Renato Naville Watanabe, Marcos Duarte
Laboratory of Biomechanics and Motor Control
Federal University of ABC, Brazil
In Mechanics we are interested in the study of motion (including deformation) and forces (and the relation between them) of anything in nature.
As a good rule of thumb, we model the phenomenon of interest as simple as possible, with just enough complexity to understand the phenomenon.
For example, we could model a person jumping as a particle (the center of gravity, with no size) moving in one direction (the vertical) if all we want is to estimate the jump height and relate that to the external forces to the human body. So, mechanics of a particle might be all we need.
However, if the person jumps and performs a somersault, to understand this last part of the motion we have to model the human body as one of more objects which displaces and rotates in two or three dimensions. In this case, we would need what is called mechanics of rigid bodies.
If, besides the gross motions of the segments of the body, we are interested in understanding the deformation in the the human body segments and tissues, now we would have to describe the mechanical behavior of the body (e.g., how it deforms) under the action of forces. In this case we would have to include some constitutive laws describing the mechanical properties of the body.
"The theoretical development of the laws of motion of bodies is a problem of such interest and importance that it has engaged the attention of all the most eminent mathematicians since the invention of dynamics as a mathematical science by Galileo, and especially since the wonderful extension which was given to that science by Newton."
— Hamilton, 1834 (apud Taylor, 2005).
Let's start with the study of the forces and motion in Mechanics looking at the motion of a particle using the Newtonian approach.
The Newton's laws of motion describe the relationship between the forces acting on a body and the resultant linear motion due to those forces:
These three statements are astonishing in their simplicity and how much of knowledge they empower.
Isaac Newton was born in 1943 and his works that resulted in these equations and other discoveries were mostly done in the years of 1666 and 1667, when he was only 24 years old!
Later, we will study in details how to draw a free-body diagram.
Let's see now some examples on how to find the motion of a particle
A football ball is kicked with an angle of 30 degrees with the ground (horizontal).
The mass of the ball is 0.43 kg. The initial speed of the ball is 20 m/s and the initial height is 0 m. Consider the gravitational acceleration as 9.81 $m/s^2$.
Find the motion of the ball.
Solution:
We know that:
The only force acting on the ball is the gravitational force:
$$ \vec{\bf{F}}_g = -mg \; \hat{\bf{j}} $$ So, we apply the Newton's second law: $$ \vec{\bf{F}}_g = m \frac{d^2\vec{\bf{r}}}{dt^2} \quad \rightarrow \quad - mg \; \hat{\bf{j}} = m \frac{d^2\vec{\bf{r}}}{dt^2} \quad \rightarrow \quad - g \; \hat{\bf{j}} = \frac{d^2\vec{\bf{r}}}{dt^2} $$ Now, we can separate the equation in two components (x and y): $$ 0 = \frac{d^2x}{dt^2} $$ and $$ -g = \frac{d^2y}{dt^2} $$import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
%matplotlib inline
sns.set_context('notebook', font_scale=1.2)
m = 0.43 # [kg]
x0 = 0 # [m]
y0 = 0 # [m]
vx0 = 10*np.sqrt(3) # [m/s]
vy0 = 10 # [m/s]
g = 9.81 # [m/s^2]
These equations can be easily solved by integrating both sides of each equation:
$$ 0 = \frac{d^2x}{dt^2} \quad \rightarrow v_{x0} = \frac{dx}{dt} \quad \rightarrow \quad v_{x0}t + x_{0} = x(t) $$ and $$ -g = \frac{d^2y}{dt^2} \quad \rightarrow - gt + v_{y0} = \frac{dy}{dt} \quad \rightarrow \quad -\frac{g}{2}t^2 +v_{y0}t + y_{0} = y(t) $$ So the trajectory of the ball along time is: $$ x(t) = v_{x0}t $$ and $$ y(t) = -\frac{g}{2} t^2 + v_{y0}t $$ or $$ \vec{\bf{r}}(t) = v_{x0}t \; \hat{\bf{i}} + \left(-\frac{g}{2} t^2 + v_{y0}t \right) \; \hat{\bf{j}} $$dt = 0.001 # [s]
t = np.arange(0, 2.05, dt)
x1a = vx0*t
y1a = -g/2*t**2 + vy0*t
plt.figure(figsize=(8, 4))
plt.plot(x1a, y1a, lw=4)
plt.xlim(0, 36)
plt.ylim(0, 6)
plt.title('Ball trajectory')
plt.xlabel('x')
plt.ylabel('y')
plt.show()
We can solve this problem numerically and compare both solutions.
