Title: InkMaths / Mechanics essentials
On every note of this deck, unless a note says otherwise:
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Front: SPOT THE ERROR
Say what is wrong with this claim:
At the top of its flight, a projectile's acceleration is zero.
Back: Only the vertical velocity is zero at the apex. Gravity acts all flight long, so the acceleration is 9.8 m s⁻² downwards there, the same as everywhere else.

Front: DEFINE
Define: limiting equilibrium.
Back: The state in which an object is still in equilibrium but on the point of slipping, so friction has reached its maximum value μR.

Front: STANDARD RESULT — LEARN IT
Projectile components:
u at angle θ ⇒ ?\(u \text{ at angle } θ ⇒ ?\)u at angle θ ⇒ ?
Back: u cos θ across, u sin θ up\(u \cos θ \text{ across}, \, u \sin θ \text{ up}\)u cos θ across, u sin θ up

Front: SPOT THE ERROR
Say what is wrong with this claim:
A moving object must have a force pushing it along.
Back: Newton's first law says the opposite. With no resultant force the motion simply continues. Forces change velocity, they are not needed to keep it. What slows a real object is a real backward force, friction or air resistance.

Front: RECALL
State the method for a particle in equilibrium.
Back: Resolve in two perpendicular directions and set each total to zero. Choose the directions to kill an unknown, so along and perpendicular to a slope rather than horizontal and vertical. Two equations, two unknowns, and no acceleration term anywhere.

Front: DEFINE
Define: uniform beam.
Back: A beam whose mass is evenly spread, so its whole weight can be modelled as acting at its midpoint.

Front: STANDARD RESULT — LEARN IT
Variable acceleration:
from displacement x: v = ?, a = ?\(\text{from displacement x}: v = ?, a = ?\)from displacement x: v = ?, a = ?
Back: v = dxdt, a = dvdt, and back by integrating\(v = \frac{dx}{dt}, a = \frac{dv}{dt}, \text{and back by integrating}\)v = (dx/dt), a = (dv/dt), and back by integrating

Front: STANDARD RESULT — LEARN IT
Weight on a slope:
components of mg on a slope at θ: ?\(\text{components of } mg \text{ on a slope at } θ: ?\)components of mg on a slope at θ: ?
Back: mg sin θ along, mg cos θ into\(mg \sin θ \text{ along}, \, mg \cos θ \text{ into}\)mg sin θ along, mg cos θ into

Front: STANDARD RESULT — LEARN IT
Moments:
moment = ?\(\text{moment} = ?\)moment = ?
Back: moment = force × perpendicular distance\(\text{moment} = \text{force} × \text{perpendicular distance}\)moment = force × perpendicular distance

Front: CHECK YOURSELF
Masses of 7 kg and 3 kg hang over a smooth pulley. Write the two equations of motion.
Back: 7g − T = 7a for the falling mass and T − 3g = 3a for the rising one. Adding eliminates T and gives a = 0.4g, which is 3.92 m s⁻².

Front: SPOT THE ERROR
Say what is wrong with this claim:
A block resting on a rough slope has friction of size μR.
Back: μR is the most friction the surface can supply, not the amount it is supplying. Sitting still on a slope at 10° when it could hold 30°, friction is only mg sin 10°. Write F = μR only when the question says on the point of slipping, in limiting equilibrium, or moving.

Front: CHECK YOURSELF
A 5 kg crate is pulled by 30 N against 10 N of resistance. Find its acceleration.
Back: Resultant 20 N, so a = 20/5 = 4 m s⁻². F = ma takes the resultant along the direction of motion, never one force on its own.

Front: STANDARD RESULT — LEARN IT
The friction law:
F ? μR\(F \, ? \, μR\)F ? μR
Back: F ≤ μR, equality only at the limit or when sliding\(F ≤ μR, \, \text{equality only at the limit or when sliding}\)F ≤ μR, equality only at the limit or when sliding

Front: STANDARD RESULT — LEARN IT
Weight:
W = ?\(W = ?\)W = ?
Back: W = mg, with g = 9.8 m s^−2\(W = mg, \, \text{with } g = 9.8 \text{ m s}^{−2}\)W = mg, with g = 9.8 m s^(−2)

Front: RECALL
State what each modelling word removes.
Back: A particle has no size, so no moments and no spin; air resistance is a separate assumption. Light means no mass, so a string keeps the same tension all along and a rod has no weight. Smooth means no friction. Inextensible means both ends share one speed and one acceleration. Quote the word the question used.