Mechanical advantage - screw

It's a "simple machine" we are all quite familiar with due to its widespread use in various applications. In fact, we may be familiar with what might be difficult to believe that it is even a "machine".

Well, technically, a machine can be anything that performs mechanical "work", so even has something as trivial as a screw can be one. Screw systems such as Archimedes screw are at the base of numerous mechanical devices, including those used for the conveyance of water, grain or even transporting fish from one container to another. As you can see, the concept of the screw is thus not limited to the sharp small pieces of galvanized metal used in carpentry jobs, or fix-it. This is one of the components core number in the world of mechanics.

Speaking from a point of reference technical screw nothing, but a "inclined plan" wrapped a shank or cylinder. A slipway is also one of the 6 machines simply called thus by scientists of the Renaissance, but the distinction between the screw and the slipway is that the former only in spiral around a cylinder, but also converts a linear rotational force.

Inclined plane

First, let's define a slipway. Imagine a straight line with the 2 points of endpoint at different heights and you have yourself a slipway. In practice, think of a ramp next to the stairs. Ramp jumps to the same height as the stairs, but has probably start from further back.

So much more return ramp goes, the mechanical advantage. The advantage of the mechanics of the incline is the length of the slope divided by the slope - MA slope length/slope height = height.

Screw

If the thread is wrapped a screw is essentially a plan tilted, gradually climbing tree screw. The advantage of the mechanics of the screw is calculated by MA pdm/l = where dm is equal to the average or mean diameter of the screw thread, and l is equal to the head of the screw thread.

* You need to know the actual mechanical advantage of screw system is more than the above equation, the advantage of the mechanics of the driver or similar device transforming.

Conservation of mechanical energy to education

Now, I go to the conservation of mechanical energy. Again, I'll show you how I explain the solutions of the physical problem in terms of fundamental principles. As always, the statement of principles will be the first line of the solution to the problem. The problem that I have chosen to illustrate the method requires the application of the second law of Newton and mechanical energy conservation.

Notation I used of this series described in the preceding articles, specifically "kinematics of teaching", "Teaching the second law of Newton" and "the problems of labour-power."

Problem. A small box of mass m starts rest and sliding down the surface friction of a cylinder of RADIUS r-free show box leaves the surface when the angle between the line of radiial in the box and the vertical axis is th = arccos(2/3).

Analysis. The box is just affected the frictionless cylindrical surface, exercising a normal force outwards n on it. The only other force on the box is its weight MG. Box moves on a circular path, so we apply Newton's second law in radial direction (positive outward). With the help of a free body diagram, we have

... Newton's second law

... Sum (fr) = MAr

...-MGcos (th) + N = M(-V**2), r.

Since the cylindrical surface can only push (he can not draw), box cannot remain on the surface, unless force normal n is greater than zero. As a result, box leaves the surface at the point where N = 0. Since the last equation corresponds to

... cos (th) = (V * 2) / RG.

But this tells us much if we do not know the speed v box, we are going to see what we can learn from the application of the conservation of mechanical energy. We use a framework co-ordinated by inertia with axis y vertical and the origin at the center of the circular cylinder. Assimilate us mechanical energy box at the top of the bottle and the time it leaves the cylinder. The initial position of the box is Yi = R, its muzzle velocity is Vi = 0, its final position is Yu = Rcos (th) and its final speed is given = V, speed when he leaves the surface. Now with the conservation of mechanical energy.

... Mechanical energy conservation

... (IVM * 2) / 2 + MGYi = (MVu * 2) / 2 + MGYu

... (M0 * 2) / 2 + MGR = (MV * 2) / 2 + MGRcos (th);

therefore... 2 * V = 2 GR (1 - cos (th)).

Finally, connect this result in cos equation (th) that we found earlier, we

... cos (th) = 2 GR (1 - cos (th)) /RG = 2 - 2cos (th);

and... Th = arccos(2/3).

Dr. William Moebs is physical retired professor who has taught at both universities: Indiana - Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how stressed the fundamental principles by visiting physical support.

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Conservation of mechanical energy and rolling motion

In this article, I show how it is easy to solve the problems of rotational motion of fundamentals. It is a continuation of the last two articles on rolling motion. The notation used is summarized in section "teaching dynamic rotation. As usual, I describe the method in terms of an example.

Problem. A solid ball of mass m and radius r is rolling through a horizontal speed v surface when it encounters an inclined at an angle th. Distance d along the inclined ball moves to stop and go back down? Assume that the ball moves without dragging?

Analysis. Since the ball moves without dragging, mechanical energy is conserved. We will use a frame of reference which the origin is a distance r above at the bottom of the slope. This is to centre the ball as it starts up the ramp, SW = 0 Yi. When assimilate us the mechanical energy of the ball at the bottom of the slope (where Yi = 0 and Vi = V) and to the point where it stops (Yu = h and Vu = 0), we

Mechanical energy conservation

Mechanical energy = mechanical energy final initials

M(VI**2)/2 Icm(Wi**2)/2 + MGYi = M(Vu**2)/2 Icm(Wu**2)/2 + MGYu

M(V**2)/2 + + MG (0) Icm(W**2)/2 = M(0**2)/2 Icm(0**2)/2 + MGH;

where h is the vertical displacement of the ball stops on the slope at the moment. If d is the distance the ball moves along the slope h = d sin (th). This insertion with W = V/R and Icm = 2 M(R**2)/5 in energy equation, we find, after some simplification, the ball moves the slope distance

d = 7 (V * 2) / (10Gsin (th))

to rotate and position downwards.

This solution of the problem is exceptionally easy. Once again the same message: start all solutions of problem with a fundamental principle. When you do, your ability to solve problems is greatly improved.

Dr. William Moebs is physical retired professor who has taught at both universities: Indiana - Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how stressed the fundamental principles by visiting physical support.