Markdown → MathJax (Sorgente ⟂ Rendering)

# Fare su fogli a parte DISEGNANDO I VETTORI IN UNO SPAZIO TRIDIMENSIONALE

# Momento di una forza – esercizi vettoriali

Convenzione:  (I vettori sono indicati in <b>F</b> carattere <b>BOLD (neretto)</b> : è un convenzione tipografica per semplificarne la rappresentazione, funziona con i computer non con la penna (in quel caso si usa il simbolo vettore))
$$
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}, 
\quad 
\boldsymbol{\tau} = (yF_z - zF_y,\; zF_x - xF_z,\; xF_y - yF_x).
$$

---

## Esercizio 1 – determinante (piano xy)

$$
\mathbf{r} = (0.30,\,0.40,\,0)\,\text{m}, 
\quad \mathbf{F} = (10,\,-20,\,0)\,\text{N}
$$

$$
\boldsymbol{\tau} =
\begin{vmatrix}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\
0.30 & 0.40 & 0 \\
10 & -20 & 0
\end{vmatrix}
= (0,\,0,\,-10)\,\text{N·m}
$$

---

## Esercizio 2 – determinante (caso 3D)

$$
\mathbf{r} = (0.20,\,-0.10,\,0.30)\,\text{m}, 
\quad \mathbf{F} = (15,\,8,\,-6)\,\text{N}
$$

$$
\boldsymbol{\tau} =
\begin{vmatrix}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\
0.20 & -0.10 & 0.30 \\
15 & 8 & -6
\end{vmatrix}
=(-1.8,\,5.7,\,3.1)\,\text{N·m}
$$

---

## Esercizio 3 – determinante (momento totale di più forze)

$$
\begin{aligned}
\mathbf{r}_A &= (0.50,\,0,\,0)\,\text{m}, & \mathbf{F}_A &= (0,\,40,\,0)\,\text{N}\\
\mathbf{r}_B &= (0,\,0.40,\,0)\,\text{m}, & \mathbf{F}_B &= (-30,\,0,\,0)\,\text{N}\\
\mathbf{r}_C &= (0.20,\,0.20,\,0)\,\text{m}, & \mathbf{F}_C &= (0,\,0,\,-25)\,\text{N}
\end{aligned}
$$

$$
\boldsymbol{\tau}_A=(0,\,0,\,20), \quad
\boldsymbol{\tau}_B=(0,\,0,\,12), \quad
\boldsymbol{\tau}_C=(-5,\,5,\,0)
$$

$$
\boldsymbol{\tau}_{\text{tot}} = (-5,\,5,\,32)\,\text{N·m}
$$

---

## Esercizio 4 – formula per componenti

$$
\mathbf{r}=(0.30,\,-0.20,\,0.50)\,\text{m}, 
\quad \mathbf{F}=(12,\,-4,\,9)\,\text{N}
$$

$$
\begin{aligned}
\tau_x &= yF_z - zF_y = (-0.20)(9) - (0.50)(-4) = 0.2 \\
\tau_y &= zF_x - xF_z = (0.50)(12) - (0.30)(9) = 3.3 \\
\tau_z &= xF_y - yF_x = (0.30)(-4) - (-0.20)(12) = 1.2
\end{aligned}
$$

$$
\boldsymbol{\tau} = (0.2,\,3.3,\,1.2)\,\text{N·m}
$$

---

## Esercizio 5 – formula per componenti (piano xz)

$$
\mathbf{r}=(0.60,\,0.10,\,0)\,\text{m}, 
\quad \mathbf{F}=(3.0,\,1.5,\,0.5)\,\text{N}
$$

$$
\begin{aligned}
\tau_x &= yF_z - zF_y = 0.10\cdot0.5 - 0 = 0.05 \\
\tau_y &= zF_x - xF_z = 0 - 0.60\cdot0.5 = -0.30 \\
\tau_z &= xF_y - yF_x = 0.60\cdot1.5 - 0.10\cdot3.0 = 0.60
\end{aligned}
$$

$$
\boldsymbol{\tau} = (0.05,\,-0.30,\,0.60)\,\text{N·m}
$$

---

## Esercizio 6 – formula per componenti (momento totale di due forze)

$$
\begin{aligned}
\mathbf{r}_1 &= (0,\,0.25,\,0.10)\,\text{m}, & \mathbf{F}_1 &= (20,\,-5,\,0)\,\text{N}\\
\mathbf{r}_2 &= (0.15,\,0,\,0.20)\,\text{m}, & \mathbf{F}_2 &= (-10,\,0,\,8)\,\text{N}
\end{aligned}
$$

$$
\boldsymbol{\tau}_1 = (0.50,\,2.00,\,-5.00), 
\quad \boldsymbol{\tau}_2 = (0,\,-3.20,\,0)
$$

$$
\boldsymbol{\tau}_{\text{tot}} = (0.50,\,-1.20,\,-5.00)\,\text{N·m}
$$

---

Markdown → MathJax (Sorgente ⟂ Rendering)

Sorgente Markdown

Rendering MathJax