Dynamics: Newton's laws, connected particles and friction flashcards

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Newton's first law
A particle stays at rest or moves with constant velocity unless a resultant force acts on it.
Newton's second law
The resultant force on a particle equals mass × acceleration, F = ma, and the acceleration is in the direction of the resultant force.
Newton's third law
If A exerts a force on B, then B exerts a force on A that is equal in size and opposite in direction.
Light (string, rod, tow bar)
Its mass is negligible, so the tension (or thrust) is the same all along it.
Inextensible string
It does not stretch, so particles joined by it (while taut) move with the same speed and the same size of acceleration.
Smooth pulley or peg
No friction between it and the string, so the tension is the same on both sides.
Coefficient of friction μ
The constant for a pair of surfaces such that friction on a sliding particle is μR. No units; μ = 0 for a smooth surface.
Thrust
A pushing force in a rod or rigid tow bar (compression). A string can only pull, never push.
Newton's second law in vector form
F = ma, with F and a vectors: divide each i, j component of the resultant force by m. (No — learn it)
Weight of a particle of mass m
W = mg, with g = 9.8 m s⁻². (No — learn it)
Friction on a particle sliding on a rough surface
F = μR, opposing the motion. (No — learn it)
Normal reaction on a particle on a plane inclined at α (no other force into the plane)
R = mg cos α; the weight component along the plane is mg sin α. (No — derive it)
Normal reaction when a force P pulls at θ above the horizontal on level ground
R = mg − P sin θ. (No — derive it)
Force of a string on a smooth pulley, the two parts of the string at angle θ to each other
2T cos(θ/2), along the bisector of the angle between the strings. (No — derive it)
Force diagram of a particle sliding down a rough plane
Weight vertically down; normal reaction perpendicular to the plane; friction up the plane (opposing the motion).
Force diagram of a particle projected up a rough plane
Weight vertically down; normal reaction perpendicular to the plane; friction down the plane while it moves up.
v–t graph of a particle pulled by a string that goes slack part-way
Straight-line segments: one gradient while the string is taut, a new constant gradient once the tension is zero.
Forces on a pulley at a table edge (horizontal and vertical string)
Two tensions T, one horizontal and one vertical; resultant T√2 at 45° below the horizontal.
Forces on a passenger in a lift
Weight mg down and normal reaction R from the floor up; R − mg = ma with up positive.
Why is friction on a moving particle always μR (not less)?
When the particle is sliding, friction has reached its limiting value; F ≤ μR only matters for a particle at rest (statics).
Why can the tension be ignored when finding the acceleration of a car and trailer together?
It is internal: it pulls the car back and the trailer forward equally, so it cancels for the whole system.
Why can you not treat two particles over a pulley as one body in a single F = ma?
They move in different directions; write one equation per particle, each in its own direction of motion, then add.
What happens to the tension when the hanging particle hits the ground?
The string goes slack, T = 0. The other particle keeps its speed and moves under its remaining forces only.
What links two stages of motion when a force changes?
The velocity at the instant of change: the final velocity of one stage is the initial velocity of the next.
Why is the tension less than the weight of a hanging particle that accelerates downwards?
Its resultant force mg − T must point down (= ma > 0), so T < mg.
When is a tow bar in thrust?
When the towing vehicle decelerates more than the trailer would under its own resistance, so the trailer pushes on the car.
40 kg crate, push 150 N, resistance 62 N. Acceleration?
150 − 62 = 40a, so a = 2.2 m s⁻².
Only force (6i − 9j) N on a 3 kg particle. Acceleration and its magnitude?
a = (2i − 3j) m s⁻²; |a| = √13 = 3.61 m s⁻².
A particle slides on rough level ground, μ = 0.3, no other horizontal force. Deceleration?
μmg = ma, so a = 0.3 × 9.8 = 2.94 m s⁻² (independent of m).
Particle released on a smooth plane at 35°. Acceleration?
a = g sin 35° = 5.62 m s⁻² down the plane.
Particle slides down a rough plane, tan α = ¾, μ = 0.5. Acceleration?
a = g(sin α − μ cos α) = 9.8(0.6 − 0.4) = 1.96 m s⁻².
4 kg and 6 kg hang over a smooth fixed pulley. Acceleration and tension?
6g − T = 6a, T − 4g = 4a ⇒ a = 1.96 m s⁻², T = 47.0 N.
70 kg passenger, lift moving up and slowing at 1.5 m s⁻². Force from the floor?
R − 70g = 70(−1.5) ⇒ R = 70 × 8.3 = 581 N.
The lighter particle of a pulley system is rising at 2.8 m s⁻¹ when the string goes slack. Further rise?
Now only gravity acts: 0 = 2.8² − 2 × 9.8 × s ⇒ s = 0.4 m.
Writing R = mg on a slope or with an angled pull
Wrong: resolve perpendicular to the motion. R = mg cos α on a slope; R = mg − P sin θ for a pull at θ above the horizontal.
Friction direction on the return journey down a slope
Friction reverses when the velocity reverses: redraw the diagram for each stage.
Rounding with g = 9.8
Keep 4+ significant figures in working; give final answers to 2 or 3 s.f.
A force in a tow bar comes out negative
You assumed tension; it is a thrust of that size. State it in words.
Using F = ma with only the applied force
F must be the resultant: subtract friction, resistance and weight components first.

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