Problem — Two Blocks Tied Together on a Smooth Plane

Problem

Two blocks of masses $m_1 = 3{,}0\;\text{kg}$ and $m_2 = 5{,}0\;\text{kg}$ are tied together by an inextensible rope and lie on a smooth horizontal plane. A horizontal force $F = 16\;\text{N}$ pulls block $m_2$. Calculate the acceleration of the system and the tension in the rope.

Solution

Step 1 — The system as a whole. The rope is inextensible, so the two blocks move with the same acceleration $a$. Considering the system $m_1 + m_2$ as a single body, the tensions are internal forces and the only external horizontal force is $F$:

$F = (m_1 + m_2)\,a \quad\Longrightarrow\quad a = \frac{F}{m_1 + m_2}$

$a = \frac{16\;\text{N}}{(3{,}0 + 5{,}0)\;\text{kg}} = \frac{16}{8{,}0} = 2{,}0\;\text{m/s}^2$

Step 2 — The tension in the rope. Isolate block $m_1$: horizontally, only the rope’s tension $T$ acts on it, dragging it along. Newton’s second law gives

$T = m_1\,a = 3{,}0\;\text{kg} \times 2{,}0\;\text{m/s}^2 = 6{,}0\;\text{N}$

Check. Applying the second law to $m_2$: $F - T = m_2\,a$, that is $16 - 6{,}0 = 10 = 5{,}0 \times 2{,}0$. Consistent.

$\ev{a = 2{,}0\;\mathrm{m/s^2} \qquad T = 6{,}0\;\mathrm{N}}$

Links

Topics: Dynamics Concepts: Newton’s second law · Tension Objects: Ideal rope · Block