Force and Translational Dynamics

Task 1 - The method, with angled forces

Hook

You pull a sledge across the snow with a rope held at an angle. Pulling at an angle wastes some of your effort lifting the sledge instead of moving it forward. How much of your pull actually moves it, and what happens to the force from the ground?

Intuition

An angled force does two jobs at once. Its horizontal component pulls the sledge forward. Its vertical component lifts it slightly, which means the ground no longer has to support the full weight, so the normal force drops. Splitting the force into components along your chosen axes separates the two jobs.

Axes are your choice. Rotate them below and notice that the components change but the gold net force arrow does not. Physics does not depend on how you draw your axes, only the amount of arithmetic does.

Formalise

The four-step method, written as equations for the two axes.

Worked example

A 10 kg sledge is pulled by a rope at 30° above the horizontal with a tension of 50 N. The snow is frictionless. Find the acceleration and the normal force.

1. Free body diagram. Weight $98.1\text{ N}$ down, normal force $N$ up, tension 50 N at 30°.
2. Axes. x horizontal, y vertical. The sledge only moves along x, so $a_y = 0$.
3. Components. $T_x = 50\cos 30^\circ = 43.3\text{ N}$, $T_y = 50\sin 30^\circ = 25\text{ N}$.
4. x equation. $43.3 = 10a$, so $a = 4.33\text{ m/s}^2$.
5. y equation. $N + 25 - 98.1 = 0$, so $N = 73.1\text{ N}$.

Your turn

A 15 kg lawnmower is pushed along its handle with 80 N, directed 40° below the horizontal. Ignore friction.

Task 2 - Inclined planes and friction

Hook

A box rests on a ramp. You slowly tilt the ramp steeper and steeper. For a while nothing happens, then suddenly the box starts to slide, and once it is moving it keeps accelerating even if you stop tilting. Why the sudden change?

Intuition

Tilt the axes to match the slope. Weight then splits into a part pulling down the slope, $mg\sin\theta$, and a part pressing into the slope, $mg\cos\theta$, which the normal force balances.

Static friction holds a still object in place by matching whatever force tries to move it, up to a maximum of $\mu_s N$. When the down-slope pull exceeds that maximum, the box breaks free. Once sliding, kinetic friction takes over at a fixed $\mu_k N$, which is usually smaller than the static maximum. That drop is why the box lurches into motion. Find the angle where the block below starts to slide, then press Release.

Formalise

With x along the slope pointing down it and y perpendicular to the slope.

Worked example

A block slides down a 30° slope with $\mu_k = 0.2$. Find its acceleration.

1. Divide through by m. $a = g(\sin\theta - \mu_k\cos\theta)$.
2. Substitute. $a = 9.81(0.5 - 0.2 \times 0.866) = 9.81 \times 0.327 = 3.21\text{ m/s}^2$.

The mass cancelled. A heavy block and a light block made of the same material slide down together.

Your turn

Task 3 - Connected bodies

Hook

Two masses hang either side of a pulley on a light string, 3 kg on one side and 5 kg on the other. The heavier side falls and the lighter side rises. How fast do they accelerate, and how hard is the string pulling?

Intuition

There are two ways to look at connected objects, and using both is a good check. Treat the whole thing as one system, and the tension becomes an internal force that cancels out, leaving only the external forces to drive the total mass. Or treat each object separately with its own free body diagram, where the tension appears as an external force on each one. The system view gives the acceleration quickly. The separate view is needed to find the tension.

Formalise

For this setup, called an Atwood machine, with $m_2 > m_1$.

Worked example

1. System view. The unbalanced force is the difference in weights, $(5 - 3) \times 9.81 = 19.62\text{ N}$, driving a total of 8 kg. $a = 19.62 / 8 = 2.45\text{ m/s}^2$.
2. Separate view for the 3 kg mass. It accelerates upward, so $T - 3g = 3a$, giving $T = 3(9.81 + 2.45) = 36.8\text{ N}$.
3. Check with the 5 kg mass. It accelerates downward, so $5g - T = 5a$, giving $T = 5(9.81 - 2.45) = 36.8\text{ N}$. The two views agree.

The tension sits between the two weights, 29.4 N and 49.1 N. It is larger than the light mass's weight, so that mass speeds upward, and smaller than the heavy mass's weight, so that one speeds downward.

Your turn

A 4 kg block on a frictionless table is tied by a string over a pulley at the table's edge to a 2 kg mass hanging below.

Task 4 - Forces that change: springs and drag

Hook

Hang a mass on a spring and let go. It bounces up and down and, if there is any friction or air resistance at all, eventually settles at one position. Where does it settle, and why does it overshoot first?

Intuition

A spring's force grows the further it is stretched or squashed, always pointing back toward its natural length. This is Hooke's law. Released from its natural length, the mass accelerates downward because the spring is not yet pulling hard enough. When it reaches the position where the spring force equals its weight, the forces balance, but the mass is moving at full speed, so by the first law it carries on and overshoots.

Drag is a force that opposes velocity. At low speeds it is roughly proportional to the speed, a model called linear drag. Drag takes energy out of the motion, so the bounces get smaller. Explore the three types of damping below by changing b.

