PhysicsLaws of motionClass 11

Laws of motion, friction and pulleys: interactive simulation

Every Newton’s-laws problem follows the same recipe: isolate each body, draw every force on it, write F = ma along each direction and add the constraints that tie the bodies together.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards, AP Physics 1, A-level Physics, IB Physics · 4 modes in one lab

What you can explore

  1. 1

    Incline & friction

    Free body diagram, components of weight, static vs kinetic friction and the angle of repose.
  2. 2

    Atwood & pulleys

    Fixed-pulley Atwood machine and a movable pulley with the a₁ = 2a₂ constraint.
  3. 3

    Block on block

    Two stacked blocks with friction between them — when do they slip?
  4. 4

    Apparent weight (lift)

    Normal force and scale reading when a lift accelerates up, down or falls freely.

Try this

Each part of this lab shows that recipe live: the free body diagram on an incline, the string constraint in pulleys, the friction test for two stacked blocks, and the normal force in a lift.

  1. 1

    With μs = 0.5, raise the incline slowly from 0°.

    What you should see

    The block stays at rest up to about 26.6° (tan θ = 0.5), with static friction exactly equal to mg sin θ.

  2. 2

    Set θ = 30° with μk = 0.4.

    What you should see

    It slides with a = 9.8(0.5 − 0.4 × 0.866) ≈ 1.51 m/s².

  3. 3

    In Atwood & pulleys, hang 3 kg and 2 kg on a fixed pulley.

    What you should see

    a = g/5 = 1.96 m/s² and T = 23.52 N.

  4. 4

    Switch to the movable pulley with the same masses.

    What you should see

    Now the 3 kg falls at 5.6 m/s², the pulley rises at 2.8 m/s², and T = 12.6 N.

  5. 5

    In Block on block (2 kg on 4 kg, μ = 0.3), pull the lower block.

    What you should see

    They move together up to F = 17.64 N. Pull the upper block instead and they slip above 8.82 N.

  6. 6

    In the lift, accelerate a 60 kg person upwards at 2 m/s², then let the lift fall freely.

    What you should see

    The scale reads about 72.2 kg, then 0 kg in free fall.

Formulas behind it

Newton’s second law

F⃗net=ma⃗\vec F_{net} = m\vec a

Friction

fs≤μsN,fk=μkNf_s \le \mu_s N,\qquad f_k = \mu_k N

Sliding down an incline

a=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k\cos\theta)

Angle of repose

tan⁡θr=μs\tan\theta_r = \mu_s

Atwood machine

a=(m1−m2)gm1+m2,T=2m1m2gm1+m2a = \frac{(m_1 - m_2)g}{m_1 + m_2},\qquad T = \frac{2m_1m_2g}{m_1 + m_2}

Apparent weight in a lift

N=m(g+a)N = m(g + a)

Common mistakes

  • Always writing friction as μN. Static friction only takes the value it needs, up to μsN.
  • Taking N = mg on an incline. It is mg cos θ (if nothing else pushes into the surface).
  • Assuming both masses have the same acceleration when one hangs from a movable pulley.
  • Drawing an extra “force of motion” in a free body diagram. Only real interactions — gravity, contact, tension, friction — belong there.
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FAQ

Why is static friction larger than kinetic friction?

Surfaces at rest have time to settle into each other’s microscopic bumps and form stronger contacts, so more force is needed to start sliding than to keep sliding.

Can the tension be the same on both sides of a pulley?

Yes, if the string and pulley are massless and the pulley is frictionless. A massive pulley needs different tensions to give it angular acceleration.

What does a weighing scale actually measure?

The normal force it exerts on you. In an accelerating lift that is m(g + a), which is why you feel heavier or lighter.

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