Statics of a particle flashcards

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Equilibrium of a particle
The resultant force on the particle is zero: the components of the forces in any direction add to zero.
Resultant force
The single force with the same effect as all the forces acting together; its components are the sums of their components.
Normal reaction
The contact force from a surface, perpendicular to the surface and away from it.
Thrust
A pushing force in a light rod, acting along the rod towards the particle. (A string can only pull: tension.)
Coefficient of friction μ
The constant for a pair of surfaces such that friction F ≤ μR. It has no units; μ = 0 means a smooth surface.
Limiting equilibrium
At rest but on the point of moving: friction has reached its maximum, F = μR.
Smooth surface
A surface with no friction: the only contact force is the normal reaction.
Angle of friction λ
The angle between the total contact force and the normal in limiting equilibrium: tan λ = μ.
Components of a force F at angle θ to a direction
F cos θ along the direction; F sin θ perpendicular to it. (No — learn it)
Magnitude and direction of a resultant with components X and Y
R = √(X² + Y²); tan φ = Y/X, with the quadrant taken from a sketch. (No — learn it)
Friction on a particle at rest
F ≤ μR; F = μR in limiting equilibrium. (No — learn it)
Components of weight on a plane at angle α
mg sin α down the plane; mg cos α into the plane. (No — derive it)
Condition for a particle to rest on a rough plane at angle α with no other force
μ ≥ tan α (the mass cancels). (No — derive it)
Least force to move a particle of weight mg on rough horizontal ground
P_min = mg sin λ, applied at angle λ above the horizontal, where tan λ = μ. (No — derive it)
Triangle of forces: three forces P, Q, S in equilibrium
P/sin(angle opposite P) = Q/sin(angle opposite Q) = S/sin(angle opposite S) — the sine rule. (No — learn it)
Three forces in equilibrium drawn tip to tail
They form a closed triangle; each interior angle is 180° minus the angle between the two forces meeting there.
Graph of friction against an increasing horizontal pull P on a particle at rest
F = P (a line through the origin, gradient 1) until F reaches μR; after that the particle slides.
Force diagram: particle on a rough plane about to slide UP
Weight vertically down; R perpendicular to the plane; friction μR DOWN the plane; applied force.
Force diagram: particle on a rough plane about to slide DOWN
Weight vertically down; R perpendicular to the plane; friction μR UP the plane; applied force.
Graph of the pull P needed to move a particle against the angle θ above the horizontal
Falls from μmg at θ = 0 to a minimum at tan θ = μ, then rises steeply.
Why can resolving give at most two unknowns for a particle?
Only two independent directions exist in a plane, so resolving gives two independent equations.
Why is R ≠ mg cos α when a horizontal force pushes a particle on a plane?
The horizontal force has a component P sin α into the plane, so R = mg cos α + P sin α.
Why is there a range of forces that hold a particle at rest on a rough plane?
Friction can act either way along the plane, up to μR. The two ends of the range are the two limiting cases.
Why does pulling slightly upwards make a particle easier to move on rough ground?
The upward component reduces R, and so the maximum friction μR, by more than the horizontal pull is reduced.
Which way does friction act on a particle at rest?
Along the surface, opposite to the motion that would happen if the surface were smooth.
One force is removed from a set of forces in equilibrium. What is the resultant of the rest?
Equal in size and opposite in direction to the force removed.
Find the magnitude of the resultant of (4i + 2j) N and (−i + 2j) N, and its angle with i.
Resultant (3i + 4j) N; magnitude 5 N; angle tan⁻¹(4/3) = 53.1° to i.
A 40 N weight hangs from the midpoint of a string; each half makes 35° with the horizontal. Tension?
2T sin 35° = 40, so T = 34.9 N.
A 2.5 kg particle on a smooth plane at 30° is held by a horizontal force P. Find P.
P cos 30° = 2.5g sin 30°, so P = 2.5g tan 30° = 14.1 N.
A 5 kg particle on rough horizontal ground is about to slide when a horizontal 18 N force acts. Find μ.
R = 5g = 49 N; μ = 18/49 = 0.367.
A 3 kg particle rests on rough ground (μ = 0.3) with a 5 N horizontal force on it. Friction?
5 N. The maximum, μR = 0.3 × 29.4 = 8.82 N, is not reached.
A particle on a rough plane is about to slide when the plane is at 32°. Find μ.
μ = tan 32° = 0.625.
Least force to move an 8 kg particle on rough ground with μ = 0.25?
At tan⁻¹ 0.25 = 14.0° above the horizontal: P = 8g sin 14.0° = 19.0 N.
Friction for a particle that is simply 'at rest'
Find F by resolving. Only use F = μR if the question says 'on the point of', 'about to', 'just' or 'limiting'.
R = mg cos α on every inclined plane
Wrong whenever another force has a component perpendicular to the plane: always resolve perpendicular in full.
Range of P on a rough plane: one equation for both cases
Re-resolve for each case: friction changes direction and R usually changes too.
Rounding with g = 9.8
Keep full calculator values between steps; give final answers to 2 or 3 significant figures.
Writing resolving equations
Label each one with its direction, e.g. R(↑) or R(parallel to the plane), and write every term before simplifying.

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