Problem — Three Blocks Connected by Two Ropes

Problem

Three blocks of masses $m_1 = 2{,}0\;\text{kg}$, $m_2 = 3{,}0\;\text{kg}$, $m_3 = 5{,}0\;\text{kg}$ are connected by ideal ropes on a smooth horizontal plane, as in the figure. A force $F = 20\;\text{N}$ pulls the system. Calculate the common acceleration and the tensions $T_1$ and $T_2$ in the two ropes.

Solution

Step 1 — Acceleration of the system. The ropes are ideal and inextensible: the three blocks move with the same acceleration $a$. Considering the system as a whole, the tensions are internal and the only external horizontal force is $F$:

$F = (m_1 + m_2 + m_3)\,a \quad\Longrightarrow\quad a = \frac{F}{m_1 + m_2 + m_3}$

$a = \frac{20\;\text{N}}{(2{,}0 + 3{,}0 + 5{,}0)\;\text{kg}} = \frac{20}{10} = 2{,}0\;\text{m/s}^2$

Step 2 — Tension $T_1$. The force $F$ is applied to $m_3$; rope 1 connects $m_1$ to $m_2$ and is the only horizontal force acting on $m_1$, dragging it along:

$T_1 = m_1\,a = 2{,}0\;\text{kg} \times 2{,}0\;\text{m/s}^2 = 4{,}0\;\text{N}$

Step 3 — Tension $T_2$. Rope 2 connects $m_2$ to $m_3$ and must drag along everything “behind” it, that is, the group $m_1 + m_2$:

$T_2 = (m_1 + m_2)\,a = 5{,}0\;\text{kg} \times 2{,}0\;\text{m/s}^2 = 10\;\text{N}$

Check. On block $m_3$: $F - T_2 = m_3\,a$, that is $20 - 10 = 10 = 5{,}0 \times 2{,}0$. Consistent.

$\ev{a = 2{,}0\;\mathrm{m/s^2} \qquad T_1 = 4{,}0\;\mathrm{N} \qquad T_2 = 10\;\mathrm{N}}$

Links

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