Theoretical Mechanics IPSP

The pictures in the margin show the sign of a seesaw, a playground toy that works even for people with vastly different weight and size. Figure 2.16a) shows a balanced scale. When the forces acting on the scale do not add up to zero, we pick up the scale. It moves. The according force balance for the beam of the scale is shown in Figure 2.16c). In general the beam does not stay at rest, when the two masses are not attached at the same distance from the fulcrum. The force balance, Figure 2.16c), still hods, and the beam turns, rather than being lifted. The sum of attached forces tells us if an object is displaced. In analogy we introduce the torque to describe whether it turns. When the beam is vertical there is no torque, and it takes its maximum when the beam is horizontal. In the former case the forces act parallel to the beam, and in the latter they act in orthogonal direction. Moreover, a weight that is attached at a larger distance to the fulcrum induces a larger torque, and the torque also increases with mass. This is expressed in the lever rule.

Remark 2.26 Adopting a lever where force is applied on a long arm allows one to move very heavy objects or break very stable objects. Common technological applications are the crowbar and the lever. Archimedes was so impressed by this principle that he is quoted to have remarked “Δοσ μοι που στω και κινω την γην” (Archimedes, 1878), i.e. ``Give me but one firm spot on which to stand, and I will move the earth'' (Oxford Dictionary of Quotations, 1953)

Observe the sign of the torque: In Example 2.23 it is positive for counterclockwise motion, and negative for clockwise motion. The axis of rotation is fixed by the fulcrum. However, when acting the crowbar, one applies a horizontal force to get the crowbar under the obstacle. This induces a rotation around a vertical axis. Subsequently, a vertical force is applied to lift the obstacle. It induces a rotation around a horizontal axes. The relation between the directions of the lever arm, the force, and the rotation axis is commonly illustrated by the right-hand rule (Figure 2.19): Here the arm points in the direction of the lever arm, the fingers in the direction of the applied force, and the thump along the rotation axis. This suggests to define torque as a product of two vectors, the arm $\mathbf\ell$ and the force $\mathbf F$ that provide the torque, $\mathbf T$, which is a vector of length $|\mathbf\ell| \: |\mathbf F| \: \sin\angle{\mathbf\ell, \mathbf F}$ in a direction normal to the plane defined by $\mathbf\ell$ and $\mathbf F$. This operation, $\mathbf T = \mathbf\ell \times \mathbf F$ defines the cross product. We explore its properties in a mathematical digression.

Remark 2.27 The cross product of a vector with itself vanishes \begin{align*} \forall \mathbf v \in\mathbb{R}^3 : \mathbf v \times \mathbf v = \mathbf 0 \end{align*}

Proof. Vanishing of $\mathbf v \times \mathbf v$ is a consequence of anti-commutativity: \begin{align*} \mathbf v \times \mathbf v = - \mathbf v \times \mathbf v \quad\Rightarrow 2 \: \mathbf v \times \mathbf v = \mathbf 0 \quad\Rightarrow \mathbf v \times \mathbf v = \mathbf 0 \\ \end{align*}

Proof. The identity $\mathbf u \cdot ( \mathbf v \times \mathbf w) = \varepsilon_{ijk} \, u_i \, v_j \, w_k$ follows from the compatibility with scalar product and the relation for the basis vectors $\mathbf e_i \cdot ( \mathbf e_j \times \mathbf e_k)$. The details of the proof are given as Problem 2.23.

Proof of Theorem 2.4 We show that $\mathbf e_1$, $\mathbf e_2$, and $\mathbf e_3=\mathbf e_1 \times \mathbf e_2$ form three orthonormal vectors. By assumption $\mathbf e_1$ and $\mathbf e_2$ are orthonormal. Hence, we show that $\mathbf e_3$ is a unit vector that is orthogonal to $\mathbf e_1$ and $\mathbf e_3$: \begin{align*} \mathbf e_1 \cdot \mathbf e_3 &= \mathbf e_1 \cdot (\mathbf e_1 \times \mathbf e_2) = \mathbf e_2 \cdot (\mathbf e_1 \times \mathbf e_1) = \mathbf e_2 \cdot \mathbf 0 = 0 \end{align*} \begin{align*} \mathbf e_2 \cdot \mathbf e_3 &= \mathbf e_2 \cdot (\mathbf e_1 \times \mathbf e_2) = \mathbf e_1 \cdot (\mathbf e_2 \times \mathbf e_2) = \mathbf e_1 \cdot \mathbf 0 = 0 \end{align*} \begin{align*} \mathbf e_3 \cdot \mathbf e_3 &= \mathbf e_3 \cdot (\mathbf e_1 \times \mathbf e_2) = \mathbf e_1 \cdot (\mathbf e_2 \times \mathbf e_3) = 1 \end{align*}