We start from the differential equations of each coordinate, and then break them into two first-order differential equations:
$$ 0 = \frac{d^2x}{dt^2} $$ and $$ - g = \frac{d^2y}{dt^2} $$ The first equation can be broken as: $$ \frac{dv_x}{dt} = 0 $$ $$ \frac{dx}{dt} = v_x(t) $$ And the second equation can be broken as: $$ \frac{dv_y}{dt} = -g $$ $$ \frac{dy}{dt} = v_y(t) $$ You can use any numerical integration method you want (Euler, Runge-Kutta, etc), but here we will use the Euler method. Let's see the solution for the $x(t)$ variable.The derivative of $x(t)$ is given by:
$$ \frac{dx}{dt} = \lim\limits_{\Delta t \rightarrow 0} \frac{x(t+\Delta t) - x(t)}{\Delta t} $$ Whcih can be approximated by: $$ \frac{dx}{dt} \approx \frac{x(t+\Delta t) - x(t)}{\Delta t} \quad \rightarrow \quad x(t+\Delta t) \approx x(t) + \Delta t \frac{dx}{dt} $$ So, with the initial conditions of all the variables, we can apply the equation above to find the values of the variables along time (for a revision of Ordinary Differential Equations, see the notebook [Ordinary Differential Equation](https://nbviewer.jupyter.org/github/BMClab/bmc/blob/master/notebooks/OrdinaryDifferentialEquation.ipynb)). In the cell below, we apply the Euler method for the four first-order differential equations.x = x0
y = y0
vx = vx0
vy = vy0
r = np.array([x, y])
while y >= 0:
dxdt = vx
x = x + dt * dxdt
dydt = vy
y = y + dt * dydt
dvxdt = 0
vx = vx + dt * dvxdt
dvydt = -g
vy = vy + dt * dvydt
r = np.vstack((r, np.array([x,y])))
plt.figure(figsize=(8, 4))
x1n = r[:, 0]
y1n = r[:, 1]
plt.plot(x1n, y1n, lw=4)
plt.xlim(0, 36)
plt.ylim(0, 6)
plt.title('Ball trajectory')
plt.xlabel('x')
plt.ylabel('y')
plt.show()
Now, besides the gravity, we consider the drag due to the air resistance ($b$ = 0.006 Ns/m). First we will consider the drag force proportional to the speed and opposite direction to the velocity vector.
So the forces being applied on the ball are:
$$ \vec{\bf{F}} = -mg \; \hat{\bf{j}} - b\vec{\bf{v}} = -mg \; \hat{\bf{j}} - b\frac{d\vec{\bf{r}}}{dt} = -mg \; \hat{\bf{j}} - b\left(\frac{dx}{dt} \; \hat{\bf{i}}+\frac{dy}{dt} \; \hat{\bf{j}}\right) = - b\frac{dx}{dt} \; \hat{\bf{i}} - \left(mg + b\frac{dy}{dt}\right) \; \hat{\bf{j}} $$ Writing down the Newton's second law: $$ \vec{\bf{F}} = m \frac{d^2\vec{\bf{r}}}{dt^2} \quad \rightarrow \quad - b\frac{dx}{dt} \; \hat{\bf{i}} - \left(mg + b\frac{dy}{dt}\right) \; \hat{\bf{j}} = m\left(\frac{d^2x}{dt^2} \; \hat{\bf{i}}+\frac{d^2y}{dt^2} \; \hat{\bf{j}}\right) $$ Now, we can separate into one equation for each coordinate: $$ - b\frac{dx}{dt} = m\frac{d^2x}{dt^2} \quad \rightarrow \quad \frac{d^2x}{dt^2} = -\frac{b}{m} \frac{dx}{dt} $$ $$ -mg - b\frac{dy}{dt} = m\frac{d^2y}{dt^2} \quad \rightarrow \quad \frac{d^2y}{dt^2} = -\frac{b}{m}\frac{dy}{dt} - g $$ We can solve these equations analytically, for example, by using Laplace Transform or classical methods to solve linear differential equations.The solution of a linear differential equation can be found by finding the natural (or homogeneous) solution and the forced (or non-homogeneous) solution and then adding both solutions.
