Verla
PhysicsClassical Physics

How to Find Tension in a Two-Mass Atwood Machine

For ideal masses of 2.00 kg and 4.00 kg, the heavier mass accelerates downward at 3.27 m/s², the lighter mass accelerates upward at the same rate, and the string tension is 26.1 N.

Question

An ideal Atwood machine has two hanging masses connected by a light, inextensible string over a frictionless pulley. The masses are m1=2.00 kgm_1=2.00\,\mathrm{kg} and m2=4.00 kgm_2=4.00\,\mathrm{kg}. The system is released from rest.

  1. Find the magnitude and direction of each mass's acceleration.
  2. Find the tension in the string.

Use g=9.80 m/s2g=9.80\,\mathrm{m/s^2}.

Answer

Since m2>m1m_2>m_1, mass m2m_2 accelerates downward and m1m_1 accelerates upward. Using those directions as positive for each mass,

T−m1g=m1a,m2g−T=m2a.T-m_1g=m_1a, \qquad m_2g-T=m_2a.

Adding the equations eliminates TT:

a=m2−m1m1+m2g=4.00−2.004.00+2.00(9.80)=3.27 m/s2.a=\frac{m_2-m_1}{m_1+m_2}g =\frac{4.00-2.00}{4.00+2.00}(9.80) =3.27\,\mathrm{m/s^2}.

Therefore, m1m_1 accelerates upward at 3.27 m/s23.27\,\mathrm{m/s^2} and m2m_2 accelerates downward at the same magnitude.

Substitute into either free-body equation:

T=m1(g+a)=(2.00)(9.80+3.27)=26.1 N.T=m_1(g+a)=(2.00)(9.80+3.27)=26.1\,\mathrm{N}.

The independent check

T=m2(g−a)=(4.00)(9.80−3.27)=26.1 NT=m_2(g-a)=(4.00)(9.80-3.27)=26.1\,\mathrm{N}

gives the same result.

Evidence boundary

This solution assumes a taut, massless, inextensible string and a massless, frictionless pulley, so both masses share one acceleration magnitude and the string tension is the same on both sides. A massive pulley, axle friction, string stretch, or slack would require a different model and can produce unequal tensions.

Sources

These references support the concepts and methods used in the explanation above.

How to Find Tension in an Atwood Machine | Verla