Remark 2.29 (bac-cab rule) The double cross product can be expressed in terms of scalar products. Commonly this relation is stated in terms of three vectors $\mathbf a$, $\mathbf b$, and $\mathbf c \in \mathbb{R}^3$, \begin{align*} \mathbf a \times (\mathbf b \times \mathbf c) = \mathbf b \: (\mathbf a \cdot \mathbf c) - \mathbf c \: ( \mathbf a \cdot \mathbf b ) \end{align*} and referred to as bac-cab rule.

Proof.
We express the three vectors in terms of their coordinates with respect to the orthonormal basis $\mathbf e_1$, $\mathbf e_2$, $\mathbf e_3$,
\begin{align*} \mathbf a = \sum_{i=1}^3 a_i \, \mathbf e_i \quad \mathbf b = \sum_{j=1}^3 b_j \, \mathbf e_j \quad \mathbf c = \sum_{k=1}^3 c_k \, \mathbf e_k \quad \text{with } a_i, b_j, c_k \in \mathbb{R} \end{align*} and use the rules defining the cross products and inner products
\begin{align*} \mathbf a \times (\mathbf b \times \mathbf c) &= \left( \sum_{i=1}^3 a_i \, \mathbf e_i \right) \times \left[ \left( \sum_{j=1}^3 b_j \, \mathbf e_j \right) \times \left( \sum_{k=1}^3 c_k \, \mathbf e_k \right) \right] \\ &= \sum_{i,j,k=1}^3 a_i \, b_j \, c_k \mathbf e_i \times ( \mathbf e_j \times \mathbf e_k ) \end{align*} When $j=k$ or when $j$ and $k$ are both different from $i$ then the summand vanishes due to Remark 2.27. For $i=j \neq k$ one has $\mathbf e_i \times ( \mathbf e_j \times \mathbf e_k ) = -\mathbf e_k$, and for $i=k \neq j$ one has $\mathbf e_i \times ( \mathbf e_j \times \mathbf e_k ) = \mathbf e_j$. Consequently, \begin{align*} \mathbf a \times (\mathbf b \times \mathbf c) &= \sum_{i,k=1}^3 a_i \, b_i \, c_k (-\mathbf e_k) + \sum_{i,j=1}^3 a_i \, b_j \, c_i (\mathbf e_j) \\ &= \mathbf b \: (\mathbf a \cdot \mathbf c) - \mathbf c \: ( \mathbf a \cdot \mathbf b ) \\ \end{align*}

Remark 2.30 (Jacobi identity) The cross product obeys the Jacobi identity: \begin{align*} \mathbf u \times (\mathbf v \times \mathbf w) + \mathbf v \times (\mathbf w \times \mathbf u) + \mathbf w \times (\mathbf u \times \mathbf v) = \mathbf 0 \end{align*}

Proof. This can be verified by evaluating the triple cross products by the bac-cab rule. Details are give as Problem 2.24.

Proof. For component $k$ of $\mathbf a \times \mathbf b$ we have \begin{align*} [ \mathbf a \times \mathbf b ]_k &= \hat{\boldsymbol e}_k \cdot ( \mathbf a \times \mathbf b ) = \hat{\boldsymbol e}_k \cdot \left[ \left( \sum_{i=1}^3 a_i \, \hat{\boldsymbol e}_i \right) \times \left( \sum_{j=1}^3 b_j \, \hat{\boldsymbol e}_j \right) \right] \\ &= \sum_{i,j=1}^3 a_i \, b_j \;\; \hat{\boldsymbol e}_k \cdot \left[ \hat{\boldsymbol e}_i \times \hat{\boldsymbol e}_j \right]_k = \sum_{i,j=1}^3 a_i \, b_j \; \varepsilon_{ijk} \end{align*} In the remark this is explicitly written out for $k\in\{1,2,3\}$.

