01 · Parallel pulls
Each vertical segment has tension 24 N. Find the force the string exerts on the pulley.
Hint
Add the two downward vectors.
Worked solution
48 N downward.
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Forces on a pulley and its support
Add string tension vectors to find the force on a pulley and support reaction, distinguishing parallel pulls from perpendicular pulls and including pulley weight.
Before you startVector resultants, pulley tension and Newton’s third law.
01 / Change the body in the force diagram
String forces on the pulley point away along the outgoing string segments.
First find the tension, then add its two vector contributions on the pulley.
The pulley is a different body from either connected particle. Each string segment pulls the pulley along that segment. Equal tension magnitudes do not mean the two vectors have opposite directions. Use the actual geometry.
Add the pulls on the pulleyExplore
A fixed light smooth pulley has equal string tension T on each side. Choose parallel downward string pulls or perpendicular leftward/downward pulls. Blue arrows show the string forces; orange shows their resultant. The support force on this light pulley is opposite that resultant.
02 / Two downward pulls add directly
Parallel same-direction forces have resultant 2T.
A light fixed pulley has two vertical hanging string segments, each with tension 18 N.Worked example
String force 1: 18 N downward
This acts on the pulley.
String force 2: 18 N downward
It points the same way, so there is no cancellation.
Resultant from strings = 36 N downward
For equilibrium, the support exerts 36 N upward on the light pulley.
Watch: changing the string directions changes the resultant
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Support reaction
For question 1, the pulley is light and fixed. Find the force of the support on the pulley.
Hint
Its resultant force must be zero.
Worked solution
48 N upward. The pulley exerts the opposite 48 N downward force on the support.
03 / Perpendicular pulls need Pythagoras
The resultant is not the arithmetic sum of magnitudes.
A light fixed pulley at a table edge is pulled left and down by string tensions of 18 N.Worked example
Resultant components: 18 N left and 18 N down
Choose a common vector axis for forces on this pulley.
|F| = √(18² + 18²) = 18√2 N
Magnitude ≈ 25.5 N.
Direction: 45° below the leftward horizontal
The support force is equal and opposite: up and right.
03 · Perpendicular pulls
T = 24 N on perpendicular leftward and downward segments. Find the resultant magnitude and direction.
Hint
Use equal perpendicular components.
Worked solution
24√2 N ≈ 33.9 N, at 45° below the leftward horizontal.
04 · Compare geometries
For T = 24 N, how much greater is the parallel resultant magnitude than the perpendicular one?
Hint
Compare 2T with √2T.
Worked solution
48 − 24√2 N ≈ 14.1 N.
05 · Reverse the force
For question 3, state the support’s force on the light fixed pulley.
Hint
Reverse both components.
Worked solution
24 N right and 24 N up: magnitude 24√2 N, at 45° above the rightward horizontal.
04 / Calculate tension from the connected masses before using geometry
Do not replace T by the weight of one accelerating particle.
06 · Two hanging bodies
A 3 kg and 5 kg pair over a fixed smooth pulley has acceleration 2.45 m/s², with the 5 kg mass descending. Find T and the string force on the pulley.
Hint
For the lighter body, T − 29.4 = 3 × 2.45. Both string segments are vertical.
Worked solution
T = 36.75 N. String resultant on the pulley is 73.5 N downward.
07 · Table-edge pulley
A 3 kg table particle and 2 kg hanging particle have acceleration 3 m/s², with the hanger descending. The string directions at the pulley are horizontal and vertical. Find the string resultant magnitude.
Hint
Use 19.6 − T = 2 × 3.
Worked solution
T = 13.6 N, so resultant magnitude is 13.6√2 N ≈ 19.2 N.
08 · Weight shortcut
Why would taking T = 19.6 N in question 7 be wrong?
Hint
The hanger is accelerating.
Worked solution
Its weight and tension are unequal: 19.6 − T = 6 N. Setting T equal to its weight would incorrectly impose zero acceleration.
05 / Include pulley weight only when the model includes it
Support reaction balances every external force on the fixed pulley.
09 · Weighted parallel pulley
A fixed pulley weighs 12 N. Its two vertical string segments each have tension 20 N. Find the upward support force.
Hint
Weight is an additional downward force.
Worked solution
Support force = 20 + 20 + 12 = 52 N upward.
10 · Weighted table-edge pulley
A fixed pulley weighs 15 N. The string pulls 20 N left and 20 N down. Find support-force components and magnitude.
Hint
The support balances leftward 20 N and downward 35 N.
Worked solution
Support components are 20 N right and 35 N up. Magnitude = √(20² + 35²) = 5√65 N ≈ 40.3 N.
11 · Support angle
Find the support-force angle above the rightward horizontal in question 10.
Hint
Use tan θ = vertical/horizontal.
Worked solution
θ = tan⁻¹(35/20) ≈ 60.3°. The pulley weight changes the direction; it is no longer 45°.
06 / Infer tension from a known support force
State whether the pulley is light before solving backwards.
12 · Known parallel reaction
A light fixed pulley with two downward vertical string segments has an upward support reaction of 70 N. Find T.
Hint
Use 2T = 70.
Worked solution
T = 35 N.
13 · Known perpendicular reaction
A light fixed pulley with perpendicular string pulls has support-force magnitude 30√2 N. Find T.
Hint
Use √2T = 30√2.
Worked solution
T = 30 N. The two support components each have magnitude 30 N.
07 / Do not cancel forces belonging to different bodies
A support reaction is not another string tension on a particle.
14 · A mistaken pair
A learner cancels the two string pulls on a pulley because “tensions are equal and opposite”. Diagnose the error for two vertical hanging segments.
Hint
Draw both forces on the pulley.
Worked solution
Both string pulls on the pulley are downward, so they add. Each pull has its own third-law partner acting on the string, not another opposite force on the pulley. The support supplies the upward balancing force.
For a fixed pulley, sum force vectors in one common axis system. The compatible opposite axes used to add the particles’ motion equations are not a substitute for the pulley’s own force diagram.
08 / Find T, draw the string directions, then add vectors
Support force reverses the total non-support force on a fixed pulley.
Use 2T for parallel same-direction pulls and √2T for equal perpendicular pulls. Include any stated pulley weight when finding the support reaction. Name which body receives each force and give both magnitude and direction.
Section 1 of 8 · Change the body in the force diagram