All Labs

Atwood Machine Lab

Investigate how two unequal masses connected by a string over a frictionless pulley produce a measurable acceleration. Adjust mass1 and mass2, observe the animated pulley system, and verify that the net force equals the total mass times acceleration.

Guided Experiment: Atwood Machine — Acceleration vs Mass Ratio

If you increase the heavier mass m1 while keeping m2 constant, what do you predict will happen to the acceleration and tension?

Write your hypothesis in the Lab Report panel, then click Next.

Controls

Mass 1 (heavier)1.50 kg
Mass 2 (lighter)1.00 kg
Gravity9.81 m/s²
Friction Coefficient0.00

Results

a=(1.5000−1.0000)(9.8100)1.5000+1.0000=1.9620 m/s2a = \frac{(1.5000 - 1.0000)(9.8100)}{1.5000 + 1.0000} = 1.9620 \text{ m/s}^2
Ideal Acceleration
1.9620 m/s²
Tension T
11.7720 N
Net Force
4.9050 N
Mass Ratio m1/m2
1.500
Newton's Second Law: F = ma
(m1+m2)⋅a=2.5000×1.9620=4.9050 N(m_1 + m_2) \cdot a = 2.5000 \times 1.9620 = 4.9050 \text{ N}
F_net = 4.9050 N ✓ matches
9.8100 N = m2g < T = 11.7720 N < m1g = 14.7150 N✓ valid range

Acceleration vs Mass Ratio

Current ratio = 1.500
a = (r-1)g/(r+1)Current (m1/m2, a)

Data Table

(0 rows)
#Trialm1(kg)m2(kg)m1/m2Acceleration(m/s²)Tension(N)F_net(N)
0 / 500
0 / 500
0 / 500

Reference Guide

Atwood Machine

An Atwood machine consists of two masses connected by a light inextensible string over a massless frictionless pulley. When the masses differ, the heavier side descends and the lighter side rises with uniform acceleration.

The device was invented in 1784 by George Atwood to demonstrate and measure the effect of gravity with reduced acceleration, making timing more practical.

Acceleration Formula

Apply Newton's second law to each mass and solve simultaneously. The string transmits tension T equally to both sides.

a=(m1−m2) gm1+m2a = \frac{(m_1 - m_2)\,g}{m_1 + m_2}

When m1 = m2 the acceleration is zero. As the ratio m1/m2 grows, the acceleration approaches g asymptotically.

Tension in the String

The string tension is the same throughout (massless string, frictionless pulley). It lies between the two weight values:

T=2 m1m2 gm1+m2T = \frac{2\,m_1 m_2\,g}{m_1 + m_2}

This confirms that neither mass is in free fall — T reduces each effective downward force below the true weight. When masses are equal T equals mg (no acceleration, static equilibrium).

Newton's Second Law Verification

The net unbalanced force acting on the entire system is the weight difference. Newton's second law predicts:

Fnet=(m1−m2)g=(m1+m2) aF_{net} = (m_1 - m_2)g = (m_1 + m_2)\,a

The product (m1 + m2) x a must equal (m1 - m2)g in every trial. Any deviation indicates friction, pulley inertia, or string stretch.