Formalise

Hooke's law, with $x$ the displacement from the natural length and $k$ the spring constant in N/m, and linear drag with coefficient $b$ in kg/s. A falling object with linear drag reaches terminal velocity when drag balances weight. At high speed, drag grows with the square of speed instead, where $\rho$ is the air density, $C_d$ a shape factor and $A$ the frontal area.

Worked example

1. Equilibrium of a hanging spring. A 0.5 kg mass on a spring with $k = 50\text{ N/m}$ settles where $kx = mg$. $x = 0.5 \times 9.81 / 50 = 0.098\text{ m}$, just under 10 cm. This is the gold line in the diagram.
2. Terminal velocity. An 80 kg skydiver modelled with $b = 15\text{ kg/s}$ reaches $v_t = 80 \times 9.81 / 15 = 52.3\text{ m/s}$.

Your turn

Task 5 - Circular motion

Hook

A car drives round a bend at a steady speed. Its speed never changes, but its direction does. If the velocity is changing, Newton's second law says there must be a net force. What is it, and which way does it point?

Intuition

To keep turning, an object needs an acceleration toward the centre of the circle, called the centripetal acceleration. It does not speed the object up. It only swings its velocity round. The force causing it is not a new kind of force. It is always supplied by something real, such as friction between tyres and road, tension in a string, or gravity for an orbit.

If that force disappears, the object does not fly outward. By the first law it carries straight on along the tangent, the direction it was heading at that instant.

Formalise

Worked example

A 1000 kg car takes a flat bend of radius 50 m at 15 m/s.

1. Force needed. $F = 1000 \times 15^2 / 50 = 4500\text{ N}$, supplied by sideways friction on the tyres.
2. Top speed. With $\mu_s = 0.7$ the most friction available is $0.7 \times 1000 \times 9.81 = 6867\text{ N}$. Setting $mv^2/r = \mu_s mg$ gives $v = \sqrt{\mu_s g r} = \sqrt{0.7 \times 9.81 \times 50} = 18.5\text{ m/s}$.

Above 18.5 m/s the tyres cannot supply enough sideways force and the car slides toward the outside of the bend, following a straighter path than the road.

Your turn

Task 6 - The force accumulator

A physics engine never solves a problem like the ones above with algebra. Instead each body keeps a force accumulator. Every step, each force source adds its force into the accumulator. Then the body divides the total by its mass, updates its velocity and position, and clears the accumulator ready for the next step. Gravity, springs, drag, thrusters and wind all use the same addForce call, so new forces can be added without changing the integration code.

Create a new Console App (C++) project named ForceAccumulator and replace its source file with the skeleton below, saved as ForceAccumulator.cpp. It models the 0.5 kg mass hanging on a 50 N/m spring from Task 4, with y pointing up and the spring's anchor at the origin.

1. Complete addForce so it adds f into force.
2. Complete integrate. Calculate the acceleration from the accumulated force and the mass, then update velocity and then position, scaled by dt.
3. Complete clearForces so the accumulator returns to zero.
4. In the loop in main, add the three forces before integrate is called. Gravity is (0, -m g). The spring force is -k times the body's displacement from anchor. Drag is -b times the body's velocity.

Success condition: the printed y values swing below and above about -0.098 m with shrinking swings, and by 10 s they have settled on the predicted equilibrium printed at the end. Then set b to 0.0 and confirm the swings no longer shrink.

#include <cmath>
#include <iomanip>
#include <iostream>

struct Vec2
{
    double x;
    double y;
};

Vec2 operator+(Vec2 a, Vec2 b) { return { a.x + b.x, a.y + b.y }; }
Vec2 operator-(Vec2 a, Vec2 b) { return { a.x - b.x, a.y - b.y }; }
Vec2 operator*(Vec2 a, double s) { return { a.x * s, a.y * s }; }
Vec2& operator+=(Vec2& a, Vec2 b) { a.x += b.x; a.y += b.y; return a; }

struct Body
{
    double mass;
    Vec2 position;
    Vec2 velocity;
    Vec2 force;

    void addForce(Vec2 f)
    {
        // TODO: add f into the force accumulator
    }

    void integrate(double dt)
    {
        // TODO: acceleration is the accumulated force divided by mass
        // TODO: update velocity, then position, with semi-implicit Euler
    }

    void clearForces()
    {
        // TODO: reset the force accumulator to zero
    }
};

int main()
{
    const double g = 9.81;
    const double k = 50.0;     // spring stiffness in N/m
    const double b = 1.0;      // linear drag coefficient in kg/s
    const Vec2 anchor{ 0.0, 0.0 };
    const double dt = 0.001;

    Body body{ 0.5, { 0.0, 0.0 }, { 0.0, 0.0 }, { 0.0, 0.0 } };

    std::cout << std::fixed << std::setprecision(4);
    std::cout << "t (s)    y (m)\n";

    for (int step = 0; step <= 10000; step++)
    {
        if (step % 500 == 0)
        {
            std::cout << step * dt << "   " << body.position.y << "\n";
        }

        // TODO: add gravity, (0, -m g)

        // TODO: add the spring force, -k times the displacement from the anchor

        // TODO: add linear drag, -b times the velocity

        body.integrate(dt);
        body.clearForces();
    }

    std::cout << "\nPredicted equilibrium y = " << -body.mass * g / k << " m\n";
    return 0;
}