Newton’s laws and free-body diagrams

Choose one cart and name its forces

A hand pushes a cart rightward with 12 N while the track exerts a 3 N leftward resistive force. Should we subtract these numbers, add another force because the cart is moving, or include the force the cart exerts on the hand? First choose the body: our system is the cart alone. The hand, track and Earth are outside its boundary.

We use a supplied teaching model with constant mass m = 3 kg. The cart is represented by a point for its translation; rotation, deformation and the actual locations of contact forces are outside this model. Choose fixed perpendicular axes, x rightward and y upward. Time t starts at the declared initial instant.

For 0 ≤ t ≤ 2 s, retain only the four external force components in the table. Supply downward gravitational acceleration of magnitude g = 10 m/s², so the Earth’s gravitational force on the cart has magnitude m g = 30 N. The upward support S is initially unknown. The cart stays at constant height. These are stipulated conditions, not measurements or universal values for a real track.

Connect resultant force to acceleration

Velocity describes the rate and direction of position change, in m/s. Acceleration describes the rate of velocity change, in m/s². Write these vectors as v and a. The symbol Δ means a change: over a time interval Δt, constant acceleration gives Δv = a Δt. A rightward velocity tells us that the cart is moving right; it does not by itself tell us the direction of acceleration.

Newton’s second law in our positive constant-mass particle model is F_net = m a. The resultant F_net is the vector sum of the external forces on the chosen body. In these axes, F_net,x = m a_x and F_net,y = m a_y. An ordered pair gives x first and y second, with its trailing unit applying to each entry. Force is measured in newtons: 1 N = 1 kg m/s². Dividing N by kg gives m/s², not m/s.

The selected reference frame is assumed inertial: a free particle would keep a constant velocity in it. This condition matters; merely drawing fixed axes does not establish it. An accelerating reference frame generally needs additional terms in its equations of motion. Here we use the inertial-frame balance throughout.

Constant height gives a_y = 0. With only S upward and 30 N downward, S − 30 N = (3 kg)(0 m/s²), so S = 30 N. Horizontally, F_net,x = 12 N − 3 N = 9 N and a_x = 9 N/(3 kg) = 3 m/s². Thus F_net = (9, 0) N and a = (3, 0) m/s².

Draw individual forces and a separate resultant

A free-body diagram isolates the chosen body and shows the external forces acting on it. For this point model, draw the four arrows from the same point and label each by its agent and recipient. Their common origin represents the translational calculation; it does not assert that the real forces are applied at the same physical place.

The figure uses one force scale in both panels. The 30 N support and weight arrows have equal lengths and opposite directions. The rightward 12 N arrow is four times as long as the leftward 3 N arrow. The coordinate guide identifies directions; its short lines are not forces.

The right panel shows the computed 9 N resultant separately. Do not add it as a fifth external force to the left panel. Likewise, velocity and acceleration are different quantity kinds and do not belong in this force inventory. Support S and resistance R are perpendicular components of the track’s contact force; use those two arrows without also adding their combined contact vector.

Balanced force can mean steady motion

Newton’s first law says that, in an inertial frame, zero resultant external force throughout an interval leaves a particle’s velocity constant. This includes remaining at rest, but also continuing in a straight line with a constant nonzero velocity. The initial velocity decides which.

Consider a separate variant of the cart model: replace the 12 N hand force by 3 N rightward and retain the 3 N resistance leftward, the 30 N support and the 30 N weight. All signed components sum to zero. If the initial velocity is (1, 0) m/s, it stays (1, 0) m/s for the supplied interval. This particle is in translational equilibrium while moving.

Zero velocity at one instant is a different statement. A particle can be momentarily at rest while its velocity is changing. Nor does a nonzero velocity require an extra “force of motion”: identify interactions, rather than inventing a force from the direction of travel.

Third-law pairs have two recipients

Newton’s third law pairs mutual interaction forces in this Newtonian model: the force of A on B is equal in magnitude and opposite in direction to the force of B on A. The labels exchange agent and recipient. “Action” and “reaction” name simultaneous members of the pair; they do not mean an earlier push followed by a later response.

The main cart’s upward support and downward weight are not a third-law pair. Both act on the cart. The partner of the Earth’s downward force on the cart is the cart’s upward gravitational force on the Earth. The partner of the track’s upward support on the cart is the cart’s downward force on the track.

Choose separate diagrams for A and B and place the appropriate member on each. The pair does not cancel the force on either body. If instead both bodies lie inside one combined boundary, the paired interaction is internal and cancels from that combined force sum. Other external forces must still be identified.

Use the conditions to find unknown forces

The second law does not supply every force magnitude on its own. A spring, friction, gravity or another interaction needs a suitable force model or supplied information. Our 12 N push and 3 N resistive contact are given. We have not derived a friction rule or claimed that resistance always has this magnitude or direction.

The support equals m g here because the vertical acceleration is zero and no other vertical force is retained. To see why that condition matters, change the model at one instant: keep m = 3 kg and weight 30 N, but supply upward acceleration a_y = 2 m/s² with only support and weight vertical. Then S − 30 N = (3 kg)(2 m/s²), so S = 36 N. “Normal” means perpendicular to the contact surface, not automatically equal to weight.

