Two bodies connected by a pulley

Problem

Two connected bodies. Two blocks are connected by a rope that runs without friction over an ideal pulley: $m_1 = 2{,}0\;\text{kg}$ on the smooth horizontal table, $m_2 = 3{,}0\;\text{kg}$ hanging off the edge. Calculate: (a) the acceleration of the system, (b) the tension in the rope.

Solution

Step 1 — Free bodies. The ideal pulley redirects the rope without altering its tension $T$, the same on both stretches. The inextensible rope imposes the same magnitude of acceleration $a$ on the two blocks: $m_1$ slides horizontally toward the pulley while $m_2$ descends. The weight of $m_2$ is the driving force of the system.

Block $m_1$ (horizontal, smooth table). The only horizontal force is the tension:

$T = m_1 a \quad \text{(I)}$

(Vertically the normal balances the weight of $m_1$: it does not affect the motion.)

Block $m_2$ (vertical, descending). Taking downward as positive, acting on it are the weight $m_2 g$ (driving) and the tension $T$ (braking):

$m_2 g - T = m_2 a \quad \text{(II)}$

Step 2 — Acceleration. Adding (I) and (II) eliminates $T$:

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

$a = \frac{m_2\,g}{m_1 + m_2} = \frac{3{,}0 \cdot 9{,}81}{2{,}0 + 3{,}0} = \frac{29{,}43}{5{,}0}$

$\ev{a \approx 5{,}89\;\text{m/s}^2}$

Step 3 — Tension. Substituting into (I):

$T = m_1 a = 2{,}0 \cdot 5{,}89$

$\ev{T \approx 11{,}8\;\text{N}}$

Check. The tension ($11{,}8\;\text{N}$) is smaller than the weight of $m_2$ ($P_2 = 29{,}4\;\text{N}$): correct, otherwise $m_2$ would not descend. Moreover, if there were no table ($m_1 = 0$), we would recover $a = g$ and $T = 0$, free fall. Consistent.

Links

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