Forces and equilibrium

A heavy door opens easily when pushed at the handle, but is hard to open near the hinges. A pulled block may remain at rest or start sliding. How can we predict which forces act and whether they change the body's overall motion or its rotation?

A ball falls towards Earth, and Earth is attracted towards the ball. In Newton's model all masses attract, although the force between everyday objects is tiny. For two point particles, specify masses m_1,m_2 in kg and separation r in metres. Bodies may be treated as points when their sizes are negligible. For non-overlapping spherically symmetric bodies, the point-particle law gives the attraction exactly using the distance between centres.

The experimental relation is F=Gm_1m_2/r^2, where G\approx6{,}67\cdot10^{-11}\,\text{Н}\cdot\text{м}^2/\text{кг}^2. Doubling one mass doubles the force; doubling both quadruples it. Doubling separation reduces force to a quarter: distance enters squared. Attraction weakens at large distances. Do not set r=0 for real bodies: the point model no longer describes their arrangement.

Substitute the planet's mass M and the surface distance R from its centre. For a small body, F=GMm/R^2, so a=F/m gives g=GM/R^2 and F_{\text{тяж}}=mg. The falling body's mass cancels: more mass increases both attraction and inertia in the same proportion. Assume a spherically symmetric planet and neglect rotation; far above the surface, distance from the centre is no longer R.

Hang a load from a spring: it stretches and pulls the load back. Displacing particles in the material changes their interactions, producing an elastic force that tends to reduce deformation. Let \Delta l=l-l_0, in metres, be the length change from the unloaded state. It is negative in compression and positive in extension. Recovery after unloading means elastic deformation; it does not by itself guarantee a linear force–extension relation.

For small deformations of a linear spring, doubling extension doubles the restoring force. Hooke's law describes this region: F_{\text{упр}}=k|\Delta l|. Spring constant k, in N/m, gives force per metre of deformation; a stiffer spring needs more force for the same extension. Magnitude alone gives no direction: the force acts to restore shape. At \Delta l=0 it vanishes; for large deformations, proportionality may fail even before breakage.

Earth pulls a standing person down while the floor pushes up. The normal force N acts on the person perpendicular to the support. Apparent weight P is the force the person exerts on the support or suspension because of Earth's attraction; here the recipient is the floor. Magnitudes may match, but these forces are not interchangeable on a force diagram. Name both interacting objects, then draw the arrow on the one whose motion you study.

At a horizontal contact with no tangential component, the body's force on the support and the support's force on the body form a third-law pair. Their magnitudes are equal: P=N. This remains true when the support accelerates and says nothing by itself about mg. Imagine a lift starting upwards: the floor pushes harder on the person and the person presses harder on the floor. Find the common magnitude from the person's equation of motion.

If the only vertical forces are N upwards and mg downwards, an upward axis gives N-mg=ma_y. Thus N=mg only when a_y=0. Upward acceleration increases the normal force; downward acceleration reduces it. In free fall, a_y=-g and N=0. An additional vertical pull also changes the result. So N=mg follows from specific conditions; it is not a universal property of a support.

A block slides on a table, and surface interaction resists their relative sliding. In the dry-friction model, F_{\text{тр}}=\mu N: greater normal load gives greater resistance. The dimensionless coefficient \mu depends on the surface pair and its condition. This is approximate. Friction acts along the contact opposite to relative sliding: on a moving belt, it may accelerate the block relative to the floor rather than slow it down.

Pull a stationary block first with 1 N and then 2 N without moving it. Static friction is respectively 1 and 2 N, balancing the pull. Contact has a limit: F_{\text{тр}}\leq\mu_{\text{п}}N. Equality occurs at the threshold of slipping; with no horizontal pull, friction may be zero. The static coefficient \mu_{\text{п}} need not equal the kinetic coefficient.

On a horizontal table with no vertical acceleration or additional vertical forces, N=mg. If the block is sliding, substitution gives F_{\text{тр}}=\mu mg. Doubling mass then doubles friction. But pull the block with an angled rope: part of the pull unloads the support, reducing N, so substituting mg is wrong. Check both underlying conditions before using the combined formula.

Place a block on a slope at angle \alpha to the horizontal, with axes along and perpendicular to it. The same gravitational force has components F_{\parallel}=mg\sin\alpha down the slope and mg\cos\alpha into it. These are force components, not new interactions. At \alpha=0, gravity has no along-surface pull; increasing inclination increases that component and reduces the normal one. These limits help distinguish sine from cosine.

While a block stays in contact with a stationary plane, its acceleration perpendicular to it is zero. If only the normal force and the perpendicular gravity component act on that axis, N-mg\cos\alpha=0, so N=mg\cos\alpha. A horizontal plane gives mg; approaching vertical gives a smaller normal load. An extra force pressing the block would change this balance. Merely naming an inclined plane does not determine its normal force.

Apply the sliding model to the normal force just found: F_{\text{тр}}=\mu mg\cos\alpha. A block sliding down a fixed slope experiences friction upwards. If thrown up, friction points down along with gravity's slope component. Relative velocity sets the direction, not gravity. For a stationary block, first find the force needed for equilibrium and compare it with the static-friction limit.

Return to the door. Pushing across it at the handle turns it easily, while pushing along it towards the hinges barely turns it. Both hand position and the force's line of action matter. Moment arm l is the perpendicular distance from the axis to that line in a planar problem. Measured in metres, it may be shorter than the axis-to-hand distance. If the line crosses the axis, the arm is zero even at a distant handle.

In a planar problem, torque describes a force's turning effect: its magnitude is M=Fl, in N·m. Doubling force at the same arm doubles torque; doubling the arm at the same force does likewise. At l=0, torque vanishes: the force may load the hinges without turning the door. Choose the sign by rotation direction. Although N·m has the same dimensions as a joule, torque and work answer different questions.

For an initially stationary rigid body to remain in planar equilibrium, check \sum\vec F=0 and \sum M=0. The first prevents centre-of-mass acceleration; the second prevents angular acceleration. Opposite senses of rotation have opposite moment signs. A zero sum does not stop motion that already exists. Static equilibrium also requires no initial translation or rotation.

Apply equal opposite tangent forces at opposite sides of a steering wheel. Their vector sum is zero, but both turn the wheel the same way: their moments add. This force couple shows why testing only net force is insufficient for an extended body. A point particle has no size and its rotation is not described; moving to a door or lever extends the model and requires a moment check.

A 50 kg sledge slides right on a horizontal surface. The pull is 40 N right and the friction coefficient is 0.02. Take g=10 м/с². Find acceleration.

There is no vertical acceleration or additional vertical force: N=mg=500 Н.

Kinetic friction points left: Fтр=μN=0,02×500=10 Н.

Horizontal net force: 40−10=30 Н to the right.

Acceleration a=30/50=0,6 м/с² to the right. It is smaller than 40/50 because some of the pull balances friction.

The sledge accelerates at 0.6 m/s² rightwards: under the same conditions its speed increases by 0.6 m/s each second. Of the 40 N pull, 30 N remains after friction; without friction acceleration would be 0.8 m/s².

Choose a force formula after its physical model: small deformation, dry sliding, static contact or attraction between suitable bodies.

Find normal force from motion perpendicular to the surface. N = mg and Fтр = μmg apply under specific conditions.

An extended body may rotate with zero net force. Static equilibrium requires balanced moments and a suitable initial state too.

Next we describe collisions using momentum.