Constraints · AP C/F=ma · 7 min
Constraints are physics equations, not bookkeeping afterthoughts.
01 The trap
Wrong path
Students write force equations for each object but never connect their accelerations.
Why it feels right
The force equations look complete because every object has Newton's second law written down.
02 Correct model
A string, contact, rolling condition, or pivot adds a geometry equation. Without it, there are more unknowns than physics.
03 Mini-example
Prompt
Two masses hang over one ideal pulley. If the left mass accelerates upward, what is the right mass's acceleration?
Common wrong answer
The same positive acceleration.
Correct reasoning
Same magnitude, opposite signed direction if upward is positive for both.
Diagnostic cue
If two bodies are connected, ask what their motions must share before solving for forces.
04 Guided practice
Warm-up isolates the principle. Bridge changes the context. Contest-style requires a complete setup on less familiar geometry.
Two hanging masses share one ideal string over a pulley. Up is positive for both. If the left mass has acceleration +a, what is the right mass's acceleration?
The string length is fixed.
The right mass has acceleration -a. The magnitudes match, but the signed components differ because one side lengthens when the other shortens.
Trap: Writing both accelerations as +a because they have the same magnitude.
A load hangs from a movable pulley supported by two segments of the same string. If the free end of the string is pulled down 20 cm, how far does the load rise?
Both supporting segments shorten equally.
The load rises 10 cm. Pulling 20 cm of string removes 20 cm of total length from two support segments, so each shortens by 10 cm.
Trap: Assuming the load moves the same distance as the pulled end.
A wheel rolls right without slipping. If rightward center-of-mass acceleration is positive, what sign should angular acceleration have if counterclockwise is positive?
A wheel rolling right rotates clockwise.
Clockwise is negative, so α is negative when a_cm is positive. With this sign choice, a_cm = -αR.
Trap: Writing a = αR without checking sign conventions.
A mass M is attached to a movable pulley supported by two vertical string segments. The free end is pulled downward with acceleration a. What is the upward acceleration of M?
The two support segments shorten together.
The free end supplies twice the length change of one support segment, so x = 2y. Differentiating twice gives a_M = a/2 upward.
Trap: Giving M the same acceleration as the pulled end.
A movable load is supported by three vertical segments of one inextensible string. If the free end is pulled downward 0.45 m, how far does the load rise?
A load displacement changes all three supporting segments by the same amount.
The three support segments shorten by a total of 0.45 m, so each shortens by 0.45/3 = 0.15 m. The load rises 0.15 m.
Trap: Using a two-segment relation without counting the supporting segments.
A mass m hangs from the free end of a light string. The same string passes around a movable pulley that supports a load M with two vertical string segments. All pulleys are ideal. Take downward as positive for both masses. (a) Derive the acceleration constraint. (b) For M > 2m, derive both accelerations and the tension. (c) Check the result when M = 2m.
If y_m and y_M are downward coordinates, the variable length is y_m + 2y_M.
The fixed length gives a_m + 2a_M = 0. With mg - T = ma_m and Mg - 2T = Ma_M, substitution gives a_M = (M - 2m)g/(M + 4m), a_m = -2a_M, and T = 3mMg/(M + 4m). When M = 2m, both accelerations vanish and T = mg = Mg/2.
Trap: Assigning equal acceleration magnitudes to the free end and movable pulley.
05 Mock checks