Force
Table of Contents
when no force acts on a body, it will remain at rest or maintain uniform motion
when a force is applied to a body, it will change its state of motion
Force is a vector concept
Total force (or net force) is a vector sum of individual forces acting on the body from surroundings
Consider one object (the system of interest)
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Perpendicular to the surface (surface being a plane) of contact
Here normal refers to the geometry terminology for being perpendicular
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m = 1.0; % kg
k = 100.0; % N/m
g = 9.8; % m/sˆ2
v0 = 0.0; % in m/s
time = 2.0; % s
dt = 0.0001; % s
n = ceil(time/dt);
t = zeros(n,1);
y = zeros(n,1);
v = zeros(n,1);
a = zeros(n,1);
y(1) = 0.0;
v(1) = v0;
a(1) = -k*y(1) - m*g;
for i = 1:n-1
F = -k*y(i) - m*g;
a(i+1) = F/m;
v(i+1) = v(i) + a(i+1)*dt;
y(i+1) = y(i) + v(i+1)*dt;
t(i+1) = t(i) + dt;
end
subplot(3,1,1), plot(t,a), ylabel('a')
subplot(3,1,2), plot(t,v), ylabel('v')
subplot(3,1,3), plot(t,y), ylabel('y'), xlabel('t')
Question
Consider a car at rest on a level surface. We can conclude that the downward gravitational pull of Earth on the car and upward contact force of Earth on it are equal and opposite because
$m$ is at rest $(a = 0)$
$\implies $ No total external forces $(N = mg)$
Person $A$ is pulling a string with force $F$. Tell all forces
$\implies F_1 = F \text{ [from Newton's } 2^{nd} \text{ Law] }$
$\implies \text{ A string is pulling a wall by } F \text{ [from Newton's } 3^{rd} \text{ Law]}$
Question
A body of mass $m$ is suspended from a spring with spring constant $k$ in configuration (a) and the spring is stretched $0.1$m. If two identical bodies of mass $\frac{m}{2}$ are suspended from a spring with the same spring constant $k$ in configuration (b), how much will the spring stretch? Explain your answer.
Problem 1
Problem 2
Problem 3
Find $\vec a$ for $m$ and $M$
Inertial Reference Frame
Non Inertial Reference Frame
Is Newton’s Law wrong ?
Observation (reference) frame is accelerating !!!
Frame moves with $a_B$
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Summary
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Problem 3 revisited
Find $\vec a$ for $m$ and $M$
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Example – Inertial circular frame
Example – Non inertial circular frame
Problem
Problem
Ignore (mass, moment of inertia) of a pulley $\implies$ It only changes directions
Think of a small piece of string segment
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We begin our discussion of rotational dynamics and rotational motion. We will start by introducing the concept of torque, the new manifestation of forces that produce a change in rotational motion.
To include rotational motion, the net torque applied to a rigid body will produce a change in rotational motion:
In our study of translational motion under force, the location where the force was applied to the system did not matter. The subsequent translational motion only depended on the magnitude and direction of the net force. We will see that this simplification no longer holds for rotational motion. Forces applied to a body in an attempt to produce rotation will have different effects depending on the location of application and the direction of the applied force.
Consider trying to close a door by applying a force to the door near the hinges or in the middle of the door. Getting the door to close is much easier when the force is applied far from the hinges (see force $F_3$ in the figure below). This is why we usually place door handles by the edge opposite the hinges.
A well chosen location is not enough to guarantee effective rotation from an applied force. Consider another experiment. Suppose that you stand at the edge of an open door and push the door edge toward the hinges (see the force $F_1$ in the figure below). The door will not rotate, even with a hard push. This indicates that the direction of the force also affects the rotation produced.
Definition of Torque
Consider a force $\vec{F}$ that is applied at a given point A as shown in the figure below. We say that this force produces a torque about point Q. The torque produced is a vector $\vec{\tau}$ is defined by the vector product :
$$\begin{align} \vec{\tau} = \vec{r}\times \vec{F} \end{align}$$where the vector $\vec{r}$ is the position vector of the point A measured from point Q. (see below).
How can we generate the moment?
Vector
Magnitude
Direction (not obvious)
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moment of inertia = roational mass
$$I = \sum_{i} m_i r_i^2$$
Just as mass quantifies the resistance of an object to changes in linear velocity, the moment of inertia, also called rotational inertia, quantifies the resistance of an object to changes in angular velocity. A system's moment of inertia depends on
moment of inertia for other rigid bodies
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Analogy of Rotational Dynamics with Translation Dynamics (Newton's Second Law)
By analogy to the manner in which force changes velocity
$$\sum\vec{F}=m \frac{d \vec{v}}{dt} =m\vec{a}$$the angular dynamics and net torque model has a similar equation of change, but for rotational velocity around a fixed axis. If we apply a non zero torque on an object (e.g. push or pull on a door handle), it will result in a change of rotational motion - e.g. the door will start to rotate faster or slower about its (fixed) hinges.
By analogy, the angular velocity of an object rotating about an axis that passes through a point $Q$, is changed by the net torque on the object measured about point Q:
$$\sum\vec{\tau}_{\small{Q}} = I_Q \frac{d \vec \omega}{dt} = I_Q \vec{\alpha}$$Where $\alpha$ is the angular acceleration: the rate of change of the angular speed $\omega$, and $\displaystyle I_Q$ is the moment of inertia about point $Q$, a quantity that is analogous to the inertial mass in $\sum\vec{F}=m\vec{a}$.
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