First, we solve the first differential equation ($x$ coordinate). The characteristic polynomial of the equation is:
$$ \lambda^2 + \frac{b}{m}\lambda = 0 $$The roots of this equation are $\lambda = 0$ and $\lambda = -\frac{b}{m}$, and consequently, its natural modes are:
$$ x_{n_1}(t) = Ae^{0t} = A \\ x_{n_2}(t) = B e^{-\frac{b}{m}t} $$As there is no external forces in the $x$ direction, there is no forced solution. So, the motion of the ball in the $x$ coordinate is:
$$ x(t) = A + Be^{-\frac{b}{m}t} $$To find the values of the $A$ and $B$ constants, we must use the initial conditions $x(0)$ and $v_x(0)$.
$$ x(0) = 0 = A + B $$ $$ v_x(0) = v_{x0} = \frac{dx(0)}{dt} = -\frac{Bb}{m}e^{-\frac{b}{m}0} \quad \rightarrow \quad B = -\frac{v_{x0}m}{b} \quad \rightarrow \quad A = \frac{v_{x0}m}{b} $$ So: $$ x(t) = \frac{v_{x0}m}{b} - \frac{v_{x0}m}{b}e^{-\frac{b}{m}t} = \frac{v_{x0}m}{b}\left(1-e^{-\frac{b}{m}t} \right) $$Now, we solve the differential equation for the $y$ coordinate.
First, we find the natural solution, which comprises the solution from the equation without the gravitational force (the force indepedent of y(t) and its derivatives), in this case:
The complete solution of the motion of the ball in the $y$ coordinate is the sum of the natural and forced solutions:
$$ y(t) = A + Be^{-\frac{b}{m}t} - \frac{mg}{b}t $$To find the values of the constants A and B, we must use the initial conditions $y(0)$ and $v_y(0)$.
$$ y(0) = 0 = A + B $$ $$ v_y(0) = v_{y0} = \frac{dy(0)}{dt} = - \frac{Bb}{m} - \frac{mg}{b} \\ B = -\frac{m^2g}{b^2} -\frac{v_{y0}m}{b} \\ A = +\frac{m^2g}{b^2} + \frac{v_{y0}m}{b} $$So, the motion of the ball in the y coordinate is:
$$ y(t) = \left(\frac{m^2g}{b^2} + \frac{v_{y0}m}{b}\right) - \left(\frac{m^2g}{b^2} + \frac{v_{y0}m}{b}\right)e^{-\frac{b}{m}t} -\frac{mg}{b}t \\ y(t) = \left(\frac{m^2g}{b^2} + \frac{v_{y0}m}{b}\right)\left(1 - e^{-\frac{b}{m}t}\right) - \frac{mg}{b}t $$b = 0.006 # [Ns/m]
t = np.arange(0, 2.05, 0.01)
x2a = vx0*m/b*(1 - np.exp(-b/m*t))
y2a = (vy0*m/b + g*m**2/b**2) * (1 - np.exp(-b/m*t)) - g*m/b*t
plt.figure(figsize=(8, 4))
plt.plot(x2a, y2a, lw=4)
plt.xlim(0, 36)
plt.ylim(0, 6)
plt.title('Ball trajectory')
plt.xlabel('x')
plt.ylabel('y')
plt.show()
Now, we will solve the same situation using a numerical method (Euler method).
