Forces and Newton's laws flashcards

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Free-body diagram
A diagram of one body showing every force acting on it (and no others), as labelled arrows starting on the body.
Centre of gravity
The single point at which the whole weight of a body can be taken to act. For a uniform body it is at the geometric centre.
Gravitational field strength g
The gravitational force per unit mass on a body in the field. Unit N kg⁻¹; 9.81 N kg⁻¹ at the Earth's surface.
Weight
The gravitational force on a body; W = mg. A vector (downwards), unit N.
Newton's first law
A body stays at rest or moves with constant velocity unless a resultant force acts on it.
Newton's second law (constant mass)
The resultant force on a body equals its mass × acceleration, ΣF = ma; the acceleration is in the direction of the resultant force.
Newton's third law
If body A exerts a force on body B, then B exerts a force on A that is equal in size and opposite in direction.
Terminal velocity
The constant, maximum velocity reached when the resistive forces on a body balance the driving force (for a falling body, its weight).
Equilibrium (of forces)
The state of a body with zero resultant force: it is at rest or moving at constant velocity.
The newton
The resultant force that gives a mass of 1 kg an acceleration of 1 m s⁻². 1 N = 1 kg m s⁻².
Newton's second law for constant mass
ΣF = ma (ΣF = resultant force). (Yes)
Gravitational field strength from force and mass
g = F/m. (Yes)
Weight from mass
W = mg. (Yes)
Components of weight on a slope at angle θ
mg sin θ down the slope; mg cos θ perpendicular to (into) the slope. (No — derive it)
Normal contact force on a person of mass m in a lift with upward acceleration a
R = m(g + a); use a negative a for a downward acceleration. (No — derive it)
Shape of the v–t graph of a body falling from rest to terminal velocity
Starts with gradient g, then the gradient decreases steadily until the line becomes horizontal at the terminal velocity.
v–t graph when a skydiver opens a parachute (downwards positive)
A steep drop in velocity (deceleration), with the gradient decreasing to zero at a new, lower terminal velocity.
Free-body diagram of a box at rest on a rough slope
Weight vertically down from the centre of gravity; normal contact force perpendicular to the slope; friction up the slope along the surface.
Three forces in equilibrium drawn tip to tail
They form a closed triangle — the resultant is zero.
Graph of a against F for a trolley with friction
A straight line with gradient 1/m that meets the F-axis at the value of the frictional force.
Explain why a falling body reaches terminal velocity
Speed rises → drag rises (weight constant) → resultant force and acceleration fall → when drag = weight, resultant = 0, so a = 0 and v is constant.
Four properties of a Newton's third law pair
Act on different bodies; equal magnitude; opposite directions along the same line; same type of force.
Why weight and normal contact force on a book on a table are NOT a third-law pair
They act on the same body and are different types of force. They are equal only because the book is in equilibrium (first law).
Mass versus weight
Mass (kg, scalar) is the same everywhere; weight (N, vector) is the gravitational force and changes with g.
Why do scales read more in a lift accelerating upwards?
The floor must push up with R > mg so that the resultant force R − mg = ma acts upwards. Scales read R.
Connected bodies: why can the tension be ignored when finding the acceleration of the whole system?
It is an internal force: it pulls the two bodies equally in opposite directions, so it cancels for the system as a whole.
What provides the forward force when you walk?
Your foot pushes backwards on the ground; by the third law the ground pushes forwards on you (friction).
Boat + crew 260 kg; forward force 410 N, resistance 150 N. Acceleration?
ΣF = 410 − 150 = 260 N; a = 260 / 260 = 1.0 m s⁻². (Yes)
A 5.0 kg box rests on a slope at 25°. Friction and normal contact force?
Friction = 5.0 × 9.81 × sin 25° = 21 N up the slope; R = 5.0 × 9.81 × cos 25° = 44 N.
A 60 kg person in a lift accelerating upwards at 2.0 m s⁻². Normal contact force?
R = m(g + a) = 60 × 11.81 = 709 N (≈ 710 N).

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