Two Masses, One String: The Atwood Machine and the Table Problem

Two Masses, One String: The Atwood Machine and the Table Problem

Two masses joined by a string do not move as two free particles. They share one acceleration if the string does not stretch. The Atwood machine and the mass-on-a-table problem are the same idea in different layouts. Both turn Newton’s second law into a pair of equations and one unknown $a$a.

Hang two masses over a light pulley. Call them $m_1$m1​ and $m_2$m2​, with $m_1>m_2$m1​>m2​. Gravity pulls both down. The string pulls both toward the pulley. If the pulley is smooth and the string is light, the tension $T$T is the same. The heavier mass goes down. The lighter mass goes up. The magnitude of acceleration is the same.

Write the line of motion for each mass. Down is positive for $m_1$m1​:

$$m_1g-T=m_1a.$$m1​g−T=m1​a.

Up is positive for $m_2$m2​ if we keep the same $a$a:

$$T-m_2g=m_2a.$$T−m2​g=m2​a.

Add the two equations. Tension cancels. You get

$$a=\frac{(m_1-m_2)g}{m_1+m_2}.$$a=m1​+m2​(m1​−m2​)g​.

Put $a$a back into either equation to find $T$T:

$$T=\frac{2m_1m_2}{m_1+m_2}\,g.$$T=m1​+m2​2m1​m2​​g.

Tension lies between the two weights. It equals neither. That is the usual student error.

Now keep one mass on a smooth horizontal table. The other hangs. The hanging mass $m$m falls. The table mass $M$M slides toward the edge. Gravity does not fight $M$M along the table. Tension is the only horizontal force on $M$M:

$$T=Ma.$$T=Ma.

The hanging mass still feels both gravity and tension:

$$mg-T=ma.$$mg−T=ma.

Solve and you get

$$a=\frac{m}{M+m}g,\qquad T=\frac{Mm}{M+m}g.$$a=M+mm​g,T=M+mMm​g.

The table removes one weight from the denominator’s “opposing” role. Acceleration is larger than in a balanced Atwood pair with the same two numbers if the second mass had also been hanging.

A smooth table means no friction. Real tables add a kinetic term $\mu_k Mg$μk​Mg against the slide. Then

$$a=\frac{mg-\mu_k Mg}{M+m}$$a=M+mmg−μk​Mg​

provided the hanging weight can overcome static friction at the start. If it cannot, $a=0$a=0 and the string just sits there.

The string constraint is the hidden law. If the string length is fixed, the displacements are equal. So are the speeds and the accelerations. A massive pulley changes that bookkeeping. The pulley then needs its own torque equation. Introductory courses leave the pulley massless so $T$T stays single-valued.

Always draw free-body diagrams before algebra. Mark a consistent positive direction along the string. Do not assign $g$g to the table mass along the horizontal. Do not set $T=mg$T=mg unless the system is at rest.

In short, connected bodies share $a$a and share $T$T when the string and pulley are ideal. The Atwood machine subtracts two weights. The table problem subtracts only the hanging weight from the driving side. Same method. Different force list. Write both laws, add them, and the tension drops out.