We start from the equations previously found for each coordinate:
We can separate each equation into two first order equations and apply the Euler method:
x = x0
y = y0
vx = vx0
vy = vy0
r = np.array([x, y])
while y >= 0:
dxdt = vx
x = x + dt * dxdt
dydt = vy
y = y + dt * dydt
dvxdt = -b/m*vx
vx = vx + dt * dvxdt
dvydt = -g - b/m*vy
vy = vy + dt * dvydt
r = np.vstack((r, np.array([x, y])))
plt.figure(figsize=(8, 4))
x2n = r[:, 0]
y2n = r[:, 1]
plt.plot(x2n, y2n, lw=4)
plt.xlim(0, 36)
plt.ylim(0, 6)
plt.title('Ball trajectory')
plt.xlabel('x')
plt.ylabel('y')
plt.show()
Now, we will consider the drag force due to the air resistance proportional to the square of speed and still in the opposite direction of the velocity vector.
So the forces being applied on the ball are (for a revision of Time-varying frames and the meaning of the $\hat{\bf{e_t}}$, see Time-varying frames notebook):
$$ \vec{\bf{F}} = -mg \; \hat{\bf{j}} - bv^2\hat{\bf{e_t}} \\ \vec{\bf{F}} = -mg \; \hat{\bf{j}} - b (v_x^2+v_y^2) \frac{v_x \; \hat{\bf{i}} + v_y \; \hat{\bf{j}}}{\sqrt{v_x^2+v_y^2}} \\ \vec{\bf{F}} = -mg \; \hat{\bf{j}} - b \sqrt{v_x^2+v_y^2} \,(v_x \; \hat{\bf{i}}+v_y \; \hat{\bf{j}}) \\ \vec{\bf{F}} = -mg \; \hat{\bf{j}} - b \sqrt{\left(\frac{dx}{dt} \right)^2+\left(\frac{dy}{dt} \right)^2} \,\left(\frac{dx}{dt} \hat{\bf{i}} + \frac{dy}{dt} \; \hat{\bf{j}}\right) $$ Writing down the Newton's second law: $$ \vec{\bf{F}} = m \frac{d^2\vec{\bf{r}}}{dt^2} \\ -mg \; \hat{\bf{j}} - b \sqrt{\left(\frac{dx}{dt} \right)^2+\left(\frac{dy}{dt} \right)^2} \,\left(\frac{dx}{dt} \hat{\bf{i}}+\frac{dy}{dt}\hat{\bf{j}}\right) = m\left(\frac{d^2x}{dt^2}\hat{\bf{i}}+\frac{d^2y}{dt^2}\hat{\bf{j}}\right) $$ Now, we can separate into one equation for each coordinate: $$ - b \sqrt{\left(\frac{dx}{dt} \right)^2+\left(\frac{dy}{dt} \right)^2} \,\frac{dx}{dt} = m\frac{d^2x}{dt^2} \quad \rightarrow \\ \frac{d^2x}{dt^2} = - \frac{b}{m} \sqrt{\left(\frac{dx}{dt} \right)^2+\left(\frac{dy}{dt} \right)^2} \,\frac{dx}{dt} $$ $$ -mg - b \sqrt{\left(\frac{dx}{dt} \right)^2+\left(\frac{dy}{dt} \right)^2} \,\frac{dy}{dt} = m\frac{d^2y}{dt^2} \quad \rightarrow \\ \frac{d^2y}{dt^2} = - \frac{b}{m} \sqrt{\left(\frac{dx}{dt} \right)^2 + \left(\frac{dy}{dt} \right)^2} \,\frac{dy}{dt} -g $$ These equations are very difficult to solve analytically, but they can be easily solved by using numerical methods. Below we will use the same numerical method (Euler method) to solve these equations. For that, again we must break each equation into two first-order differential equations: $$ \frac{dv_x}{dt} = - \frac{b}{m} \sqrt{v_x^2+v_y^2} \,v_x $$ $$ \frac{dx}{dt} = v_x $$ $$ \frac{dv_y}{dt} = - \frac{b}{m} \sqrt{v_x^2+v_y^2} \,v_y -g $$ $$ \frac{dy}{dt} = v_y $$ Now, we can apply the Euler method to find a solution.x = x0
y = y0
vx = vx0
vy = vy0
r = np.array([x, y])
while y >= 0:
dxdt = vx
x = x + dt * dxdt
dydt = vy
y = y + dt * dydt
dvxdt = -b/m * np.sqrt(vx**2 + vy**2)*vx
vx = vx + dt * dvxdt
