http://youtu.be/0zQDEabagO4

http://youtu.be/pTgUo8YZDPA

http://youtu.be/uO9NTAxGxSg

http://youtu.be/ZYzZ7un03uA

Imagine a force of 70 Newtons acting on a large smooth block of ice resting on a smoothly frozen lake or skating rink.

• If this is the only force being applied in the horizontal direction, then the block will certainly accelerate in the direction of this force.
• If the force is exerted at an angle of 30 degrees relative to the positive x axis, then the acceleration of the block will be in this direction and if the block is originally at rest, it will move in this direction as it accelerates at a rate of a = 70 N / (block mass).
• If the block is already moving in some other direction, its direction of motion will become closer to the direction of the 70 Newton force and its speed will change accordingly.

This force is completely equivalent to the application of the two component forces Fx = 70 N cos (30 deg) = 60 N (approx.) in the x direction and Fy = 70 N sin (30 deg) = 35 N in the y direction.

• This means that is some guy is applying the 70 N force, he can be fired and replaced by two other guys, one pulling with a force of Fx = 60 N in the x direction and the other with a force of Fy = 35 N in the y direction. The result will be identical.
• The original force is completely replaced by its components, which have exactly the same effect as the original force.

If a second force having magnitude 40 N and angle of 115 degrees from the positive x axis, is also applied, its effect will be equivalent to that of its components Fx = -17 N (approximately) and Fy = 36 N, as shown below.

The result of applying the two original forces is therefore the same as that of a 43 Newton force in the x direction and a 71 Newton force in the y direction.

http://youtu.be/NxK_Er16eOM

We can of course replace these two forces by a single force.

• This single force will be represented by a vector from the common point of application of the two component forces to the opposite corner of the rectangle having the two components as sides.
• This rectangle is sketched in the figure below.

• The resultant vector F divides the rectangle into two right triangles.
• The lower of these triangles can be used to show the angle of the resultant force with the x axis.

• The 43 Newton x component is in the direction of the x axis and we can see from the right triangle that the tangent of the desired angle must be tan(`theta) = opp / adj = 71 N / 43 N, so that the angle must be `theta = tan^-1 (71 N / 43 N).
• This inverse tangent is easily found using a calculator. Being careful to be sure that we are in the degree mode, we obtain `theta = 59 deg (approx.).

We can represent the original forces and the resultant force as in the figure below.

• It turns out that the 83 Newton force at the 59 degree angle represents the vector needed to span the diagonal of the parallelogram defined by the two original force vectors (the two remaining sides of the parallelogram are shown as dotted lines).

Vectors can also be applied to velocities, momenta, and a variety of other quantities other than force.

• For example, a projectile has both x and y velocities, as we have seen.
• At any instant the magnitude and angle of its velocity can be found by determining the resultant of its x and y velocities.

Consider the problem of finding the magnitude and angle of the velocity of a projectile 1 second after it is projected in the horizontal direction with a velocity of 5 m/s.

We can easily determine the x and y components of its velocity at this instant.

• The x component is 5 m/s, while the y component is -9.8 m/s (since it was projected in the horizontal direction its initial y velocity was 0; since it has been accelerating for one second at an acceleration of -9..8 m/s/s, its velocity has changed by -9.8 m/s).

• The magnitude of the velocity is easily found from the Pythagorean Theorem to be approximately 11 m/s, as shown in the figure below.
• The angle is similarly found to be -63 degrees.
• We usually wish to express our angles as positive angles, which we do in this case by adding 360 degrees (a full circle, which does not change the location of the angle) to our -63 degrees to obtain the positive angle `theta = 297 degrees.

We thus conclude that after 1 second the velocity of the object will be 11 m/s at 297 deg.

http://youtu.be/lePmqRQ2Rs0

Vectors are also used to represent momentum, which always has a direction (the direction of the momentum of any moving object is the same as that of its velocity).

Consider the situation shown below, where a 5 kg object is moving in the x direction at 10 meters per second, toward a collision with a 12 kg object moving at seven meters per second and angle of 212 degrees has measured from the positive x axis.

• We will assume an inelastic collision and determine the magnitude in direction of the velocity of the combined objects after collision.

The figure below shows the two momenta in purple, and the components of the two momenta in red.

At the top of the figure below we show only the components of the two momenta.

• It should be clear that the two x momenta add up to - 21 kg m/s, while the net y component will just be the -44 kg m/s momentum of the second object.

• Since the inverse tangent function always gives angles between -90 degrees and 90 degrees, corresponding to positive x coordinates, when the x coordinate of the vector is negative we will always add 180 degrees to the inverse tangent.

• Applying this rule to the present situation gives us in angle of 64.5 deg + 180 deg = 224.5 deg.
• Note that adding 180 degrees to the angle of a vector always gives us an equal and opposite vector.

http://youtu.be/AQvvFwi1syc

http://youtu.be/QBEGmQlWpp0

The magnitude of the total momentum is easily calculated by the Pythagorean Theorem.

