The tension is constant throughout a rope that is in equilibrium, but what happens if the rope is accelerating? For example, FIGURE 7.20a shows two connected blocks being pulled by force $\vec{F}$. Is the string’s tension at the right end, where it pulls back on B, the same as the tension at the left end, where it pulls on A?

FIGURE 7.20 Tension pulls forward on block A, backward on block B.

FIGURE 7.20b shows the horizontal forces acting on the blocks and the string. The only horizontal forces acting on the string are $\vec{T}{}_{\text{A on S}}$ and $\vec{T}{}_{\text{B on S}}$, so Newton’s second law for the string is

where $m_{\text{S}}$ is the mass of the string. If the string is accelerating, then the tensions at the two ends can not be the same. The tension at the “front” of the string must be greater than the tension at the “back” in order to accelerate the string!

Often in physics and engineering problems the mass of the string or rope is much less than the masses of the objects that it connects. In such cases, we can adopt the massless string approximation. In the limit $m_{\text{S}}\to 0$, Equation 7.5 becomes

In other words, the tension in a massless string is constant. This is nice, but it isn’t the primary justification for the massless string approximation.

Look again at Figure 7.20b. If $T_{\text{B on S}}=T_{\text{A on S}}$, then

That is, the force on block A is equal and opposite to the force on block B. Forces $\vec{T}{}_{\text{S on A}}$ and $\vec{T}{}_{\text{S on B}}$ act as if they are an action/reaction pair of forces. Thus we can draw the simplified diagram of FIGURE 7.21 in which the string is missing and blocks A and B interact directly with each other through forces that we can call $\vec{T}{}_{\text{A on B}}$ and $\vec{T}{}_{\text{B on A}}$.

FIGURE 7.21 The massless string approximation allows objects A and B to act as if they are directly interacting.

In other words, if objects A and B interact with each other through a massless string, we can omit the string and treat forces $\stackrel{\mathbf{⃗}}{\mathit{F}}{}_{\mathbf{A}\mathbf{\text{on}}\mathbf{B}}$ and $\stackrel{\mathbf{⃗}}{\mathit{F}}{}_{\mathbf{B}\mathbf{\text{on}}\mathbf{A}}$ as if they are an action/reaction pair. This is not literally true because A and B are not in contact. Nonetheless, all a massless string does is transmit a force from A to B without changing the magnitude of that force. This is the real significance of the massless string approximation.

EXAMPLE 7.6 Comparing two tensions

Blocks A and B in FIGURE 7.22 are connected by massless string 2 and pulled across a frictionless table by massless string 1. B has a larger mass than A. Is the tension in string 2 larger than, smaller than, or equal to the tension in string 1?

FIGURE 7.22 Blocks A and B are pulled across a frictionless table by massless strings.

FIGURE 7.22 Full Alternative Text

MODEL The massless string approximation allows us to treat A and B as if they interact directly with each other. The blocks are accelerating because there’s a force to the right and no friction.

SOLVE B has a larger mass, so it may be tempting to conclude that string 2, which pulls B, has a greater tension than string 1, which pulls A. The flaw in this reasoning is that Newton’s second law tells us only about the net force. The net force on B is larger than the net force on A, but the net force on A is not just the tension $\vec{T}{}_1$ in the forward direction. The tension in string 2 also pulls backward on A!

FIGURE 7.23 shows the horizontal forces in this frictionless situation. Because the string is massless, forces $\vec{T}{}_{\text{A on B}}$ and $\vec{T}{}_{\text{B on A}}$ act as if they are an action/reaction pair.

FIGURE 7.23 The horizontal forces on blocks A and B.

FIGURE 7.23 Full Alternative Text

From Newton’s third law,

$$T_{\text{A on B}}=T_{\text{B on A}}=T_2$$

where $T_2$ is the tension in string 2. From Newton’s second law, the net force on A is

$$(F_{\text{A net}}){}_x=T_1-T_{\text{B on A}}=T_1-T_2=m_{\text{A}}{a}_{\text{A}x}$$

The net force on A is the difference in tensions. The blocks are accelerating to the right, making $a_{\text{A}x}>0$, so

$$T_1>T_2$$

The tension in string 2 is smaller than the tension in string 1.

REVIEW This is not an intuitively obvious result. A careful study of the reasoning in this example is worthwhile. An alternative analysis would note that $\vec{T}{}_1$ accelerates both blocks, of combined mass $(m_{\text{A}}+m_{\text{B}})$, whereas $\vec{T}{}_2$ accelerates only block B. Thus string 1 must have the larger tension.