GRE Physics Companion: Tension and Massless Strings

This companion is designed for rapid review after M02-05. Most tension questions become straightforward once the system choice and string assumptions are identified before the algebra begins.

1 Fast triage

For a body attached to a taut string, draw the tension force along the string and away from the body.

For a massless ideal string segment with no other tangential force,

(1)

For an inextensible string, use the fixed length relation to connect the motions of the attached bodies.

For multiple connected bodies, it is often fastest to find the common acceleration from the full system first and then isolate one body to find the tension.

2 Common traps

A hanging body does not imply T = mg. Instead,

(2)

A string that goes slack has

(3)

Two different strings do not generally have equal tension.

Equal tension magnitude along one ideal massless string does not mean the tension vectors on two different attached bodies point in the same direction.

Massless and inextensible are different assumptions: massless controls force balance on the connector, while inextensible controls the geometric length constraint.

3 Worked GRE example 1: two connected blocks

Two blocks of masses m and 2m lie on a frictionless horizontal surface and are connected by a massless inextensible string. A horizontal force F pulls the 2m block. Find the tension.

For the full system,

(4)

Thus

(5)

Now isolate the block of mass m. Its only horizontal force is tension:

(6)

Therefore

(7)

The common acceleration came from the total external force; the internal tension came from one block’s equation.

4 Worked GRE example 2: nearly horizontal support cables

A Weight W is supported symmetrically by two cables, each making angle 𝜃 above the horizontal. Find the tension in each cable and describe what happens as 𝜃 becomes small.

Vertical equilibrium gives

(8)

Hence

(9)

As 𝜃 → 0, sin 𝜃 → 0 and the required tension grows very large. A nearly horizontal cable must provide the required vertical support through a small vertical component of a large tension.

5 GRE speed questions

1. A 5 kg mass hangs motionless from a light string. The string tension is (A) zero (B) less than mg (C) equal to mg (D) greater than mg.
2. A hanging mass accelerates downward with magnitude a < g. Its string tension is (A) m(g − a) (B) mg (C) m(g + a) (D) zero for every downward acceleration.
3. Two blocks of masses m and 3m are connected on a frictionless table. A force F pulls the 3m block. The common acceleration is (A) F∕m (B) F∕(2m) (C) F∕(3m) (D) F∕(4m).
4. In Question 3, the tension in the connector is (A) F∕4 (B) F∕3 (C) 3F∕4 (D) F.
5. Which assumption gives the fixed length relation between positions of bodies connected by an ideal string? (A) massless (B) inextensible (C) frictionless (D) weightless.
6. A flexible ideal string becomes slack. Its tension is (A) negative (B) zero (C) mg (D) unchanged from the taut value.

6 Answers and rationales

1. C. Zero acceleration gives T − mg = 0.
2. A. Taking upward as positive, a_y = −a, so T = m(g − a).
3. D. The total mass is 4m, so a = F∕(4m).
4. A. The block of mass m is accelerated only by tension, so T = ma = F∕4.
5. B. Inextensibility fixes the total string length and therefore supplies the kinematic constraint.
6. B. A flexible string cannot sustain compression; when slack, the ideal tension is zero.

References