Worked exercise — two boxes in equilibrium

Problem

Box 1 ($m_1 = 3\;\text{kg}$) rests on the floor, with box 2 ($m_2 = 8\;\text{kg}$) on top of it. Find the reaction from the floor.

Solution

Both boxes are in equilibrium, so the resultant force on each is zero. We use $g = 9{,}81\;\text{m/s}^2$.

Method 1 — body by body.

Body 2 (weight and reaction from box 1): $\vv{P}_2 + \vv{R}_{12} = \vv{0}$, from which

$R_{12} = m_2 g = 8 \cdot 9{,}81 = 78{,}5\;\text{N}$

Body 1: acting on box 1 are its own weight, the reaction $\vv{R}_{21}$ that box 2 presses onto it (by the third law $R_{21} = R_{12} = 78{,}5\;\text{N}$) and the reaction from the floor $\vv{R}_{\text{pav}}$. At vertical equilibrium:

$R_{\text{pav}} = m_1 g + R_{21} = 29{,}4 + 78{,}5$

$\ev{R_{\text{pav}} = 107{,}9\;\text{N}}$

Method 2 — system (1+2). Treating the two boxes as a single system, the internal forces $\vv{R}_{12}$ and $\vv{R}_{21}$ cancel out and only the weights and the floor’s reaction remain:

$R_{\text{pav}} = (m_1 + m_2)\,g = 11 \cdot 9{,}81$

$\ev{R_{\text{pav}} = 107{,}9\;\text{N}}$

Both methods give the same result: reasoning on the system is faster, but only the body-by-body analysis reveals the internal contact forces.

Links

Topics: Dynamics Concepts: Newton’s second law · Normal force · Newton’s third law Skills: Free-body diagram · Interaction diagram Objects: Block