Pulleys and connected objects Knowledge-Based Study

Physics · Forces & Newton's Laws · Physics · Applications of Newton's Laws

Physics topic explanations on Pulleys and connected objects.

Study notes

  1. Connected Objects and Constraints

    In classical mechanics, when two or more objects are linked together by ropes, cables, or strings over pulleys, they form a connected system. The motion of one body directly determines or constrains the motion of the others.

    To analyze these systems accurately, physicists often begin with idealizing assumptions:

    • Light (massless) string: The mass of the string is negligible compared to the connected objects. Consequently, the magnitude of the tension TT is uniform along any continuous segment of the string between two objects or across an ideal pulley.
    • Inextensible string: The string does not stretch. This kinematic constraint ensures that if two objects are connected directly by a taut string passing over fixed pulleys, their speeds and the magnitudes of their accelerations must be identical: a1=a2=aa_1 = a_2 = a.
    • Ideal pulley: The pulley is assumed to be frictionless and massless. It alters only the direction of the tension force, not its magnitude.

    Two Hanging Masses over a Fixed Pulley

    An Atwood machine consists of two masses, m1m_1 and m2m_2, connected by a light, inextensible cord passing over a massless, frictionless pulley. Assume m2>m1m_2 > m_1, so m2m_2 accelerates downward while m1m_1 accelerates upward at the same rate aa.

    Isolating Each Body

    We apply Newton's second law, ΣF=ma\Sigma F = ma, to each mass along its direction of motion:

    • For mass m1m_1 (upward motion as positive):
      T−m1g=m1aT - m_1 g = m_1 a
    • For mass m2m_2 (downward motion as positive):
      m2g−T=m2am_2 g - T = m_2 a

    Solving for System Acceleration and Tension

    Adding the two equations eliminates the internal tension TT:

    (m2−m1)g=(m1+m2)a(m_2 - m_1)g = (m_1 + m_2)a

    Thus, the magnitude of acceleration is:

    a=m2−m1m1+m2ga = \frac{m_2 - m_1}{m_1 + m_2} g

    Substituting this acceleration back into either equation yields the string tension:

    T=2m1m2m1+m2gT = \frac{2 m_1 m_2}{m_1 + m_2} g

    Common Misconception: The tension TT is neither equal to m2gm_2 g nor m1gm_1 g. For m2m_2 to accelerate downward, TT must be strictly less than m2gm_2 g, and for m1m_1 to accelerate upward, TT must be strictly greater than m1gm_1 g.

    Mass on a Surface Connected to a Hanging Mass

    Consider mass m1m_1 resting on a horizontal table and connected by a horizontal string over a frictionless pulley at the table edge to a vertically hanging mass m2m_2.

    Frictionless Surface

    Along the horizontal axis for m1m_1, the only force is tension: T=m1aT = m_1 a. For the hanging mass m2m_2, moving downward: m2g−T=m2am_2 g - T = m_2 a.

    Combining these gives:

    a=m2gm1+m2a = \frac{m_2 g}{m_1 + m_2}

    T=m1m2gm1+m2T = \frac{m_1 m_2 g}{m_1 + m_2}

    Surface with Kinetic Friction

    If kinetic friction fk=μkN=μkm1gf_k = \mu_k N = \mu_k m_1 g opposes the motion of m1m_1, Newton's second law for each body becomes:

    • For m1m_1: T−μkm1g=m1aT - \mu_k m_1 g = m_1 a
    • For m2m_2: m2g−T=m2am_2 g - T = m_2 a

    Summing both equations yields:

    a=(m2−μkm1)gm1+m2a = \frac{(m_2 - \mu_k m_1)g}{m_1 + m_2}

    Motion occurs only if the gravitational pulling force m2gm_2 g exceeds the maximum static friction force fs,max⁡=μsm1gf_{s,\max} = \mu_s m_1 g.

    Treating Connected Objects as a Combined System

    While drawing separate free-body diagrams for each mass is always reliable, you can also treat the entire connected chain as a single composite object of total mass M=ΣmiM = \Sigma m_i.

    • Internal Forces: Forces between connected parts, such as tension in the connecting string, cancel out when summing forces over the entire system.
    • Net Driving Force: The acceleration is found directly by dividing the net external force acting along the line of motion by the total mass: a=ΣFext, parallelmtotala = \frac{\Sigma F_{\text{ext, parallel}}}{m_{\text{total}}}
    • Finding Internal Tension: Once aa is determined, isolate a single mass and apply ΣF=ma\Sigma F = ma to compute the tension.

    Step-by-Step Procedure:

    1. Define a consistent coordinate direction along the path of motion (for example, clockwise or along the string).
    2. Identify all external forces acting on each object (gravity, normal force, friction).
    3. Set up equations of motion for each body or for the entire system.
    4. Verify that calculated accelerations satisfy 0≤a<g0 \le a < g for free-hanging systems and that tensions are non-negative.