Friction, tension, drag, springs, and contact forces

Start with a pull that does not move the body

A 2 kg body rests on a fixed horizontal floor. Choose right as +x and up as +y in a fixed inertial frame. Supply g = 10 m/s² for this teaching model, so its weight is 20 N downward. A horizontal pull of 6 N to the right need not make it slide: the floor can exert 6 N of static friction to the left. The horizontal forces then cancel.

For this floor–body pair, supply the dry-contact coefficients μ_s = 0.4 and μ_k = 0.25. These dimensionless numbers are stipulated, not measurements of a named material. The floor's normal force is 20 N in this first case, so the static friction bound is (0.4)(20 N) = 8 N. The required 6 N fits inside that bound; the actual friction need not be 8 N.

The published Units, standards, and physical constants lesson prepares quantity values and units. Algebra, functions, graphs, and logarithms prepares signed arithmetic and substitution; no logarithm is needed here. Vectors and vector spaces is an optional return path for component notation. The force sums used below are introduced locally, so an unpublished mechanics lesson is not required.

Draw only forces on the chosen body

A free-body diagram isolates the chosen body and represents interactions acting on it. In the floor cases, list its weight, the floor's normal force, the floor's friction and the applied pull. Weight is gravitational; the floor, string, spring and fluid forces considered here arise through contact. Forces the body exerts on the floor or string belong to those other bodies' diagrams.

Force is a vector, measured in newtons, with 1 N = 1 kg m/s². For a constant-mass particle model in an inertial frame, the external force sum obeys ΣF_x = m a_x and ΣF_y = m a_y. The subscripts denote signed components along our perpendicular axes. A body's velocity is not an extra force. A zero force sum gives zero acceleration; it permits constant velocity, including rest.

At a contact surface, the normal component is perpendicular to the surface and friction is parallel to it. On our horizontal floor the normal force is upward. Write N for its nonnegative magnitude and f_x for signed friction. These are two components of the floor's contact force; do not add a third 'total contact force' after already including them.

The diagrams use a point approximation for translational motion. They do not specify force application points or settle an extended body's torque balance, tipping or rotation. We assume the horizontal support contact can be maintained in the cases where that condition is stated.

Separate static adjustment from sliding

In the stated static model, |f_x| ≤ μ_s N. First calculate the force needed to avoid relative sliding, then check this inequality. For rest on this stationary floor with only a horizontal pull P_x besides friction, balance requires f_x = −P_x. At P_x = 2 N, friction is −2 N. At P_x = 8 N, friction is −8 N and reaches the inclusive bound. Equality alone does not require the model to switch to sliding.

If P_x = 10 N, maintaining rest would need |f_x| = 10 N, exceeding 8 N. Rest is therefore impossible under these assumptions. This test does not license using the static maximum as the actual friction of an already-sliding body. Choose the force model that matches the relative state.

For the separate condition of a body already sliding right relative to the same floor, supply the kinetic law f_x = −μ_k N. A 10 N rightward pull then competes with friction −5 N, leaving +5 N and a_x = (5 N)/(2 kg) = +2.5 m/s². Sliding left would reverse the friction direction. At zero relative speed, reassess the state; do not assign the right-sliding sign automatically.

Static means no relative sliding at the contact, not necessarily rest in every frame. Its direction is determined by the motion constraint and other forces. The dry-friction formulas are useful contact approximations, not universal laws for all surfaces, speeds or loads. The two coefficients chosen here are part of this supplied model.

Let tension change the normal force

An ideal taut flexible string pulls the body along the string; it does not push like a rigid rod. Uniform tension is justified for a straight massless segment with only its two collinear endpoint forces: their sum must be zero. A massive string, another force along the segment or a frictional pulley requires further analysis. A slack string cannot provide the stipulated nonzero tension.

For case C, supply one tension vector with T_x = 6 N and T_y = 8 N. Its magnitude is √[(6 N)² + (8 N)²] = 10 N. These components describe the same force. Draw one diagonal tension arrow, not that arrow plus two additional component forces.

