pure translation has its IC at infinity. The IC generally has a non-zero acceleration, so it cannot be used for accelerations.

A wheel of radius r rolling without slipping has v = ωr and a = αr at its centre. Its contact point is the IC: momentarily at rest, though it accelerates upward at ω²r. The top of the wheel moves at 2v, forward; a point level with the centre moves at √2 v at 45°. A 0.3 m wheel with centre speed 6 m/s turns at 20 rad/s, and its top moves at 12 m/s.

5. Dynamics of rigid bodies in plane motion

The equations of plane motion are ΣF = m a_G and ΣM_G = I_G α, or, for rotation about a fixed axis O, ΣM_O = I_O α with I_O = I_G + md² (parallel axis). A uniform rod of length L pinned at one end and released from the horizontal has I_O = mL²/3 and a gravity moment mgL/2, so α = 3g/(2L) at release — 9.81 rad/s² for L = 1.5 m — and the end starts downward at 3g/2, faster than free fall.

Rolling without slipping down an incline θ: a = g sin θ/(1 + k²/r²)
Body	k²/r²	Acceleration
Solid sphere	2/5	(5/7) g sin θ
Solid cylinder or disc	1/2	(2/3) g sin θ
Thin ring or hollow cylinder	1	(1/2) g sin θ
Block sliding without friction	0	g sin θ

The more of its mass a body carries far from the axis, the more of the available energy goes into rotation and the slower it descends: on a 30° slope a solid cylinder accelerates at (2/3)(9.81)(0.5) ≈ 3.27 m/s². Work–energy (T₁ + V₁ + W = T₂ + V₂, with T = ½mv_G² + ½I_Gω²) gives speeds without accelerations, and impulse–momentum handles impacts; choose the principle whose quantities the question names.

Key takeaways

Plane equilibrium gives three equations; a two-force member carries equal, opposite, collinear forces.
Check whether it moves before using μN; rolling resistance ≈ Wa/r; a belt on the point of slip has T₁/T₂ = e^μθ, θ in radians, and P = (T₁ − T₂)v.
A plane truss is determinate when m = 2j − 3; spot zero-force members at unloaded two-member joints.
Rolling without slip: v = ωr, contact point at rest, top at 2v; the IC has zero velocity, not zero acceleration.
ΣM_O = I_O α; a rod pinned at one end starts with α = 3g/(2L); down an incline a = g sin θ/(1 + k²/r²).

Practice questions (11)

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.

