CHAPTER 7

Scalar Product of Vectors

Given two arbitrary vectors $\vec{A}$ and $\vec{B}$ we define their scalar (or dot) product as

$\vec{A}\cdot\vec{B} = AB\cos\theta$

where $\theta$ is the angle beteen the two vectors. Note that this can be interpreted either as $A(B\cos\theta)$ or as $(A\cos\theta )B$: i.e. either as the magnitude of A times the magnitude of the projection of B onto A or as the magnitude of B times the magnitude of the projection of A onto B. From the definition it also follows that

• the result of the scalar product of two vectors is a scalar, i.e. a simple number (which of course can be positive or negative)
• in fact, the result will be negative when $90^{0} < \theta < 270^{0}$
• if $\vec{A},\vec{B}$ are parallel $\vec{A}\cdot\vec{B} = AB$ (or (-AB) if anti-parallel)
• if $\vec{A},\vec{B}$ are orthogonal $\vec{A}\cdot\vec{B} = 0$
• $\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}$
• $\vec{A}\cdot(\vec{B} + \vec{C}) = \vec{A}\cdot\vec{B} + \vec{A}\cdot\vec{C}$
• $\vec{A}\cdot\vec{B} = A_{x}B_{x} + A_{y}B_{y}$

From the last result, it also follows that

$\cos\theta = \vec{A}\cdot\vec{B} /(AB) = (A_{x}B_{x} + A_{y}B_{y})/AB = (A_{x}B_{x} + A_{y}B_{y})/ (\sqrt{(A_{x}^{2}+A_{y}^{2})(B_{x}^{2}+B_{y}^{2})}$

WORK

In Physics, work takes place when a force acts on a moving object. Specifically, if the force $\vec{F}$ acts on an object during a displacement $\vec{d}$, the work done by the force is

$W = \vec{F}\cdot\vec{d}$

from this, we see immediately that

• no motion = no work
• force perpendicualr to displacement = no work
• force and displacement at more than 90⁰ = negative work

NB : until now, the sign of physical quantities (velocity, acceleration, force, etc.) was determied by our choice of the frame of reference. But note that when dealing with work, the choice of sign is not arbitrary any more, but it is uniquely determined by the relative orientation of force and displacement

In any particular situation, there can be more than one force acting on an object, and we can estimate the work done by each force. We can also have the case of zero net force : in this case, our object moves with constant velocity. The net work is zero, even though each of the individual forces does do some work.

Example : lifting a weight against gravity. How much work do you do? And how much work does gravity do?

and what if, rather than lifting it, I push the same weight up an inclined slope?

Calculating the total work is easy if, during the displacement, the force is constant (both in magnitude and direction). How can one handle the case of a variable force? Limiting ourselves for simplicity to motion in the x direction, if a force $\vec{F}$ acts during a displacement $\Delta x$, the work done by the force is

$W = F_{x}\Delta x$

(what about the y component of the force?).

But what if the force is not constant ? You should know the trick : you take small intervals $\Delta x$,during which the force is almost constant, compute the work for all the small $\Delta x$'s and sum all the terms. The smaller you make your $\Delta x$ the more accurate the result will be. Therefore, you make them infinitesimally small, with the result

$W = \int_{x1}^{x2} F_{x}(x)~dx$

Equivalently, you can also say that the total work is given by the area subtended by the F_x vs. x plot.

Important applications :

1.

gravity, F = G m₁m₂/r² = k/r². (if r changes by a large enough amount, we cannot assume that gravity is constant).

$W = \int^{r2}_{r1} k/r^{2} = .....$

2.

work involving stretched or compressed springs. The starting point is that when a spring is stretched or compressed by an amount x, it exerts a force F = -kx, k = spring constant, units : N/m. The work done by a spring when stretched or compressed by a total amount x is then

$W = \int^{x}_{0} -kx~dx = -\frac{1}{2}kx^{2}$

What is the work done by the compressiong (or stretching) agent?

Question : consider the motion of an object under the sole effect of a spring restoring force. What will its equation of motion be? (i.e. what will Newton's 2nd law tell us?)