A complete force calculation states which quantities are supplied and which are solved, then checks that the result respects the model’s constraints. Neither the drawn arrow lengths nor the arithmetic establishes that a real cart has exactly these forces.

Keep direction, body and time consistent

Start each calculation by naming the body and the instant or interval. List the external interactions, select signs, and add matching vector components. Divide the resultant by that body’s positive constant mass. Use an initial velocity only when the question also asks how velocity changes.

The 12 N and 3 N horizontal magnitudes do not add to a 15 N rightward force: the signed components are +12 N and −3 N. Similarly, a third-law force on the hand cannot be added to the cart’s force balance. Mixing recipients changes the system without declaring it.

Force and acceleration have the same direction in F_net = m a with m positive. Velocity can have another direction. At a separately supplied instant, a 3 kg cart moving right at 2 m/s can have a 12 N rightward push and 15 N leftward resistance. Its horizontal resultant is −3 N and a_x = −1 m/s². It is moving right while slowing at that instant; this example specifies no later resistance law or trajectory.

Keep the Newtonian model in scope

These examples concern translation of a constant-mass particle in an assumed inertial frame. A real extended body can also rotate or deform. Zero resultant force alone does not establish its full mechanical equilibrium; rotational balance is a separate question. Our common-origin point diagrams omit that question explicitly.

Newtonian mechanics is an approximation to use when relativistic and quantum effects can be neglected for the question and required accuracy. Motion near the speed of light and behaviour requiring a quantum description need other models. A single speed or size cutoff does not decide validity for every situation.

You can work through the cases using signed arithmetic, vectors and Δv = a Δt for constant acceleration; no calculus is required. The next lesson, Friction, tension, drag, springs, and contact forces, examines force models that were supplied here. It can refine an inventory without adding a force merely because something moves.

Worked examples

Try it yourself

1. At one instant, a 3 kg cart has v_x = +2 m/s, a +12 N hand force and a −15 N resistive force. Other force components balance. Find a_x and say whether the direction of motion must already be leftward.

   Show the answer and reasoning

   a_x = −1 m/s². The cart is moving rightward at that instant.

   The horizontal resultant is 12 N − 15 N = −3 N; dividing by 3 kg gives −1 m/s². Velocity and acceleration have different signs. Later motion requires an interval model, which is not supplied.

2. Someone adds the separately drawn 9 N resultant to the main cart’s four forces and obtains 18 N rightward. What has gone wrong?

   Show the answer and reasoning

   The same interactions have been counted twice. The correct resultant remains 9 N rightward.

   The 9 N arrow is the computed sum of P, R, S and W, not a fifth interaction. The cart-only inventory has four external force arrows.

3. At a different supplied instant, m = 3 kg, weight is 30 N downward and a_y = +2 m/s². Only weight and normal support act vertically. Find the upward support S.

   Show the answer and reasoning

   S = 36 N upward.

   S − 30 N = (3 kg)(2 m/s²) = 6 N. Therefore S = 36 N. The main example’s constant-height condition has been replaced, so S = 30 N cannot be reused.

4. In an inertial frame, a positive constant-mass particle has zero resultant force throughout an interval and initial velocity (1, 0) m/s. Is it at rest? Conversely, does observing v = (0, 0) m/s at one instant establish zero resultant force?

   Show the answer and reasoning

   It is not at rest: its velocity stays (1, 0) m/s. Instantaneous zero velocity alone does not establish zero force.

   The interval force condition fixes acceleration to zero, not the initial velocity. At a single instant a zero velocity can be changing, so an acceleration or force condition is still needed.

5. In the main cart model, which force is the third-law partner of the track’s 30 N upward support on the cart? Is the Earth’s 30 N downward force on the cart that partner?

   Show the answer and reasoning

   The partner is the cart’s 30 N downward force on the track. The Earth’s force on the cart is not that partner.

   A third-law pair exchanges the same two bodies as agent and recipient. Support and weight in the cart’s diagram both have the cart as recipient and arise from different interactions.

6. Use the 2 kg A and 4 kg B interaction case. A learner says both bodies have acceleration magnitude 3 m/s² because their interaction forces have equal magnitude. Correct the claim.

   Show the answer and reasoning

   A has acceleration −3 m/s² and B has acceleration +1.5 m/s² along x. Their acceleration magnitudes differ.

   Each 6 N magnitude is divided by its recipient’s mass. Equal and opposite interaction forces do not specify equal masses or equal accelerations. The statement concerns the supplied instant only.

What to carry forward

• Choose a body and name each external force by its agent and recipient before summing components.
• In a positive constant-mass Newtonian particle model and an inertial frame, F_net = m a connects resultant force to acceleration.
• The resultant is a computed vector, not another interaction to add to the same inventory.
• Balanced forces over an interval allow constant nonzero velocity; motion alone creates no extra force.
• Third-law partners act on distinct bodies. Equal force magnitudes can give different acceleration magnitudes.
• Constraints and force models determine unknowns; point diagrams and Newtonian approximations have stated limits.