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Lesson 3 · 45 min

Forces, weight and moments

Use net force, distinguish weight from mass and calculate moments.

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Forces, weight and moments

A is a push or pull that can change an object’s motion or shape. is a vector, so both magnitude and direction matter. Several forces may act together; their vector sum is the resultant or net . Opposite forces of equal magnitude can balance, producing zero net even though the individual forces are not zero.

Newton’s first law states that a body remains at rest or moves with constant in a straight line unless a net external acts. Inertia is to a change of motion, and mass is a measure of inertia. A moving object does not require a continuing forward net to maintain constant ; in everyday situations, often hides this principle.

Key definition

Inertia

Resistance to a change of motion, and mass is a measure of inertia.

Exam keywords:resistancechangemotionmeasure

For a constant-mass body, Newton’s second law is. The F in this equation is the net , not necessarily a single applied . A 4 kg trolley experiencing a 12 N forward pull and 4 N backward has a net forward of 8 N. Its is in the forward direction.

Key equation
F=ma
84=2ms²
Where: F: resultant force (N); m: mass (kg); a: acceleration (m/s²).

Mass measures the amount of inertia and is measured in kilograms. Weight is the gravitational on that mass and is calculated as, in newtons. The value of g depends on location and the model in the question. If is specified, a 3 kg object weighs 30 N. Its mass remains 3 kg even if the gravitational field changes.

Key equation
W=mg
g=10ms²

is mass multiplied by . In an isolated system with no net external , total is conserved. Newton’s third law describes the interaction forces between two bodies: they have equal magnitudes, opposite directions and act on different bodies. Because the forces act on different bodies, they are not a balanced pair acting on one object.

A can also produce turning. The moment of a about a point equals the times its perpendicular distance from that point. For rotational , total clockwise moment equals total anticlockwise moment. Draw a diagram and identify the pivot before calculating. A distance measured along a tilted rod is not automatically the perpendicular lever arm needed by the formula.

Drawing a complete diagram

Isolate the body under discussion and replace its surroundings with forces acting on it. For a book on a level table, weight acts downward and the normal contact upward. For a pulled trolley, include the pull, , weight and normal reaction. Draw arrows from a simple box or point and label their directions. A arrow is not an extra . Nor should a that the body exerts on its surroundings be included on its own diagram. Resolve sloping forces into perpendicular components before forming separate equations. Balanced vertical forces do not imply that horizontal forces are balanced as well.

The scope of Newton's laws

Newton's first law identifies inertial motion: zero resultant means zero , which allows either rest or constant nonzero . The second law can be written as resultant equals rate of change of ; for constant mass this becomes. The third law applies to each interaction and produces forces on two different bodies. Earth's pull on a stone and the stone's pull on Earth form one pair. The stone's weight and a table's upward support are not a third-law pair, because both act on the stone. The forces can balance on the stone while having entirely different reaction partners.

Key equation
F=ma
Where: F: resultant force (N); m: mass (kg); a: acceleration (m/s²).

, tension and connected objects

Dry opposes relative sliding or the tendency to slide at a contact. Static adjusts up to a limiting value; it is not always equal to μR. During idealised sliding, the model uses the given and normal reaction. The normal reaction is not always mg, especially when a has a vertical component or the support is inclined. In an ideal light, inextensible string over a smooth pulley, connected bodies share the same magnitude and the tension is uniform. Write a separate equation for each body, choosing directions that match the proposed motion, then solve the equations together.

Key equation
F=μR
F=ma
Where: F: resultant force (N); m: mass (kg); a: acceleration (m/s²).

Impulse and during collisions

Impulse is the change in : for a constant or average , and for a constant-mass object. On a –time graph the signed area gives impulse. A longer stopping time reduces the average stopping for the same change, which explains the role of cushioning and bending knees when landing. Include signs when an object rebounds. A0.2 kg ball changing from +10 m/s to −5 m/s has. The is directed opposite to the original motion, and its magnitude depends on contact duration.

Key equation
J=FΔt
J=mv−mu
Δp=0.2(−5−10)=−3Ns,not−1Ns

Conservation of and

For an isolated pair during a brief collision, internal interaction forces cancel in the total-system equation. Thus total before equals total after, provided external impulse is negligible. If two bodies stick together, their common follows (m₁u₁ + m₂u₂)/(m₁ + m₂). Kinetic is conserved only in an elastic collision, not in every collision. In an inelastic collision some initial kinetic becomes internal , deformation or sound, although total remains conserved. Recoil of a gun or separation of two initially stationary skaters also illustrates conservation: their momenta are equal and opposite, but their speeds need not be equal.

Turning effects and balance

The moment about a pivot uses the perpendicular distance to the line of action, not any convenient measured length. A whose line passes through the pivot has zero moment about that pivot. A couple consists of equal, opposite, parallel forces acting along different lines, producing a turning effect without a net . Its moment is either multiplied by the perpendicular separation. For a beam in complete static , both net and net moment must vanish. Choose a pivot that removes an unknown support from the moment equation, then use vertical balance to find it afterwards.

Key definition

Its moment

Either force multiplied by the perpendicular separation.

Exam keywords:eithermultipliedperpendicularseparation

Using units and experimental evidence

has unit kg m/s, equivalent to N s, while a moment has unit N m. Although N m also has the dimensions of a joule, moments and work describe different physical quantities and are not interchangeable answers. A balanced metre-rule experiment can compare known weights and distances from a knife edge, including the rule's own weight through its centre of gravity. Repeated readings and a stable horizontal position improve confidence. If an object does not accelerate, do not claim no forces are present: demonstrate that their resultant is zero. In a written explanation, connect the observation to the relevant law and identify the body or system precisely.

Self-check with reasoning

A2 kg cart moving at6 m/s collides with a stationary1 kg cart, and they remain together. Initial is12 kg m/s; combined mass is3 kg, so final is4 m/s. Initial kinetic is36 J and final kinetic is24 J, so the collision is inelastic. The missing12 J has been transferred into deformation, internal and sound. In a second problem, a10 N downward acts0.4 m to one side of a pivot. A at0.2 m on the opposite side must be20 N for rotational balance. That alone does not determine the support reaction; the vertical- equation is still required.

Topic revision bank

Key rules & principles

  • Third-law forces act on different bodies.
  • Moments use perpendicular distances.
  • Third-law partners act on different bodies.
  • Impulse is signed change.
  • conservation does not guarantee kinetic- conservation.

Worked examples

A 5 kg body has a net force of 15 N. Find acceleration.
  1. Use F = ma.
  2. Rearrange: a = F/m.
  3. a = 15/5.

Answer: 3 m/s²

A0.15 kg ball rebounds from +12 m/s to −8 m/s in0.05 s. Find average force.
  1. Δp = 0.15(−8 −12) = −3 kg m/s.
  2. F = Δp/Δt = −3/0.05.
  3. The sign identifies direction relative to the chosen positive direction.

Answer: −60 N, or60 N opposite to original motion.

A3 kg moving cart at4 m/s sticks to a stationary1 kg cart. Find final velocity.
  1. Assume negligible external impulse.
  2. Initial = 3(4) +1(0) =12 kg m/s.
  3. Joined mass is4 kg, so4v =12.

Answer: 3 m/s in the original motion direction.

Related videos and resources

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Newton's Laws: Crash Course Physics #5

CrashCourse

Newtonian mechanics. Both YouTube metadata and the publisher's PBS episode page were found.

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