Flashcards · H240 · Topic 3

Mechanics flashcards

29 cards on Topic 3, subtopics 3.01 to 3.04. Pick a subtopic or study them all.

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3.01 Quantities and units in mechanics

The three base quantities in mechanics and their SI units: Length: metre (m). Time: second (s). Mass: kilogram (kg).
SI units of velocity, acceleration, force, weight and moment: Velocity m s⁻¹; acceleration m s⁻²; force newton N = kg m s⁻²; weight N (a force); moment N m.
Convert 72 km h⁻¹ to m s⁻¹.: Divide by 3.6: 72 ÷ 3.6 = 20 m s⁻¹.
How can units show that a formula is wrong?: Every term must have the same units. In s = ut + ½at² each term is a length (m s⁻¹ × s and m s⁻² × s²). s = u + at² fails, because u is a velocity but at² is a length.
A runner covers 450 m in 3 minutes. Find the average speed in m s⁻¹.: Convert first: 3 minutes = 180 s. 450 ÷ 180 = 2.5 m s⁻¹.

3.02 Kinematics

The constant-acceleration (suvat) formulae: v = u + at s = ut + ½at² s = vt − ½at² s = ½(u + v)t v² = u² + 2as
What do the gradient and the area of a velocity–time graph represent?: Gradient = acceleration. Area between the graph and the t-axis = displacement.
Variable acceleration in one dimension: differentiation and integration: v = ds/dt, a = dv/dt = d²s/dt². s = ∫ v dt, v = ∫ a dt, each with a constant from a given condition. At rest: v = 0. Maximum velocity: a = 0.
How do you find the distance travelled when the velocity changes sign?: Find when v = 0, split the time interval there, and add the magnitudes of the displacements for each part.
Motion under gravity: Acceleration g = 9.8 m s⁻² downwards, so a = −g taking up as positive (in vectors a = −gj).
Projectile launched at speed u at angle α above the horizontal: components and equations: Horizontal: u cos α, constant, so x = (u cos α)t. Vertical: u sin α, acceleration −g, so y = (u sin α)t − ½gt².
Time of flight, range and greatest height of a projectile landing at its launch level: T = 2u sin α/g R = u² sin 2α/g H = u² sin²α/(2g)
Vector kinematics with calculus: v = dr/dt and a = dv/dt; r = ∫ v dt and v = ∫ a dt. Work with the i and j components separately; the constant of integration is a vector.

3.03 Forces and Newton's laws

Newton’s first law: A body stays at rest or moves with constant velocity unless a resultant force acts on it: constant velocity means zero resultant force.
Newton’s second law: Resultant force = ma, in the direction of the acceleration. In vector form F = ma, component by component.
Newton’s third law: When two bodies interact, the forces on each are equal in size and opposite in direction, acting on different bodies.
A 25 kg passenger is in a lift accelerating upwards at 0.4 m s⁻² (g = 9.8). Find the normal reaction.: R − 25 × 9.8 = 25 × 0.4, so R = 245 + 10 = 255 N.
Modelling terms: smooth, light and inextensible: Smooth: no friction, only a normal contact force. Light: mass ignored. Inextensible string: same speed and acceleration at both ends; a light string has the same tension throughout.
Friction: the inequality and when it is limiting: F ≤ μR, where μ is the coefficient of friction (no units). F = μR when the body is on the point of moving (limiting) or sliding.
Magnitude of the total contact force from a rough surface: S = √(R² + F²), combining the normal reaction R and the friction F.
Components of a force F at angle θ to a line: Along the line: F cos θ. At right angles to the line: F sin θ. On a slope at α, the weight has mg sin α down the plane and mg cos α into it.
How do you solve an equilibrium problem with several forces on a particle?: 1. Draw every force. 2. Resolve in two perpendicular directions chosen to remove unknowns. 3. Set each total to zero and solve. A negative answer means the force acts the other way.
Why is the normal reaction not always mg?: R comes from resolving perpendicular to the surface. On a slope, or when a pulling force has a vertical component, R ≠ mg.

3.04 Moments

Moment of a force about a point: Force × perpendicular distance from the point to the line of action of the force. Unit N m. A force through the point has zero moment.
An 18 N force acts 0.5 m from O along a rod, at 60° to the rod. Find its moment about O.: 18 × 0.5 × sin 60° = 7.79 N m (3 s.f.).
Conditions for a rigid body to be in equilibrium: The resultant force is zero (resolve horizontally and vertically) and the resultant moment about any point is zero.
Where does the weight act on a uniform rod and on a uniform rectangular lamina?: Uniform rod: at its midpoint. Rectangular lamina: at its centre (the point of symmetry). For a non-uniform rod the point is given, or found by moments.
A beam on two supports is about to tilt about one of them. What does that tell you?: The reaction at the other support is zero. Take moments about the support still in contact.
A uniform beam AB of length 4 m and weight 100 N rests on supports at A and B. A 50 N load is 1 m from A. Find the reactions.: Moments about A: 4R_B = 100 × 2 + 50 × 1 = 250, so R_B = 62.5 N. Vertically: R_A = 150 − 62.5 = 87.5 N.