• As shown below we obtain a magnitude of approximately 49 kg meters/second.
• The total momentum, 49 kg meters/second at an angle of 244.5 deg, will be the same immediately before as immediately after collision.

If we wish to find the common velocity of the objects after collision, we need only divide their total momentum by their total mass.

• Since the total mass of the two objects is 12 kg, the magnitude of their velocity will be (49 kg m / sec) / (12 kg) = 4.08 m/s (approximately).
• The direction of their velocity will be identical to the direction of their momentum.
• The two objects therefore end up moving at 4.08 m/s at an angle of 244.5 degrees with respect to the positive x axis.

We now consider what happens when a coffee filter is dropped. As we saw in class when the filter is dropped it very quickly reaches a constant (terminal) velocity and drifts down to the floor at that velocity.

• Our common experience is that when we hold our hand out the window of a car, we experience a greater air resistance at a greater speed.
• Similarly as the coffee filter speeds up, it experiences greater and greater air resistance. This is depicted in the picture below.
• In each 'frame' the vector W, depicting the weight of the coffee cup, is the same.

• This is because the mutual attraction between the coffee cup and the Earth does not change appreciably as the coffee cup falls.

• From one 'frame' to the next we also see that the velocity of the coffee cup increases.
• The vector R depicts the force due to air resistance, which acts upward and has the effect of opposing the gravitational force W.

The result is that as velocity increases, the net force acting downward on the coffee filter decreases and the acceleration of the filter therefore decreases.

• This is shown in the two 'frames' below, where the greater air resistance R in the second frame results in a smaller net resultant when the two forces (weight and air resistance) are combined.
• The acceleration in each case is easily found from Newton's Second Law to be Fnet / m, where Fnet is the (decreasing) net force.
• As long as the magnitude of the weight is greater than that of the air resistance, the acceleration will be in the direction of the weight, or downward.
• This process will never give us an air resistance whose magnitude is greater than the weight, since at the instant the magnitude of the air resistance is equal to that of the weight acceleration will be zero and no further change in velocity will occur.
• If all the forces change very very smoothly this instant will, theoretically, never occur--we will approach terminal velocity but never reach it.  However, after a time things like pressure variations and thermal fluctuations make the difference between actual and terminal velocity meaningless.

We also note that mathematically, we will never actually reach terminal velocity. Terminal velocity will be approached as an asymptote. In physical terms weekend and do in this case approached terminal velocity so close that no possible physical measurement will show the difference.

http://youtu.be/J41h-FE6gQE

The top picture below shows several 'frames' in which we see the increasing air resistance R, constant gravitational force W and increasing air resistance R.

• The air resistance increases slowly at first then more and more rapidly as the filter speeds up, then less and less rapidly as the speed more and more gradually approaches terminal velocity.
• This gives rise to the acceleration vs. time graph shown below.
• As the acceleration approaches zero, change in velocity approaches zero and velocity approaches terminal velocity.

http://youtu.be/bsVwXH5W92g

The figure below depicts an Atwood's machine, consisting of two weights suspended over a light low-friction pulley by a light string.

• By 'light' we mean light enough that the masses of the pulley and the string can be neglected.

If one mass is greater than the other, experience tells us that the system will tend to accelerate in the direction of the greater mass.

• We have a problem when we attempt to decide what the positive direction should be for the system.
• The problem is that we cannot choose 'up' or 'down' as the positive direction, because when one object is going up the other is going down.

We solve this problem by considering the system in this situation to consist of the weights and the string.

• We will say that the system is moving in the positive direction if the pulley is thereby caused to rotate in the counterclockwise direction.
• This positive direction is indicated in the figure below. In this figure, the mass m2 appears larger than the mass m1, so its weight W2 = m2 g will be greater than the weight W1 = m1 g and the system will tend to rotate in the counterclockwise, or positive, direction.

We will next consider what forces act on the system.

If there is no friction in the pulley the total force on the system will just be the sum of the weight forces.

• When assigning directions to these forces, we don't use 'up' and 'down' as directions but 'clockwise' (negative) or 'counterclockwise' (positive).
• We see then that W2 is positive while W1 is negative, and we see that the net force must be W2 - W1.
• If there is a frictional force f due to friction in the pulley, it will act in the direction that opposes the motion.
• In this case since W2 is the greater, motion will be in the direction of W2 so that the frictional force will be in the negative direction.
• So in this case we say that the net force is -W1 - f + W2.

To determine the acceleration of the system, we divide the net force on the system by the mass of the system.

• As we have seen the net force is -W1 - f + W2.
• The mass is just the sum of the masses m1 and m2.
• The acceleration of the system will therefore be -W1 - f + W2.

http://youtu.be/WPL8JJNYRSc