We use simple machines to help us do work. We use them to amplify force or amplify displacement, but not both at the same time. (Doing both at the same time would violate conservation of energy.) Examples of simple machines are
We will analyze them using conservation of energy. We put energy in and assume the work we get out will be equal to the energy we put in. This assumes that the machine is 100% efficient; not a good assumption, but it will help us see how machines can amplify forces or displacements.
The simplest machine is an incline plane. Say we want to move a heavy drum of oil, weight = 1000N, from the ground to a platform 1m high. To just lift it up I would have to exert a force of 1000N (1000N = 220 pounds). That is hard to do. I have another alternative. I can make a ramp that goes from the ground up to the platform and roll the barrel up the ramp. If the ramp is 4m long, I'll find that I can push the barrel up by applying a 250N force to the barrel. Why?
One can analyze this by assuming that I have to do the same amount of work to get the barrel up onto the platform either by lifting it up directly or pushing it up the ramp. To lift it up directly, I apply a 1000N force as it rises 1m, so W = 1000N x 1m = 1000J. (Another way of thinking about it is that I must increase its gravitational PE by 1000J, so I have to do 1000J of work.) If I do 1000J of work when I roll it up the incline, the force is given by Fx4m = 1000J, of F = 250N. This is much easier.
Looking at it another way, the ramp's normal force is partly canceling some of the weight of the object, so I don't have to supply 1000N of force to lift it. But the normal force is perpendicular to the direction of motion, so it does no work. I have to supply all the work. In effect, the incline amplifies the effect of the force I apply. Plows, wedges and screws are basically incline planes.
Levers involve the idea of rotational motion. When you have something like a wheel or bar mounted on a shaft or axel, it can rotate around that shaft or axel. The shaft or axel is called the axis of rotation. The rotational motion is described in terms of how many complete rotations it makes in one second (or minute for rpm), which is the rotations per second. If left to itself, the rotational motion of an object will not change (due to Rotational Inertia).
|What changes rotational motion? It requires a force, but it depends on how that force is applied. The force must produce a torque about the axis of rotation. The bar at the right is free to rotate about the axis. The force at the right produces a torque about that axis, Torque = Fd. This torque tends to produce more rotation in the counterclockwise direction as shown. (I usually call this a positive, +, torque.) If it tended to increase the rotation motion in the clockwise direction I would call it a negative, -, torque. Thus for rotational motion it is torques that produce a change in the rotational motion. If the total torque = 0, there is no change. If there was an equal||
torque in the other direction (torque = -Fd), then the torques would cancel and the bar would not change its rotational motion.
|The figure shows a lever, e.g. a board, supported at the center by a fulcrum. If one person pushes down on the left side with a force F1 a distance d1 from the fulcrum, and another person pushes down on the right side with a force F2 at a distance d2 from the fulcrum, they will be in balance if F1d1 = F2d2 because they would have opposite signs and cancel, i.e. the total torque = F2d2 - F1d1 =0. In that case we say it is in rotational equilibrium.|
If the net torque is 0, one has
F1d1 = F2d2
This relationship can be rearranged to give
If d2 > d1, then F1 > F2. There are two ways of looking at it. If I want to amplify force, a small force F2 can produce a larger force, F1 , but the small force will have a larger displacement than the large force in any rotational motion of the bar. This is the principle behind pry bars, wrenches, pliers, scissors, fingernail clippers, etc. Most of the time we want to amplify force, so the apply a small force a long distance from the fulcrum to produce a large force a small distance from the fulcrum. If I want to amplify displacement, it takes a larger force F1 acting through a smaller displacement, d1, to produce a smaller force F2 acting thru a larger displacement d2. However, many of the joints in our body do just the reverse. They sacrifice the magnitude of the force to get a larger range of motion. Ankles and elbows are examples.
