Forces, moments and equilibrium

Free-body diagrams, Newton's laws of motion, weight, and moments acting on bodies in equilibrium.

Be able to draw and interpret free-body force diagrams to represent forces on a particle or on an extended but rigid body, using the concept of centre of gravity of an extended body

Be able to use the equation $\Sigma F = ma$ , and understand how to use this equation in situations where $m$ is constant (Newton's second law of motion), including Newton's first law of motion where $a$ , objects at rest or travelling at constant velocity (use of the term terminal velocity is expected)

Be able to use the equations for gravitational field strength $g = \frac{F}{m}$ and weight $W = mg$

Know and understand Newton's third law of motion and know the properties of pairs of forces in an interaction between two bodies

Be able to use the equation for the moment of a force, moment of force $= Fx$ , where $x$ is the perpendicular distance between the line of action of the force and the axis of rotation

Be able to use the concept of centre of gravity of an extended body and apply the principle of moments to an extended body in equilibrium

A box of mass $5.0\,\text{kg}$ is released on a ramp inclined at $30^\circ$ to the horizontal. A constant frictional force of $12\,\text{N}$ acts up the slope.

Calculate the acceleration of the box down the slope.

Component of weight down the slope $= 5.0 \times 9.81 \times \sin 30^\circ = 24.5\,\text{N}$ (1)
Resultant force $= 24.5 - 12 = 12.5\,\text{N}$ (1)

A uniform plank AC is $3.0\,\text{m}$ long and has a weight of $120\,\text{N}$ . It rests horizontally on two supports: one at end A and one at point B, which is $2.0\,\text{m}$ from A. A load of weight $200\,\text{N}$ is placed at end C.

(a) Calculate the force exerted on the plank by the support at B.

A skydiver jumps from a hovering helicopter. Some time later she opens her parachute and eventually lands.

Describe and explain how the forces on the skydiver, and hence her velocity, change from the moment she jumps until just before she lands. You may refer to a velocity–time sketch in words.

Indicative content:

At the start air resistance is zero, so the resultant force equals her weight and acceleration is $g$ (1)
As speed increases, air resistance (drag) increases, so the resultant force and acceleration decrease (1)
When drag equals weight the resultant force is zero and she travels at (first) terminal velocity (1)
Opening the parachute greatly increases the surface area, so drag becomes greater than weight (1)
The resultant force is upwards, so she decelerates; as speed falls, drag falls (1)
Drag again equals weight at a new, lower terminal velocity, which is maintained until landing (1)

Award 1 mark per relevant point, max 6.

Takes moments about A: $120 \times 1.5 + 200 \times 3.0 = F_B \times 2.0$ (1)
$F_B = 390\,\text{N}$ (upwards) (1)
Resolves vertically: $F_A + 390 = 120 + 200$ , so $F_A = -70\,\text{N}$ (1)
The force at A is $70\,\text{N}$ downwards, i.e. the plank must be held/pinned down at A or it would tip about B (1)

Specification points

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Sample questions