Newton's laws actually find their most practical and immediate applications in the design of simple machines. One which we've already seen is the inclined plane, using the inclined plane device shown. As before with the previous blog part 2, viz.

**http://brane-space.blogspot.com/2011/05/introduction-to-basic-physics-mechanics_03.html**

We note that in the absence of friction, given a trolley of weight mg resting on a plane inclined at angle Θ with the horizontal, the force which just tends to start it rolling down the plane is F = mg sin Θ, while a force F' = F would be needed to pull it up at a uniform velocity. From the geometry sin Θ = h/L so that:

F/mg = sin Θ = h/L or F = (mg x h)/ L

The theoretical M.A. or "mechanical advantage" is defined: M.A. = mg/F = W/F, while the actual mechanical advantage is defined:: M.A.(ac) = L/h. Meanwhile work is defined as the force x distance moved, so the work done in pulling the trolley up the plane is: w = F' x L, and this is also equal to the gain in potential energy, W x h. The efficiency of any given machine is defined:

Efficiency = work output/ work input or Wh / F'L = mgh/ F'L.

Figure 2 illustrates the condition in which friction isn't omitted and we let W be the weight of the object resting on the inclined plane. This object, say trolley, or block, tends to slide down the plane under the action of force F_g = mg sin Θ. But, if now a force F' is applied to pull the object up then we need the magnitude of F' to be that to overcome the F_g component and also the force of friction, F_f, between the object and the plane. Thus: F' = F_g + F_f

We know: F_g = mg sin Θ and F_f = uN = umg cos Θ, so that:

F' = mg sin Θ + umg cos Θ, or F' = mg(sin Θ + u cos Θ).

The principle of work states that in any machine the useful work done is equal to the energy input minus any losses, say doe to heat-friction.

Another simple machine we've already seen is a basic pulley system, set up as an Atwood machine. In another version, we see that depicted in Fig. 3(a) alongside a modification in Fig. 3(b). The pulley shown in 3(a) is a single movable pulley, in contrast to the Atwood machine which is a single fixed pulley. In operating such a pulley, say to lift a weight w, the force applied (F) must move twice as far as the weight w = mg. The mechanical advantage (assuming no friction) is s/d which is the displacement of the applied force, how much it moves, divided by the distance the weight is moved. Since for Fig. 3(a) if the weight w is moved 1' then the force F is moved 2'. Thus, s = 2' and d = 1' so: w/F = s/d or F = wd/s = ½ w.

In Figure 4 a variety of different pulley systems is depicted including a variant called the "wheel and axle" (A) and one of grouped pulleys (C) and multiplied strings(B). The wheel and axle is of particular interest in that it makes use of two different radii, an inner small one, r and a larger outer one R. If the depicted wheel (Fig. 4(A)) moves through one complete revolution, the distance the force will move is just d = 2πr. Meanwhile, the distance the force moves will be s = 2πR. If we take the mechanical advantage:

M.A. = s/d = (2πR)/ (2πr) = R/r, then:

mg/ F = w/F = R/r and so: F = (r/R)w which is the law of the wheel and axle.

Lastly, we come to perhaps the most famous machine of all, the lever. Archimedes the ancient Greek physicist is quoted as saying: "Give me a lever long enough and I will move the Earth!". A basic depiction of a workable lever is shown in Figure 5. Basically a load L is placed at one end which we wish to lift by applying a force F. Let the load be a distance a from the pivot, and the applied force acts at a distance, b. Then:

Force x distance from axis = load (mg) x distance from axis or:

F x b = L x a or F = (a/b) L = (a/b) mg.

This is called the "law of the lever". It helps to illustrate using a simple problem how it works:

Sample problem:

A 50 kg concrete block has to be moved from the ground to a wheelbarrow and a workman is provided with a board 5 m in length. If the workman pivots the block at 3.5 m from one end and lifts from the other (assume g = 10 m/s^2) What applied force is needed to lift the block? What is the work done?

We have the effort distance, a = 5.0 m - 3.5 m = 1.5 m, and the load is:

w = mg = 50 Kg (10 m/s^2)= 500 N, with load distance a = 3.5m. Then, since:

F x b = L x a, we have:

F = (a/b) w and F = (1.5 m/ 3.5m) 500 N = 214 N.

The work done is Fs = (mg)d but (d/s) = (a/b) so Fs = (a/b) mg x 1.5 m = 321 J.

Problems:

1) Show, with reasons, that the pulley system shown in Fig. 3(b) has a mechanical advantage of 2. If the weight w is lifted 1m, how much distance must the force F cover?

2) An inclined plane has an angle of Θ = π/6. The coefficient of sliding friction u = 0.3.How much force, parallel to the incline, is needed to push a 100 N object up the incline at constant speed?

3) The work done in sliding a 50 kg mass up an incline 1 m high and 5m long is 640 Joules. What is the frictional force in N?

4) In the wheel and axle device (Fig. 4 (A)) the radius r = 1 cm and R = 23 cm. Find the mechanical advantage and the applied force needed to lift a load of 80 N.

5) In the grouped pulley system depicted in Fig. 4 (C) the force applied F will move 6 times as far as the load w. If the load has a mass of 40 kg, and assuming g = 9.80 m/s^2, find the applied force. Thence or otherwise, obtain the mechanical advantage of the system. If the force F is applied through 10 m what is the work done?

6)A man raises a uniform plank 12' long and of weight 40 lbs. until it is horizontal. His left hand is on one end of the plank and his right hand is 3' from the same end. Assuming both hands exert vertical forces, find the forces exerted by each hand to support the plank.

7) In the sample lever problem it is feasible to reduce the work done to only 125 J by re-arranging the lever distances (effort and load distance). Using a sketch show how could this could be done and give the new applied force in this scenario.

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