Chapter 5 Class Notes This week, we return to forces, and consider forces pointing in different directions. Previously, in Chapter 3, the forces were parallel, but in this chapter the forces can be pointing at any angle. Additionally, a new force called friction is introduced. Finally, we will talk about circular dynamics and centripetal acceleration. But lets start with Friction. Friction The force of friction, f, depends only on the normal force, N, and is equal in magnitude, (but not in direction) to µ N, where µ is called the coefficient of friction. The direction of the frictional force is always to oppose the motion causing it. Consider the following example. There is a pulling force, P, which generates a frictional force, f, in the opposite direction. There are two types of frictional forces; static and kinetic. You have common experience with these two types of frictional forces although perhaps did not realize it. Have you ever tried to slide a heavy cardboard box along a smooth floor? It takes a lot of pushing force to get the box moving, but once it s moving, it s a lot easier to push. That s because the static frictional force is greater than the kinetic frictional force. 1
If you plot a graph of the frictional force versus the applied force it would look something like this, with the frictional force actually growing with the applied force until it reaches some maximum value, called f s (max), after which the box starts moving and the frictional force immediately decreases to a smaller value, called f k. The s and k stand for static and kinetic respectively. As you probably already realize, when the pushing (or pulling) force P exceeds the static frictional force, f s, the box starts accelerating according to Newton s second law. But before we can express this situation mathematically, we need two things. You guessed it - a coordinate system and a sign convention! 2
It s always best to align one axis of the coordinate system along the direction of the acceleration vector, a. Writing down Newton s second law, by resolving, or equivalently, summing the forces in the x- direction we have P - f = m a x Now, if the box is already moving, for the frictional force, we use f = µ k N where µ k is the coefficient of kinetic friction. The coefficients are usually less than 1, and as the above graph shows, µ k < µ s. Next, we need to find N and we do that by resolving the forces in the y-direction N - W = m (0) a y = 0 since the box is not accelerating in the vertical direction. Solving for N, we obtain N = W 3
and W = mg of course Substituting for N and f in the first two equations above, and solving for a x, we obtain a x = P - µ k g m Now, of course, there will be a situation, when we first start pushing the box, when P is insufficient to overcome the static friction, then we write, P < f s and then, at the instant just prior to the box moving, we have P = f s and then as described above, when the box is moving. Blocks on Inclined Planes Usually, to make things more interesting, and a bit more realistic, the blocks are situated on inclined planes. The picture below depicts a block sliding down an inclined plane. You know this because the friction always opposes the motion. 4
We are trying to find the acceleration down the slope, but the forces are acting at different angles. We will need to resolve these forces, but in which direction? This is where a coordinate system becomes very important. Choose a Cartesian coordinate system where one of the axes is aligned in the same direction as the acceleration vector, like this; The weight vector, W, is the only vector that is not aligned along one of the axes. So, a little bit of geometry is required to find the x and y components of the weight. If you study the next diagram carefully, you can see that the x component of the weight, W x, is W x = - W cos ( 90 - θ ) which is equivalent to W x = - W sin θ and the y component of the weight, W y, is W y = - W cos θ 5
Now, we can resolve, or equivalently, sum the forces and write down Newton s second law for the x direction; x-direction f - W sin θ = m a x Recalling that f is the frictional force, we can write µ k N - W sin θ = m a x We get the normal force, N, by resolving the forces in the y - direction; y-direction N - W cos θ = m (0) a y = 0 since the block is not accelerating in the y-direction. Thus, N = W cos θ 6
which after substituting into the previous equation, we obtain µ k W cos θ - W sin θ = m a x if we write W = mg, then we find that the masses cancel µ k mg cos θ - mg sin θ = m a x to yield a x = g (µ k cos θ - sin θ ) We find that the acceleration is independent of the mass, as expected. Lets substitute some representative values. Let θ = 22 o and µ k = 0.1, a slippery slope indeed! What we find is a x = - 2.76 m/s 2 where the minus sign tells us that the block is accelerating down the slope. The Normal Force When objects are resting on horizontal surfaces, the normal force produced by the supporting surface just balances the weight. Of course, if you now press down on the object, the supporting surface must summon a larger normal force to balance the weight plus the additional force pressing down on it. When objects are resting on slopes, as we saw in the previous example, it is only the component of the weight normal to the supporting surface that has to be supported (N = W cos θ ). As the slope of the supporting surface is increased, cos θ approaches zero, and the normal force gets smaller and smaller until the angle of the slope is 90 o then there is no normal force at all as the surface is not supporting anything which is why the block falls to the floor at this point! If we combine the ideas outlined in the previous two paragraphs you begin to realize that the normal force is not always equal to the weight. The normal force can be less than the weight if the object is 7