A member pinned at both ends and loaded only at its pins, with its weight neglected, carries in equilibrium:
1. two equal, opposite and collinear forces along the line joining the pins
2. two equal forces perpendicular to the member
3. a force and a couple at each pin
4. three concurrent forces
Show answer
Answer: A — two equal, opposite and collinear forces along the line joining the pins
With only two forces, ΣF = 0 makes them equal and opposite, and ΣM = 0 makes them collinear, so both act along the line of the pins. A pin transmits no couple.
A plane truss has 6 joints. For it to be statically determinate and just rigid, the number of members must be ____.
Numerical answer — type the value.
Show answer
Answer: 9
m = 2j − 3 = 12 − 3 = 9: two equilibrium equations per joint give 12 equations for 9 member forces and 3 support reactions. Fewer members leave a mechanism, more make it redundant.
A block weighing 100 N rests on a horizontal floor with coefficient of static friction 0.3. It is pulled by a force P inclined at 30° above the horizontal. The value of P at which sliding impends, correct to one decimal place, is ____ N.
Numerical answer — type the value.
Show answer
Answer: 29.5
N = 100 − P sin 30° = 100 − 0.5P, and at impending motion P cos 30° = 0.3N: 0.866P = 30 − 0.15P, so P = 30/1.016 ≈ 29.53 N. Taking N = 100 ignores the lift from the inclined pull and gives 34.6 N.
A flat belt passes half-way round a pulley (angle of lap 180°) with coefficient of friction 0.3. The slack-side tension is 200 N and the belt speed is 10 m/s. On the point of slipping, the power transmitted, correct to two decimal places, is ____ kW.
Numerical answer — type the value.
Show answer
Answer: 3.13
T₁/T₂ = e^μθ = e^0.3π = e^0.9425 ≈ 2.566, so T₁ ≈ 513.3 N and P = (T₁ − T₂)v = 313.3 × 10 ≈ 3133 W = 3.13 kW. Using T₁ × v gives 5.13 kW — the slack side does work against the pulley too.
Two identical truss members meet at an apex that carries a vertical load of 10 kN; each member is inclined at 60° to the horizontal and pinned at its base. The force in each member, correct to two decimal places, is ____ kN.
Numerical answer — type the value.
Show answer
Answer: 5.77
Vertical equilibrium at the apex: 2F sin 60° = 10, so F = 10/(2 × 0.866) = 5.77 kN, compressive; the horizontal components cancel by symmetry. Using cos 60° gives 10 kN, the answer for members at 30°.
At an unloaded joint of a plane truss, exactly two non-collinear members meet. The forces in these members are:
1. both zero
2. equal and opposite, but not zero
3. equal to half the support reaction
4. indeterminate without the method of sections
Show answer
Answer: A — both zero
Resolving perpendicular to one member leaves only the other’s component, which must vanish; then the first must vanish too. Two non-collinear forces cannot balance unless both are zero.
A wheel of radius 0.3 m rolls without slipping, its centre moving at 6 m/s. The speed of the topmost point of the wheel is ____ m/s.
Numerical answer — type the value.
Show answer
Answer: 12
The contact point is the instantaneous centre, and the top is 2r from it: v_top = ω × 2r = 2v = 12 m/s, with ω = 6/0.3 = 20 rad/s. Adding ωr to v gives the same 12, and subtracting gives the contact point’s zero.
A solid uniform cylinder rolls without slipping down a plane inclined at 30°. Taking g = 9.81 m/s², its acceleration, correct to two decimal places, is ____ m/s².
Numerical answer — type the value.
Show answer
Answer: 3.27
a = g sin θ/(1 + k²/r²) with k²/r² = 1/2 for a solid cylinder: a = (2/3) × 9.81 × 0.5 = 3.27 m/s². A frictionless slide would give 4.905; a ring 2.45.
A uniform rod 1.5 m long is pinned at one end and released from rest in the horizontal position. Taking g = 9.81 m/s², its angular acceleration at the instant of release is ____ rad/s².
Numerical answer — type the value.
Show answer
Answer: 9.81
ΣM_O = I_O α: mg(L/2) = (mL²/3)α, so α = 3g/(2L) = 3 × 9.81/3 = 9.81 rad/s². Using I_G = mL²/12 about the pin instead of mL²/3 gives four times as much.
Which statements about rolling resistance are true?
1. It is independent of the wheel radius
2. The force needed is roughly proportional to the load and inversely proportional to the wheel radius
3. The coefficient of rolling resistance has the dimension of length
4. It is usually much smaller than sliding friction for the same load
Show answer
Answer: B — The force needed is roughly proportional to the load and inversely proportional to the wheel radius; C — The coefficient of rolling resistance has the dimension of length; D — It is usually much smaller than sliding friction for the same load
(A) False: F ≈ Wa/r falls as the radius grows, which is why large wheels roll easily over soft ground. (B) The same relation. (C) a is the forward offset of the normal reaction, a length. (D) a/r is typically a small fraction of μ, which is why wheels replaced sledges.
Which statements about the instantaneous centre of zero velocity of a rigid body in plane motion are true?
1. For a wheel rolling without slipping it is the point of contact
2. The velocity of any point is ω times its distance from it, perpendicular to the line joining them
3. Its acceleration is always zero
4. For a body in pure translation it lies at infinity
Show answer
Answer: A — For a wheel rolling without slipping it is the point of contact; B — The velocity of any point is ω times its distance from it, perpendicular to the line joining them; D — For a body in pure translation it lies at infinity
(A) The contact point has zero velocity when there is no slip. (B) That is what momentary rotation about it means. (C) False: the contact point of a rolling wheel accelerates at ω²r towards the centre; the IC is a velocity construction only. (D) With ω = 0 all points share one velocity, and no finite point is at rest.