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 the string cannot stretch, so while it is taut both ends move together along it, sharing one speed and one acceleration in the set-ups the course uses. Quote the word the question used.