Force & Energy

Newton's Laws

Newtonian mechanics is built on these three laws

1. The velocity of an object is constant, i.e. the object should continue at rest or moving uniformly in a straight line, unless a net force is added on it
2. The net force on an object is proportional to the rate of change of the velocity, i.e. acceleration, and the mass of the object
3. When an object A exerts force on the other object B, object A must experience an equal magnitude but opposite direction force by object B

Newton's first law defines what an inertial frame is. Newton's second law only holds in an inertial frame. Newton's third law guarantees a system will not experience a net force when its component exerts force on another component.

Fictitious Force

The car is accelerating at a acceleration $\textbf{a}$. The car exerts force on the stone in the car by friction $\textbf{f}$. As observed by an inertial observer outside the car, the sum of force acting on the stone is $$\textbf{F} = \textbf{f} = m\textbf{a}$$ So, the stone is accelerating at $\textbf{a}$ But by an observer inside the car, which is an accelerating frame, adding all the real forces $$\textbf{f} = m\textbf{a}.$$ But the observer should observe the stone staying motionless in his frame. Thus, a fictitious force is observed by him $$\textbf{F}_{fictitious} = -m\textbf{a}$$ such that in his frame, the sum of forces is $$\begin{align} \textbf{F}' &= \textbf{f} + \textbf{F}_{fictitious} \\ &= m\textbf{a} - m\textbf{a} \\ &= 0 \end{align}$$ Therefore, for an observer in a non-inertial frame, a fictitious force must be included.

Example

A block of mass $m$ is placed on a frictionless incline of mass $M$, which is placed on a frictionless horizontal surface. Both the block and the incline are at rest. At time $t=0$, they are released to move freely. Using the X-Y frame fixed on the incline, find the acceleration of the block.

Solution:

The horizontal acceleration of the incline is $$Ma_{Mx}=N\sin\theta \implies a_{Mx}=\frac{N\sin\theta}{M}$$

As the incline is accelerating horizontally, a fictitious force $f$ is included in the X-Y frame. In the Y direction, force is balanced $$N+f\sin\theta-mg\cos\theta=0,$$ where $$f=ma_{Mx}$$ So, $$N=\frac{mg\cos\theta}{1+\frac{m}{M}\sin^2\theta}$$ and $$f=\frac{m^2g\sin\theta\cos\theta}{M+m\sin^2\theta}.$$ The acceleration of the block on X direction is $$\begin{align} ma_{m\text{X}} &= mg\sin\theta + f\cos\theta\\ &=mg\sin\theta + \frac{m^2g\sin\theta\cos\theta}{M+m\sin^2\theta}\cos\theta\\ a_{m\text{X}} &= \frac{(M+m)g\sin\theta}{M+m\sin^2\theta} \end{align}$$ The acceleration of the incline $$a_{Mx}=\frac{N\sin\theta}{M}=\frac{mg\sin\theta\cos\theta}{M+m\sin^2\theta}$$ The acceleration components of the block in $x-y$ frame are $$a_{my}=\frac{(M+m)g\sin^2\theta}{M+m\sin^2\theta},$$ and $$\begin{align} a_{mx} &= \frac{(M+m)g\sin\theta\cos\theta}{M+m\sin^2\theta}-\frac{mg\sin\theta\cos\theta}{M+m\sin^2\theta}\\ &= \frac{Mg\sin\theta\cos\theta}{M+m\sin^2\theta} \end{align}$$

Kinetic Energy

If $F(x)$ is the only force applying on an object along the path $x$, the integral of $F(x)$ with respect to x is $$\begin{align} \int^{x_f}_{x_i} F(x) dx &= \int^{x_f}_{x_i} m\frac{dv}{dt} dx \\ &= \int^{x_f}_{x_i} m\frac{dv}{dx}\frac{dx}{dt} dx \\ &= \int^{x_f}_{x_i} mv\frac{dv}{dx} dx \\ &= \int^{x_f}_{x_i} m\frac{d}{dx}\left(\frac{1}{2}v^2 \right) dx \\ &= \frac{1}{2}m(v(x_f))^2 - \frac{1}{2}m(v(x_i))^2 \end{align}$$ can tell us the final velocity if initial velocity is known. $$\int^{x_f}_{x_i} F(x) dx$$ is called the work done by the force $F(x)$ on the object. $$\frac{1}{2}mv^2$$ is called the kinetic energy. Since work done on the object is equal to the increase for its kinetic energy, we say that energy is conserved.

Potential Energy

If the integral of a force $\textbf{F}(\textbf{r})$ with respect to $\textbf{r}$ only depend on the initial and final position $$\int_{C_1}\textbf{F}(\textbf{r})\cdot d\textbf{r} = \int_{C_2}\textbf{F}(\textbf{r})\cdot d\textbf{r}$$ as long as $C_1$ has the same initial and final points as $C_2$. $\textbf{F}(\textbf{r})$ is said to be path independent. These kind of forces are called conservative force.

If a force is conservative, any closed integral must be zero $$\oint \textbf{F}\cdot\textbf{r} = 0$$ By Stoke's theorem, $$\oint \textbf{F}\cdot\textbf{r} = \int_A \nabla\times\textbf{F}\cdot d\textbf{a}$$ So, curl of the force must be zero. One can determine whether a force is conservative by checking its curl.

Because the force is path independent, by choosing $d\textbf{r}$ as the path to integrate, one can define the integral by position. $$V(\textbf{r}) = -\int^r_O\textbf{F}(\textbf{r}) \cdot d\textbf{r}$$ where $O$ is the reference point. This is called the potential energy.

The potential difference between point $a$ and point $b$ is $$V(b) - V(a) = \int^b_a \textbf{F}(\textbf{r}) \cdot d\textbf{r}$$ When $\textbf{F}(\textbf{r})$ is the only force applying on an object $$\begin{align} \int^{b}_{a} F(\textbf{r}) \cdot d\textbf{r} &= \frac{1}{2}m(v_f)^2 - \frac{1}{2}m(v_i)^2 \\ V(a) + \frac{1}{2}mv_i^2 &= V(b) + \frac{1}{2}mv_f^2 \end{align}$$ The initial potential energy plus the kinetic energy is equal to final potential energy plus the kinetic energy. This is known as the conservation of energy.