Worked Exercise — System with Two Masses and a Pulley

Problem

A mass $m_1 = 10\;\text{kg}$ hangs from a rope connected, by way of a pulley, to the end of a rigid horizontal bar $L = 4\;\text{m}$ long. At the other end of the bar hangs a mass $m_2 = 20\;\text{kg}$. The bar is held up by a force $\vv F$ applied at $60^\circ$. Find $F$ for equilibrium.

Model adopted. The text leaves the geometry partly unspecified; we adopt the most natural model, consistent with the data. The bar is hinged to the wall at its left-hand end $O$; at the right-hand end $B$ (at $L = 4\;\text{m}$ from $O$) both the weight of $m_2$ and the tension of the rope that, redirected by the pulley placed at $B$, supports $m_1$, act downward. The supporting force $\vv F$ is applied at $B$ and makes an angle of $60^\circ$ with the bar. The bar’s own weight is negligible.

Solution

Step 1 — The rope holding $m_1$. The rope holding $m_1$ has, for equilibrium of the hanging block, tension equal to its weight; the pulley only changes its direction, not its magnitude:

$t_1 = m_1 g = 10\cdot 9{,}81 = 98{,}1\;\text{N}$

Step 2 — The downward loads at $B$. At the end $B$, both the tension $t_1$ and the weight of $m_2$ act downward:

$P_2 = m_2 g = 20\cdot 9{,}81 = 196{,}2\;\text{N}, \qquad F_{\downarrow} = t_1 + P_2 = 98{,}1 + 196{,}2 = 294{,}3\;\text{N}$

Step 3 — Torque equilibrium (pole at $O$). Choosing the pole at the hinge $O$ eliminates the constraint reactions. The weight and the force $\vv F$ are both applied at $B$, with arm $L$; what matters is the vertical component of $\vv F$, namely $F\sin 60^\circ$:

$F\sin 60^\circ \cdot L - F_{\downarrow}\cdot L = 0 \;\Rightarrow\; F\sin 60^\circ = 294{,}3\;\text{N}$

Step 4 — Computing $F$.

$F = \frac{294{,}3}{\sin 60^\circ} = \frac{294{,}3}{0{,}866} \approx 339{,}8\;\text{N}$

$\ev{F \approx 3{,}4\cdot 10^{2}\;\text{N}}$

Horizontal check. The horizontal component $F\cos 60^\circ = 339{,}8\cdot 0{,}5 \approx 169{,}9\;\text{N}$ is balanced by the horizontal reaction of the hinge at $O$: the system is thus in complete equilibrium.

Note: if instead one assumed that the tension $t_1$ does not act at the end $B$ but is applied elsewhere, the numerical value of $F$ would change; the method (balance of forces and torques with a clever choice of pole) remains identical.

Links

Topics: Statica ed equilibrio Concepts: Equilibrio statico · Momento di una forza · Tensione · Forza peso Skills: Bilancio dei momenti · Scelta del polo · Scomposizione vettoriale Methods: Scomposizione in componenti cartesiane Objects: Puleggia · Asta