The cross product and the scalar triple product have distinct geometrical interpretations. The geometric meaning of the cross product $\mathbf a \times \mathbf b$ can best be seen by adopting a basis where the first basis vector is parallel to $\mathbf e_1 = \mathbf a / |\mathbf a|$, and the second basis vector $\mathbf e_2$ lies orthogonal to $\mathbf e_1$ in the plane spanned by $\mathbf a$ and $\mathbf b$. The third basis vector will then be $\mathbf e_3 = \mathbf e_1 \times \mathbf e_2$. The angle between $\mathbf a$ and$\mathbf b$, and hence also of $\mathbf e_1$ and$\mathbf b$ is denoted as$\theta$. Thus, $\mathbf b$ can be written as $\mathbf b = b_1 \, \mathbf e_1 + b_2 \, \mathbf e_2 = |\mathbf b| \, ( \cos\theta \, \mathbf e_1 + \sin\theta \, \mathbf e_2)$, cf.Figure 2.20. For this choice of the basis we find

\begin{align*} \mathbf a \times \mathbf b &= |\mathbf a| \, \mathbf e_1 \times ( b_1 \, \mathbf e_1 + b_2 \, \mathbf e_2 ) = |\mathbf a| \, b_1 \, \mathbf e_1 \times \mathbf e_1 + |\mathbf a| \, b_2 \, \mathbf e_1 \times \mathbf e_2 \\ &= |\mathbf a| \, |\mathbf b| \, \sin\theta \, \mathbf e_3 \end{align*}

Figure 2.20 illustrates that $|\mathbf a| \, |\mathbf b| \, \sin\theta$ amounts to the area of the parallelogram spanned by the vectors $\mathbf a$ and $\mathbf b$. Hence, the cross product amounts to a vector that is aligned vertically on the parallelogram, with a length that amounts to the area of the parallelogram.

In order to evaluate also the product $( \mathbf a \times \mathbf b ) \cdot \mathbf c$ we introduce the coordinate representation of $\mathbf c$ as $\mathbf c = c_1 \, \mathbf e_1 + c_2 \, \mathbf e_2 + c_3 \, \mathbf e_3$ (Figure 2.21), and observe \begin{align*} ( \mathbf a \times \mathbf b ) \cdot \mathbf c &= |\mathbf a \times \mathbf b| \, \mathbf e_3 \cdot ( c_1 \, \mathbf e_1 + c_2 \, \mathbf e_2 + c_3 \, \mathbf e_3 ) \\ &= |\mathbf a \times \mathbf b| \, c_3 = a_1 \, b_2 \, c_3 \end{align*}

This amounts to the product of the area of the parallelogram spanned by $\mathbf a$ and $\mathbf b$ multiplied by the height of the parallelepiped spanned by the vectors $\mathbf a$, $\mathbf b$, $\mathbf c$. Due to the special choice of the basis this volume amount to $a_1 \, b_2 \, c_3$ because all other contributions to the general expression $\sum_{ijk} a_i \, b_j \, c_k \, \varepsilon_{ijk}$ vanish. The symmetry of the scalar triple product, property d) in Definition 2.14, is understood from this perspective as the statement that the volume of the parallelepiped is invariant under (cyclic) renaming of the vectors that define its edges. As a final remark, we emphasize that the geometric interpretation that we have given to the cross product holds in general — in spite of the special basis adopted in the derivation. It is a distinguishing feature of vector spaces that the scalar numbers that are derived from vectors take the same values every choice of the basis. It is up to the physicist to find the basis that admits the easiest calculations.

The cross product equips us with the mathematical notions to define the torque on a body.

Remark 2.32 The value of the torque depends on the choice of the reference position $\mathbf q_0$.

Remark 2.33 In general, the torques induced by different forces point in different directions. They are added as vectors. We will further discuss this below in Example 2.24.

Problem 2.23:
Fill in the details of the proof for Remark 2.28.

Problem 2.24:
Fill in the details of the proof for Remark 2.30.

Problem 2.25: Turning a wheel
Two forces of magnitude $4\text{N}$ are acting on a wheel of radius $r$ that can freely rotate around its axis. What magnitude should a third force, $\mathbf F$, have that is attacking at a distance $r/2$ from the axis, such that there is no net torque acting on the wheel?

Problem 2.26: Nutcrackers
A common type of nutcrackers employs the principle of lever arms to crack nuts with a reasonable amount of force. We idealize the nut as a spring with spring constant $k = 1\text{kN/}mm$ and assume that it breaks when it is compressed by $\Delta= 0.6mm$. The nut is mounted at a distance of $l = 3cm$ from the joint of the nutcracker and the hand exerts a force $\mathbf F$ at a distance $L$.

1. Demonstrate that a force of magnitude $\displaystyle F = \frac{l k \Delta}{L}$ is required to crack the nut.
2. Calculate the numerical value of $F$.
3. If you try to crack the nut by placing it under a heavy stone: which mass should that stone have in order to crack the nut?