One can see that this must be true, because if the force F1 pushes the lever down slowly a distance x1, then the lever must push force F2 up a distance x2.; so the work done by F1 is W1 = F1 x1 , and that done by F2 is W2 = -F2 x2. If there is no change in the KE of the system, then net work must be 0 or W1 = - W2. Because x1 is an arc of a circle of radius d1 and x2 is an arc of a circle of radius d2, x1 = x2 (d1/d2) so one has F1d1 = F2d2 .
|A winch uses the principle of torque described above to amplify force. The geometry is shown at the right. A lever or crank is attached to a cylinder that can rotate about a shaft. The length of the lever is R. The radius of the cylinder is r. As I turn the lever, it rolls up a cable onto the cylinder. You might use such an arrangement to pull a small boat out of the water and onto a trailer. If I turn the cylinder with at a constant rate, the torques about the axis through the center of the cylinder add to zero. As shown, I am exerting a counterclockwise torque = RFme. The tension in the cable exerts a clockwise torque rFcable . If these cancell, RFme - rFcable =0. or||
If R = 3r, then Fcable = 3Fme, effectively multiplying the force I exert by a factor of 3. This is also a basic lever system. The tension in the cable is the force exerted on the boat to pull it out of the water. The work done on the boat by the cable is equal to the work I do turning the lever or crank. For one rotation of the cylinder I do a work = Fme2pR, since the end of the crank moves through a circle of circumference 2pR. The tension in the cable does work 2prFcable. If I substitute the expression above for Fcable, I find that the two works are equal. (If there is some friction on the shaft through the cylinder, I have to do a little extra work to overcome that.)
|Other systems that uses the principle of torques are a belt and pulley system, gears and chain and sprocket systems. (Chain and sprocket systems are found on bikes.) The figure at the right shows a belt and pulley system. Each pulley is mounted on an axel or shaft which can rotate. I crank on one of the pulleys, either with a lever like the one on the winch above, or by turning on of the shafts. The other pulley's axel is connected to something I want to rotate, like a tire or drill bit. I typically use this system to amplify rotation rate (or distance) or to amplify torque (or force)|
The nature of the belt is that the belt applies the same magnitude of force to the rim of each pulley. The magnitude of the force that produces a torque on the rim of pulley 1 is the same as the magnitude of the force on the rim of pulley 2. Therefore the torques are not equal unless the diameters are equal. The diameter of pulley 1 is D1 and it's D2 for pulley 2. The torque on pulley 1 is 1/2(FD1) and on 2 it's 1/2(FD2) where I've used F for both forces since they are the same. If D2 > D1, say twice as big, the torque on 2 is twice that on 1. If I want to amplify the torque, I crank pulley 1 and use it to drive pulley 2. If I want to amplify distance or rotation rate, I crank pulley 2 and use it to drive pulley 1. For a bicycle, you usually crank on the larger pulley or sprocket (2) and the chain connects to the smaller sprocket (1). In this case the tire, which is connected to sprocket 1, turns as the same rate as sprocket 1. If the ratio of diameters is 2 to 1, then the tire has twice the rotation rate as the pedal I'm cranking. However, if the radius of the tire is large, I have to exert more force on the pedal than the tire exerts on the road to propel the bike.
For a car, you usually do the opposite. The engine produces a large rotation rate on the shaft which is coupled through a gear system to the tires which rotate at a slower rate, especially in the lower gears. This will increase the force the tires can exert on the road to propel the car.
When I use machines like this to help me do work, the work I get out cannot exceed the work I put in. In an ideal system, the work I get out is the same as the work I put in. However, If I slide a box up an incline plane, I'll find that I have to put in more work than should be required to merely lift the barrel up to the platform. Somehow, some energy appears to have been "lost". What happens is not that the energy is lost, it just goes somewhere else. When I slide the box up, friction opposes that motion and does negative work on the box and I have to put in extra work to supply the energy that the friction has taken. The energy the friction has taken does not disappear, but winds up increasing the temperature of the box and the incline, among other things. The concept of efficiency reflects that some of the energy I put in did not go where I intended. The efficiency of a machine or process is defined as the useful (intended) work I get out divided by the work or energy I put in.
The efficiency is often expressed as a percentage instead of a fraction. If I have to do 2000J of work to push a 600N box up an incline to a height of 2m, its potential energy has increased by 1200J. Theoretically, I should have only had to do 1200J of work. Therefore my efficiency is
e = 1200J/2000J = 0.6 = 60%
In this case 40% of the work I did was "wasted", i.e. did not go where I wanted it to go.
Power is the rate at which we do work. It is defined as the Work we do in a time t divided by that time t. or
If I do 100J of work in 10s, my average power = 100J/10s = 10 (J/s). The units of J/s are called Watts, or W for short. A 100W light bulb uses 100J of energy every second. A 1400W hair drier uses 1400J of energy every second. If the drier is on for 1 minute or 60s, it uses
Energy used = 60s (1400J/s) = 84,000J
1 horsepower = 746W. A 1/2 horsepower engine uses 373W or 373J every second.