AP Physics 1: TensionAP Physics 1: Tension

APPHYSICS1 · UNIT2 · F O R C E S A N D N E W T O N ' S L AW S

Tension

Ropes, pulleys, and vertical and horizontal tension problems

Big idea. Tension is the pulling force a rope, string, or cable exerts along its length. A rope can only pull, never push. For a light rope (negligible mass) the tension is the same everywhere in it. Rules for ropes and pulleys

The tension force on an object points along the rope, away from the object.

A massless rope over a frictionless pulley has the same tension on both sides. Objects connected by a taut rope have the same acceleration magnitude.

Treat each object with its own FBD, then connect them with the tension and the shared acceleration. Case 1: hanging object

An object hangs from a rope. FT up, Fg down. Let up be positive: ΣF = FT − mg = ma. Motion of the hanging mass Tension For m = 5 kg

At rest or constant velocity FT = mg 49 N

Accelerating up at a FT = m(g + a) a = 2: 5(11.8) = 59 N

Accelerating down at a FT = m(g − a) a = 2: 5(7.8) = 39 N

Test trap. Tension is NOT always equal to the weight. It equals the weight only when there is no vertical acceleration. Accelerating up means the rope pulls harder than gravity. Case 2: horizontal tension (pulling blocks on a frictionless surface)

A 6 kg block and a 4 kg block are tied together. A 50 N force pulls the 6 kg block. System: a = 50 / (6 + 4) = 5 m/s².

Tension pulls only the 4 kg block: FT = m a = 4x5 = 20 N.

Case 3: table plus hanging mass (pulley)

A 3 kg block on a frictionless table is tied over a pulley to a 2 kg block hanging. Hanging block (down +): m2g − FT = m2a

Table block: FT = m1a

Add: m2g = (m1 + m2)a, so a = 19.6 / 5 = 3.92 m/s², and FT = 3(3.92) = 11.8 N.

Case 4: Atwood machine (two hanging masses)

2 kg and 3 kg over a pulley: a = (m2 − m1)g / (m1 + m2) = (1)(9.8)/5 = 1.96 m/s². For the 2 kg (going up): FT = m1(g + a) = 2(11.76) = 23.5 N. Notice it is between 19.6 N and 29.4 N. Case 5: angled cables (equilibrium)

A 10 kg sign hangs from two cables, each 30° above horizontal. The horizontal parts cancel. Vertical: 2FT sin 30° = mg = 98 N, so FT = 98 N in each cable. The flatter the cables, the larger the tension. Strategy. When the system accelerates, write ΣF = ma for each object, then solve the two equations together. Add them if the tension cancels.

Practice

1. An 8 kg mass hangs at rest. Find the tension.

2. The same 8 kg mass rises at constant velocity. Find the tension.

3. It accelerates upward at 1.5 m/s². Find the tension.

4. It accelerates downward at 1.5 m/s². Find the tension.

5. Atwood machine with 4 kg and 6 kg. Find a and FT.

6. 5 kg block on a frictionless table tied to a 5 kg hanging block. Find a and FT. 7. A rope breaks at 200 N. What is the largest mass it can lift at constant speed? 8. Two people each pull 100 N on opposite ends of a rope held at rest. What is the tension? 9. A 20 kg sign hangs from two cables at 30° above horizontal. Find the tension in each. 10. Why does a rope pulled across a pulley have the same tension on both sides? Answers

1. 78.4 N.

2. 78.4 N (no acceleration).

3. 8(9.8 + 1.5) = 90.4 N.

4. 8(9.8 − 1.5) = 66.4 N.

5. a = 2(9.8)/10 = 1.96 m/s². FT = 4(9.8 + 1.96) = 47.0 N.

6. a = 49/10 = 4.9 m/s². FT = 5(4.9) = 24.5 N.

7. 200/9.8 = about 20.4 kg.

8. 100 N (not 200 N).

9. 2FT(0.5) = 196, so FT = 196 N.

10. For a massless rope over a frictionless pulley, there is nothing to change the force along it.