on a slope, but it can also be larger than the weight if there is some other force pushing down on it. Lets consider another example, this time a block on an inclined plane, but being pushed up the slope by a pushing force P; Now diagrams become extremely important as there are a lot of vectors to keep track of. Notice that we have identified all the forces. The frictional force is now pointing down the slope, opposing the motion up the slope, as indicated by the acceleration vector a. We have a weight force W, and the reaction force, N, provided by the supporting surface. Plus, the new pushing force, P, which, if it is greater than the frictional force, will push the block up the slope. There is also a coordinate system aligned so that one of the axes is pointing up the slope, in the same direction as the acceleration vector, because we wish to find the acceleration up the slope. Lastly, we need a sign convention to tell us which signs to assign to the vectors. The next step is to resolve the forces in the x and y direction. x-direction P cos θ - f W sin θ = m a x 8
y-direction P sin θ + W cos θ = N Note that the force P must be resolved parallel and perpendicular to the slope, and, that the latter component increases the apparent weight of the block, N. Substituting f = µ k N, and N we obtain; P cos θ - µ k (P sin θ + W cos θ ) W sin θ = m a x after some factoring and re-arranging we get; a x = P (cos θ - µ k sin θ ) - g (µ k cos θ + sin θ ) m Let θ = 22 o and µ k = 0.1, as before and P = 10N and m = 1kg. What we find is a x = 4.3 m/s 2 up the slope. Note that it takes quite a large force, 10N, to overcome the weight and the increased frictional force. Centripetal Acceleration All objects moving in circular paths do so because they are continually being accelerated towards the center of the circle. To accelerate a mass that is on a circular trajectory requires a force called the centripetal force, not to be confused with the centrifugal force, which is a fictitious force, invoked only in the rotating frame of reference. To an outside observer that is not rotating, there is only one force, the centripetal force, and it is directed towards the center of the circle. Lets take a few moments to think about why there would be an acceleration at all. It has to do with the change in direction of the velocity vector as an object moves on a curved path. Consider an extreme example of a bicyclist making a right angled turn; 9
The acceleration vector, a, has a direction equal to V f - V o (from the equation a = [ V f - V o) / t ] ) which, if you look carefully at the vector diagrams above, points directly towards the center of the turn. Note that the velocity does not change in size, only in direction. This is a new concept, that changes in the direction of the velocity vector can cause an acceleration. Nothing is speeding up or slowing down here. The acceleration is entirely the consequence of a change in the direction of the velocity vector. You have common experience with this phenomenon. When you are riding a bicycle and you want to make a turn, you lean towards the inside of the turn, in so doing, you increase the frictional force which provides the necessary centripetal force to help you turn; 10
In fact, the centripetal force can be generated in a wide variety of different ways. For a satellite in orbit around a planet, it is the force of gravity. For a plane in flight, it is the horizontal component of the lift. For a car on an inclined race track, like Nascar, for example, it is the horizontal component of the normal force plus friction, and so on. The central idea is to produce some component of force that is always directed towards the center of the turn. For an object of mass m, the centripetal force will sustain a turn with a radius r and a velocity v according to the equation centripetal force = mv 2 r You put whatever is providing the force, be it friction, gravity, the normal force, etc, on the left hand side of the equals sign. Let s do an example involving an airplane; In this case, it is the horizontal component of the lift, L sin θ, that provides the centripetal force, so we write L sin θ = mv 2 r 11
Lets imagine that the plane is in level flight, which means that the plane is not accelerating upwards. So we can equate the vertical component of the lift to the planes weight like this; L cos θ = mg Dividing these two equations, we obtain Tan θ = v 2 which is an interesting result as it relates the angle of bank θ, to the planes velocity, v and the radius of turn r. A jet moving at v = 100 m/s and banking at 30 o requires an enormous radius of turn equal to about 1.8 km! rg Artificial Gravity The centripetal force is also an interesting way of providing artificial gravity in space. For if an astronaut is standing on the inside edge of a rotating wall, the wall will support the astronaut by providing a normal force, N; 12
Since it is the normal force that causes the astronaut to move in a circular path, the astronaut apparently has some weight, equal to N where N = mv 2 The phenomenon exists only so long as the astronaut is in contact with the rotating wall. If the astronaut jumps off the rotating wall, he will be weightless again, so it would be necessary to have velcro carpets and slippers so that astronauts would not accidently float away! There is a big problem with the idea of artificial gravity, however. The problem is that to generate a normal force that is comparable in magnitude to earth s gravity would require a very large structure indeed, as I ll explain in more detail in class. r 13