Units for work :

Work = Force x displacement = N x m = joule = J = [ML²T^-2]

ENERGY

The concept of energy is constantly present in everyday's life. We are aware that energy comes in many forms: mechanical, thermal, electrical, chemical, nuclear, etc. A more in depth study would show that deep down there are only two types of energy :

1.

kinetic, i.e. energy associated with motion

2.

potential, i.e. energy associated with one the fundamental force fields of Nature (gravitational, electro-magnetic, strong nuclear and weak nuclear).

For the time being, we will just concentrate on mechanical energy, and we will define energy as the capability of doing work. Obviously this defintion is a bit vague, but it is a fact that, in spite of its widespread presence and importance, energy is not too easily defined.

Given that a certain amount of energy is related to a certain amount of work, energy will be measured in the same units as work , i.e. joules (J).

Often used analogy : energy is like wealth, it exists in many forms (cash, stocks, real estate, etc.) and it continuously transforms from one form to another. In this context, work can be thought of as a financial transactions : as the transaction transfers wealth from one form to another (or from one person to another), so work has the effect of transforming energy from one form to another or from one body to another.

Coming back to mechanical energy, let's suppose that a net force $\vec{F}$ acts on an object of mass m, causing it to move from $\vec{r_1}$ to $\vec{r_2}$. As usual, it is easier to analyze the situation separately in the x and y (and z) directions. In the x projection, the total work will be :

$W_{x} = \int^{x2}_{x1} F_{x}dx = \int^{x2}_{x1} ma_{x} dx$

a_x = dv_x/dt = (dv_x/dx) (dx/dt) = v_x dv_x/dx

$W_{x} = \int^{x2}_{x1} mv_{x}dv_{x}/dx~dx = m\int^{v_{x2}}_{v_{x1}} v_{x}~dv_{x} = \frac{1}{2}mv_{x2}^{2}- \frac{1}{2}mv_{x1}^{2}$

similarly, one would have found

$W_{y} = \frac{1}{2}mv_{y2}^{2}-\frac{1}{2}mv_{y1}^{2}$

and, for the total work,

$W = W_{x}+W_{y} = \frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2} = \Delta K$

where $K = \frac{1}{2}mv^{2}$ = kinetic energy. We then see that

when a net force acts on a body, the amount of work done by the force is equal to the change in the body's kinetic energy

Example : spring loaded gun

What if the net force is zero ? The correct guess is that there is no change in kinetic energy (as it should be : no net force = no acceleration = no change in velocity). But if the net zero force is the consequence of several forces canceling each other, it is still true that each force does do some amount of work, therefore there must be some energy transaction involved.

Example : you push a crate with constant velocity along a surface with friction : you do a certain amount of work, and friction does an equal and opposite amount. Is there any energy transaction involved ?

Yes : some of your muscular (=chemical) energy is transformed into thermal energy (friction results into heating of the surfaces).

And what if you lift the crate with constant velocity up to a height h ? What happens to your muscle energy? find the answer in the next episode

POWER

In the common language, force, energy and power are often used interchangeably, but in Physics they are clearly distinct entities. We have seen force and energy, let us now introduce power.

If you reach the top of a building either by leisurely walking or by running up the stairs, the total work you do in either case is the same, but you will not feel the same. The difference is in the amount of time over which the work was done. We can then define average and instantaneous power

$P_{av} = W/\Delta t$

$P_{inst} = lim_{\Delta t\rightarrow 0} W/\Delta t = dW/dt$

Given that a certain amount of work results in some sort of energy transfer, it is useful to think of power as the rate at which energy is being transferred.

One also has

$P = dW/dt = \vec{F}\cdot d\vec{s}/dt = \vec{F}\cdot\vec{v}$

Power is measured in watts, 1 W = 1 J/s, or in horsepowers (1 hp = 760 W), while the kilowatt hour is a unit of energy.

Question : how many horsepowers in one manpower ? to find the answer, time yourself while running up a few flight of stairs. You will find that, for a short enough period you can work almost as hard as a horse (but a horse can work like a horse for a full day...)

Section 7.6 : read and enjoy
Section 7.7 : ignore