Maintained contact with the fixed horizontal floor gives a_y = 0. With only normal force, tension and weight in the vertical direction, N + T_y − mg = 0, hence N = 20 N − 8 N = 12 N. The normal force is condition-dependent; it is not automatically equal to weight. If the body is already sliding right, friction is −(0.25)(12 N) = −3 N, so the horizontal net is 6 N − 3 N = +3 N and a_x = +1.5 m/s².

This nonadhesive floor can push but cannot pull the body downward: N must be nonnegative. A calculation demanding N < 0 invalidates maintained contact. For example, a purely upward 24 N pull on this 20 N weight would demand N = −4 N. Instead the body separates, N becomes zero, and the remaining vertical forces must be used to calculate its upward acceleration. Dry-contact friction also vanishes when this contact is absent.

Compare three force diagrams

Predict which force changes when a pull gains an upward component, then compare cases B and C. Every arrow uses the same force scale in both directions and across all three cases. Weight stays 20 N, while normal force falls from 20 N to 12 N. The right-sliding friction magnitude consequently falls from 5 N to 3 N, but the horizontal pull also falls from 10 N to 6 N. The horizontal acceleration is therefore smaller in C.

Case A is a feasible rest condition; B and C are separately supplied right-sliding conditions. They are not consecutive times in one observed motion. Arrow directions and lengths represent forces, not velocity, travelled paths or the body's size. The table lists signed components; the net force is their sum, not an additional interaction to draw.

Use a signed spring deformation

For a separate horizontal model, attach a spring to a body on a smooth track and choose right as positive. Let q be its signed deformation from the relaxed position, in metres. Supply the linear law F_s,x = −k q, with k = 30 N/m and the declared range −0.15 m ≤ q ≤ 0.15 m. The sign gives a restoring force toward the relaxed position.

At q = +0.10 m the spring force is −3 N, toward the left. At q = −0.10 m it is +3 N, toward the right. This attached spring can act under both extension and compression. Its signed x component can point either way. An ideal taut string instead pulls along the string, with a nonnegative tension magnitude. The positive coefficient k is a stiffness with N/m units, not a force by itself.

Linear spring behavior is a model used over a suitable range; a real spring's response and limits need evidence. Outside the declared interval, no spring force prediction is supplied here. Other attachment geometry, large deformations, spring mass or internal damping would require additional information.

Make drag relative to the fluid

Let v_body and v_fluid be body and fluid velocity components in the same frame. Their relative velocity is u = v_body − v_fluid. Supply a one-dimensional linear drag model F_d,x = −b u, where b = 1 kg/s = 1 N s/m and |u| ≤ 3 m/s. For u = −2 m/s the drag is +2 N. It opposes the relative motion through the fluid, which need not match the body's ground-motion direction.

Combine this drag with the separate spring case at q = +0.10 m. For a stationary fluid and v_body = −2 m/s, the spring contributes −3 N and drag +2 N. Their net is −1 N, giving a_x = −0.5 m/s² for the 2 kg body. The body is moving left and accelerates left at this instant, so its speed increases despite the opposing drag. On this smooth horizontal track, weight and normal force cancel vertically; no dry friction or other horizontal force is supplied.

A different supplied model could be quadratic: F_d,x = −c u |u|, with c = 0.5 kg/m = 0.5 N s²/m² on the same declared |u| ≤ 3 m/s interval. At u = −2 m/s it also gives +2 N, but at u = +1 m/s it gives −0.5 N while the linear model gives −1 N. Agreement at one velocity does not identify a force law. The coefficient units differ, so the same numerical coefficient cannot simply be transferred between laws.

These are stipulated force relations for practice, not a claim that either law fits every fluid or speed. Selecting an actual drag model needs information about the body, fluid and regime. An instantaneous force calculation alone does not supply a full trajectory or terminal speed.

Check the state before using a force law

Start each calculation by naming the chosen body, frame, directions, contact state and interactions. Then resolve the forces, apply the stated translational balance, and test the model conditions. A required static force beyond its bound, a negative nonadhesive normal force or a slack string with nonzero tension signals an inconsistent assumption.

