PHYS2250 Chap.3 Newtonian Dynamics and Constraints
Newtonian Dynamics and Constraints
Isolate the body before writing force equations
Newtonian dynamics starts by drawing only external forces on one selected body. Component equations then follow from the vector law, while string length, contact and pulley geometry provide independent constraints.
The chapter distinguishes third-law pairs from balanced forces and explains why a negative normal reaction means the assumed contact has already ended.
Reasoning route and exam method
Newtonian dynamics fails most often before any equation is written.
A free-body diagram contains the forces acting on the selected body, not forces that body exerts elsewhere, not velocity arrows, and not a force labelled merely motion; this relationship belongs in the diagnostic checklist for the chapter. Newton’s second law is the vector equation ΣF = ma; this is a useful checkpoint before any numerical substitution.
The first law identifies inertial frames in which that equation has its ordinary form, and the third law pairs forces on different bodies; the same result should survive the relevant limiting-case test.
Third-law partners never cancel on a one-body diagram because they do not act on the same object; that distinction determines which quantities may be combined.
A light inextensible string fixes a relation between the motions of its ends, while a smooth pulley changes direction without changing the tension in the ideal model; this observation fixes the physical interpretation of the algebra.
Write the fixed total length before differentiating it; the statement supplies a direct check on the final result. That method works even when a knot moves, a pulley rides on a wedge or a segment changes direction; this condition should remain visible beside the governing equation.
Contact supplies a normal force only while contact persists; if a calculation demands a negative normal force, the assumed contact has already ended and the motion must be re-solved after separation.
A 70 kg passenger stands on a scale in a lift and the scale reads 770 N. Taking upward as positive, N − mg = ma. With g = 9.8 m s⁻², mg = 686 N, so a = (770 − 686)/70 = 1.20 m s⁻² upward.
The same reading could occur while the lift moves downward and slows; the scale determines acceleration, not velocity. A diagram that adds an upward force called motion would double-count the conclusion.
Answer. The contact assumption is invalid because the surface cannot pull. Separation occurred earlier, at N = 0, and subsequent motion is projectile motion.
Prompt.
Two variable string segments have lengths x and 2y while all other parts are fixed. Relate the accelerations.
Answer. x + 2y = constant, so after differentiating twice, aₓ + 2aᵧ = 0 with signs tied to the chosen positive directions.
DYNAMICS ROUTE Order of work. Select the body. Draw external forces. Choose axes. Write one component equation per axis. Add the geometrical constraint.
Only then solve.
The isolated block carries weight, a normal reaction, friction and the applied pull. Each arrow begins on the selected body, so the diagram can be projected directly onto horizontal and vertical axes.
! Do not put a reaction pair on one diagram
What this chapter covers
- 01
Free-body diagrams
- 02
Newton Laws
- 03
Reference Frames
- 04
String Constraints
- 05
Contact and Separation
Worked example · free
Interpret a lift scale reading
- 2Draw weight and normal reaction.
- 2Apply the vertical force equation.
- 2Interpret the sign independently of velocity.
Key terms
- Inertial Frame
- A reference frame in which Newton's laws take their ordinary form.
- Constraint Equation
- A geometrical relation limiting the possible motions of connected bodies.
- Normal Reaction
- A contact force perpendicular to a surface that exists only while the bodies remain in contact.
Newtonian Dynamics and Constraints FAQ
Do action and reaction cancel?
They act on different bodies, so they never cancel on a one-body free-body diagram. They cancel internally only when both bodies belong to the chosen system and total momentum is considered.
What does a negative normal force mean?
It means the assumed contact cannot persist because an ordinary surface cannot pull the body. Solve for the separation condition at zero normal force, then change the motion model for subsequent motion.
Why begin with string length?
A fixed total length provides the missing relation between connected displacements. Differentiating it gives velocity and acceleration constraints with reliable signs for every connected body in the chosen coordinates.
Exam move
Draw each body separately before combining equations. Write the length or contact constraint independently, then count unknowns and equations before solving. Assemble a small set that changes only one modelling choice at a time: a lift, two masses on a string, a movable pulley and a body leaving a curved surface. For each, circle every third-law partner that belongs on a different diagram.
State whether the frame is inertial before using the ordinary force equation. Differentiate the full string-length relation rather than guessing equal accelerations from appearance. In contact problems, solve for the normal reaction and identify the parameter value at which it reaches zero. End by explaining the computed acceleration in words without referring to the sign alone.
This forces the diagram, constraint and physical interpretation to agree. Redraw the complete multi-body system at the end and verify that internal tensions appear in opposite pairs while only external forces remain in the total balance.
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