dvydt = -b/m * np.sqrt(vx**2 + vy**2)*vy - g
vy = vy + dt * dvydt
r = np.vstack((r, np.array([x, y])))
plt.figure(figsize=(8, 4))
x3n = r[:, 0]
y3n = r[:, 1]
plt.plot(x3n, y3n, lw=4)
plt.xlim(0, 36)
plt.ylim(0, 6)
plt.title('Ball trajectory')
plt.xlabel('x')
plt.ylabel('y')
plt.show()
plt.figure(figsize=(8, 4))
plt.plot(x1n, y1n, lw=4, c='g', label='$g$')
plt.plot(x2n, y2n, lw=4, c='b', label='$g+bv$')
plt.plot(x3n, y3n, lw=4, c='r', label='$g+bv^2$')
plt.xlim(0, 36)
plt.ylim(0, 6)
plt.title('Ball trajectory')
plt.xlabel('x')
plt.ylabel('y')
plt.legend(loc='best', frameon=False)
plt.show()
(Example 13.1, Hibbeler's book) A $50 kg$ crate rests on a horizontal surface for which the coefficient of kinetic friction is $\mu_k = 0.3$. If the crate is subjected to a $400 N$ towing force applied upward with a $30^o$ angle w.r.t. the horizontal, determine the velocity of the crate after $3 s$ starting from rest.
(Example 13.2, Hibbeler's book) A 10 kg projectile is fired vertically upward from the ground, with an initial velocity of 50 $m/s$. Determine the maximum height to which it will travel if (a) atmospheric resistance is neglected; and (b) atmospheric resistance is measured as $F_D=0.01v^2$ $N$, where $v$ is the speed of the projectile at any instant, measured in $m/s$.
(Exercise 13.6, Hibbeler's book) A person pushes on a $60 kg$ crate with a force $F$. The force is always directed down at $30^o$ from the horizontal as shown, and its magnitude is increased until the crate begins to slide. Determine the crate’s initial acceleration if the coefficient of static friction is $\mu_s=0.6$ and the coefficient of kinetic friction is $\mu_k=0.3$.
(Exercise 13.46, Hibbeler's book) The parachutist of mass $m$ is falling with a velocity of $v_0$ at the instant she opens the parachute. If air resistance is $F_D = Cv^2$, show that her maximum velocity (terminal velocity) during the descent is $v_{max}=\sqrt{{mg}/{C}}$. Solve the problem numerically considering $m=100 kg$, $g=10 m/s^2$, and $C=10 kg/m$. Plot a simulation of this numerical solution and show that indeed the parachutist approaches the terminal velocity.
Consider a block with mass of $1 kg$ attached to a spring hanging from a ceiling (the spring constant $k = 100 N/m$). At $t = 0 s$, the spring is stretched by $0.1 m$ from the equilibrium position of the block + spring system and then it's released (the initial velocity is not specified). Find the motion of the block.
Solve exercises 12.1.16, 12.1.19, 12.1.24, 12.1.29, 12.1.30, 12.1.31(a, b, d) and 12.1.32 from Ruina and Pratap's book (2019).
(https://youtu.be/N6IhkTjWrd4) A $15 kg$ box rests on a frictionless horizontal surface attached to a $5 kg$ box as shown below.
a. What is the acceleration of the system?
b. What is the tension in the rope?
c. Now consider a coefficient of kinetic friction of $0.25$ between the horizontal surface and the box. What are the acceleration of the system and the tension in the rope?