Keep the two tasks distinct: balance tells you the force required by a proposed state, while a force model tells you whether that force is available. When the state changes, reconsider the relevant law. A kinetic-friction formula does not determine static friction, and an assumed linear spring or drag law does not establish its own physical range.

Work, energy, and power will relate force to displacement and transfer. The present diagrams answer which interactions act on one body and what instantaneous acceleration the stated models permit; they do not calculate work or prove an experimental force law.

Worked examples

Try it yourself

1. On case A's fixed floor, replace the horizontal pull by +2 N and require the body to remain at rest. Find signed friction and compare it with the 8 N bound.

   Show the answer and reasoning

   Static friction is −2 N, with magnitude 2 N below the 8 N bound.

   Horizontal balance gives +2 N + f_x = 0. The bound permits this value; it does not require friction to take its maximum magnitude.

2. With the same purely horizontal geometry, test proposed rest under pulls of +8 N and +8.1 N. Does the equality case alone require slipping?

   Show the answer and reasoning

   Rest is admissible at +8 N with friction −8 N, but inadmissible at +8.1 N. Equality alone does not require slipping.

   The required magnitudes are 8 N and 8.1 N, while μ_s N remains 8 N. The static inequality includes equality. The second proposed state demands more force than this model permits.

3. A straight massless string segment has only two endpoint forces: +9 N and −6 N along x. Can this be the stated ideal segment with uniform tension?

   Show the answer and reasoning

   No. The endpoint forces sum to +3 N rather than zero.

   A massless segment with no other forces requires a zero force sum; equal opposite endpoint forces would give uniform tension. The supplied unequal values contradict that idealization. They are not a third-law pair acting on two different bodies: both forces here act on the chosen segment.

4. For the (+6, +8) N pull in case C, test a proposed rest condition using static rather than kinetic friction. Why is comparing 6 N with case A's old 8 N bound incorrect?

   Show the answer and reasoning

   The normal force is now 12 N and the static bound 4.8 N. Rest would require friction −6 N, so it is inadmissible.

   The upward pull reduces normal force from 20 N to 12 N. Recalculate μ_s N = (0.4)(12 N) before comparing it with the required 6 N magnitude. Case A's bound belongs to a different vertical force balance.

5. The same 2 kg body is pulled purely upward by 24 N while initially at rest on the nonadhesive floor. With no other force besides weight and possible floor contact, what replaces the invalid maintained-contact result N = −4 N?

   Show the answer and reasoning

   The body separates: N = 0 N, floor friction is zero and a_y = +2 m/s².

   This floor cannot pull downward. Once contact is absent, the vertical force sum is 24 N − 20 N = +4 N; dividing by 2 kg gives +2 m/s². Do not retain a negative normal force or use it in a dry-friction formula.

6. For the linear drag model alone, let v_body = +1 m/s and v_fluid = +3 m/s in the same frame. Find u and drag with b = 1 kg/s. Is this force necessarily opposite the body's ground-motion direction?

   Show the answer and reasoning

   u = −2 m/s and drag is +2 N. It points right, the same direction as the body's ground motion in this case.

   Subtract fluid velocity from body velocity: +1 m/s − 3 m/s = −2 m/s, inside the declared range. Then −b u = +2 N. The force opposes body–fluid relative motion; ground-motion direction alone is insufficient.

What to carry forward

• A free-body diagram contains interactions acting on one chosen body; motion itself does not add a force.
• Static friction adjusts to the no-slip requirement within a bound; a maximum is not automatically the actual force.
• Sliding friction in the supplied model opposes relative sliding, with magnitude μ_k N.
• Normal force follows the perpendicular balance and contact conditions; a nonadhesive floor requires N ≥ 0.
• Tension pulls along a taut ideal string, and resolving its components does not create extra forces.
• A spring uses signed deformation; drag uses relative fluid velocity. Both laws need stated domains and coefficient units.