Two Blocks Tied Together with an Applied Force

Problem

Two blocks of masses $m_1 = 4{,}0\;\text{kg}$ and $m_2 = 6{,}0\;\text{kg}$ are connected by an inextensible rope on a smooth horizontal plane. A force $F = 25\;\text{N}$ pulls block $m_2$ in the horizontal direction. Calculate the acceleration of the system and the tension in the rope.

Solution

Step 1 — The system as a whole. The plane is smooth, so the only external horizontal force acting on the system formed by the two blocks is $F$. The rope is inextensible, so the two blocks move with the same acceleration $a$. Applying Newton’s second law to the total mass $m_1 + m_2$:

$F = (m_1 + m_2)\,a$

$a = \frac{F}{m_1 + m_2} = \frac{25}{4{,}0 + 6{,}0} = \frac{25}{10}$

$\ev{a = 2{,}5\;\text{m/s}^2}$

Step 2 — Free body of block $m_1$ alone. On block $m_1$, only the rope’s tension $T$ acts horizontally, and it is solely responsible for its acceleration:

$T = m_1\,a = 4{,}0 \cdot 2{,}5$

$\ev{T = 10\;\text{N}}$

Check. On block $m_2$, $F$ acts forward and $T$ backward: $F - T = 25 - 10 = 15\;\text{N} = m_2\,a = 6{,}0 \cdot 2{,}5 = 15\;\text{N}$. Consistent.

Links

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