01 · Smooth table
With the same masses and no resistance, find a and T.
Hint
Use T = 3a and 19.6 − T = 2a.
Worked solution
a = 19.6/5 = 3.92 m/s²; T = 11.76 N.
Understand · explore · practise
A table particle connected to a hanging mass
Combine horizontal and vertical equations for a table particle and hanging mass, include resistance and infer unknown forces from motion.
Before you startConnected particles, pulley assumptions, Newton’s second law and SUVAT.
01 / Choose positive directions along the string
One particle moves horizontally and the other vertically.
Take A toward the pulley and B downward as the positive directions.
For a fixed smooth pulley and taut light inextensible string, the scalar accelerations along these compatible coordinates are equal and the tension magnitudes match. The acceleration vectors are perpendicular, so this is not one combined particle moving along one vector.
One horizontal body, one vertical bodyExplore
A (3 kg) moves right on a horizontal table toward a fixed smooth pulley. B (2 kg) moves downward on the other end of a taut light inextensible string. Both initially have speed 2 m/s. A has a specified constant leftward resistance R; g = 9.8. The equations apply while that motion and the connection persist. No impact or reversal is included.
02 / Write horizontal and vertical equations separately
The table particle’s weight is balanced by its normal reaction.
A is 3 kg and B is 2 kg. A’s resistance is 4.6 N while A moves right and B descends.Worked example
A horizontally: T − 4.6 = 3a
Weight and normal reaction do not enter this horizontal equation.
B vertically: 19.6 − T = 2a
Use downward positive.
Add: 15 = 5a, so a = 3 m/s²
Both scalar accelerations have this value.
T = 13.6 N; normal reaction on A = 29.4 N
Check 19.6 − 13.6 = 2 × 3. The horizontal string has no vertical component at A.
Watch: combine equations along perpendicular motion directions
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02 · More resistance
Use R = 9.6 N with the same masses. Find a and T while B moves downward.
Hint
Add the two equations before substituting.
Worked solution
19.6 − 9.6 = 5a gives a = 2 m/s². T = 3 × 2 + 9.6 = 15.6 N.
03 / Separate vertical balance from horizontal dynamics
Normal reaction equals weight here because the string at A is horizontal.
03 · Table reaction
Find the normal reaction on the 3 kg table particle.
Hint
There is no vertical acceleration and no other vertical force.
Worked solution
N − 3g = 0 gives N = 29.4 N.
04 · A different geometry
If the string at A instead pulls upward at an angle, is N still automatically 3g?
Hint
Include the vertical component of tension.
Worked solution
No. The upward tension component contributes to vertical balance, so N is less than 3g while contact remains. The horizontal-string equation does not apply unchanged.
05 · Equal tension
Why can the tension magnitudes be set equal on the two sides in this model?
Hint
Name the assumptions about both string and pulley.
Worked solution
The string is light and the pulley is smooth. Tautness and inextensibility constrain the motion; they do not alone justify equal tension around a rough pulley.
04 / Infer resistance from acceleration or travel data
Find a before solving the force equations.
06 · Measured acceleration
The same 3 kg and 2 kg particles have acceleration 2.4 m/s², with B downward. Find R and T.
Hint
Use B first for tension, then A for resistance.
Worked solution
T = 19.6 − 2 × 2.4 = 14.8 N. R = 14.8 − 3 × 2.4 = 7.6 N.
07 · Travel from rest
The same system is released from rest and B travels 1.8 m in 1.2 s before any contact event. Find R and T.
Hint
Use s = ½at², then the force equations.
Worked solution
a = 2 × 1.8/1.2² = 2.5 m/s². T = 19.6 − 5 = 14.6 N; R = 14.6 − 7.5 = 7.1 N.
08 · Unknown hanging mass
A has mass 3 kg on a smooth table. Acceleration is 2.8 m/s² toward the pulley. Find hanging mass M.
Hint
T = 3a, then Mg − T = Ma.
Worked solution
T = 8.4 N. M(9.8 − 2.8) = 8.4 gives M = 1.2 kg.
05 / Calculate motion only during the original force stage
Check which body first reaches a boundary.
09 · Speed and travel
For the worked example with a = 3 m/s², release from rest and find speed and distance after 0.5 s.
Hint
Assume both particles remain clear of boundaries.
Worked solution
v = 1.5 m/s; s = ½ × 3 × 0.5² = 0.375 m for each.
10 · Ground first
B starts 1.5 m above the ground, and A has at least 2 m available before reaching the pulley. For the same from-rest example, find the first event’s time and speed.
Hint
Compare available travel and use s = 1.5.
Worked solution
B reaches the ground first: 1.5 = ½ × 3t² gives t = 1 s and speed 3 m/s. A has also moved 1.5 m. Further motion needs a new stage.
11 · Table edge first
Instead A has only 0.6 m available and B is 1.5 m above ground. Which boundary occurs first?
Hint
Their travel magnitudes are equal while taut.
Worked solution
A reaches its boundary after 0.6 m, before B can descend 1.5 m. Stop the original model there; do not continue to the later ground event using the same geometry.
06 / A negative acceleration can describe slowing motion
Do not reverse the resistance simply because a is negative.
12 · Large resistance
With masses 3 kg and 2 kg, R = 24.6 N and B initially moving downward at 2 m/s, find a, T and time to first rest.
Hint
Keep downward positive for B while it is moving downward.
Worked solution
19.6 − 24.6 = 5a gives a = −1 m/s². T = 19.6 − 2(−1) = 21.6 N. From v = 2 − t, first rest is at 2 s, provided no boundary occurs earlier.
13 · Can it start this way?
Can the R = 24.6 N model be released from rest and assumed immediately to move B downward under these same opposing-resistance equations?
Hint
At small positive t, the predicted signed velocity would be negative.
Worked solution
No. That contradicts the assumed motion used to set the resistance direction. A static-friction law or a revised motion model is needed; the prescribed moving-resistance value alone does not determine what happens from rest.
07 / Check the assumed resistance and its direction
A result must fit the physical meaning of each force.
14 · Impossible resistance data
The 3 kg and 2 kg system is said to accelerate B downward at 5 m/s² with only a nonnegative opposing resistance on A. Test the claim.
Hint
Use R = 19.6 − 5a.
Worked solution
R = 19.6 − 25 = −5.4 N. That would be an extra driving force rather than the stated opposing resistance, so the assumptions and measurement are inconsistent.
The model supplies a resistance magnitude, not a universal friction law. At rest, after a boundary event or after direction reversal, reconsider the forces before reusing its equations.
08 / Combine compatible scalar equations, then check each body
Keep geometry and event limits in the calculation.
Balance vertical forces on the table particle separately. Use T − R = ma horizontally and Mg − T = Ma vertically for the hanging particle. Solve acceleration and tension, then calculate motion only up to the first event changing the model.
Section 1 of 8 · Choose positive directions along the string