Two masses hang over a frictionless, massless pulley: mass A (3 kg) on the left and mass B (5 kg) on the right. The tension T in the rope is:
AEqual to 5 × 9.8 = 49 N, the weight of the heavier mass
BEqual to 3 × 9.8 = 29.4 N, the weight of the lighter mass
CEqual to (3 × 5 × 2 × 9.8) / (3 + 5) = 36.75 N, less than either weight
DEqual to (3 + 5) × 9.8 / 2 = 39.2 N, the average of both weights
From the Atwood machine equations: T = 2·m_A·m_B·g / (m_A + m_B) = 2×3×5×9.8/8 = 36.75 N. This is less than either weight (29.4 N and 49 N). The tension cannot equal the heavier weight — if it did, the heavier mass would be in equilibrium and wouldn't accelerate. The rope must be pulling each mass partially against gravity while also allowing them to accelerate, so T ends up between the two weights but less than either one separately.
Question 2 Multiple Choice
Why is tension constant throughout an ideal massless string, even when different forces are applied to each end?
ABecause strings are made of elastic material that distributes force evenly
BBecause Newton's Third Law requires forces to be equal and opposite at every point
CBecause any segment of a massless string has zero mass, so the net force on it must be zero, meaning both ends pull with equal force
DBecause the string is inextensible, preventing any variation in force along its length
Apply Newton's Second Law to any small segment of the string: F_net = m_segment × a. For a massless string, m_segment = 0, so F_net = 0 regardless of acceleration. This means the force pulling one end of any segment must exactly equal the force pulling the other end — the tension is the same on both sides of any cross-section. Inextensibility (option D) constrains acceleration, not force distribution. Newton's Third Law (option B) applies to force pairs between objects, not to internal force distribution along a rope.
Question 3 True / False
In an ideal (massless, inextensible) string, the tension is the same at every point along the string.
TTrue
FFalse
Answer: True
This follows directly from Newton's Second Law applied to any segment: zero mass means zero net force, so the tension forces at both ends of any segment must be equal. This idealization is what makes rope problems tractable — you can refer to 'the tension T' as a single value rather than tracking how it varies along the length. When the idealization breaks down (massive rope, frictional pulley), tension is no longer uniform.
Question 4 True / False
The tension in the rope of a pulley system generally equals the weight of the heavier object.
TTrue
FFalse
Answer: False
Tension in a pulley system is strictly less than the weight of either object when both are accelerating. If the tension equaled the heavier weight, that object would experience zero net force and wouldn't accelerate — contradicting the assumption that the system moves. Tension represents the force the rope exerts, which must be less than the heavier weight (to allow downward acceleration) and greater than the lighter weight (to pull it upward). The formula T = 2·m_A·m_B·g / (m_A + m_B) confirms this.
Question 5 Short Answer
Why can a string only pull its endpoints toward each other and never push them apart, and how does this asymmetry affect how you draw free-body diagrams?
Think about your answer, then reveal below.
Model answer: Strings transmit tension — a pulling force — because their structure only allows them to resist being stretched. When pulled taut, the intermolecular bonds along the string carry the load. When compressed, the string simply goes slack and transmits no force. This asymmetry means tension arrows on free-body diagrams always point away from the object (toward the string), never toward the object. A ball on a string hanging from a ceiling has a tension arrow pointing upward along the rope from the ball; the same string cannot push the ball downward.
This directional rule — tension always pulls, never pushes, always away from the object — is the single most important discipline in drawing tension free-body diagrams. Confusing the direction of tension (pointing toward the ceiling versus toward the ball) is the most common error in setting up Newton's Second Law equations for string-connected systems. Rigid rods can push and pull; strings can only pull.