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Simple Machines - The Movable Pulley

This article was first published on 3 Sep 2008.

One of the functions of simple machines is to lighten our workload. A small effort can move a greater load. What is sacrificed is that the distance moved by the effort is greater than the distance moved by the load.

As an example, we will take a look at the movable pulley. So how does a movable pulley help to lighten our work?

Remember that in order to be able to move a load that is greater than the effort, the distance moved by the effort has to be greater than the distance moved by the load.



With reference to the diagram above:-

Pulley A - The distance from the pulley to the ceiling is 2m. This means that the cable attached to the pulley is also 2m.

Pulley B – The distance moved by the effort is 2 m. However, the load moves only 1 m from the original position of the pulley. This is confirmed by …. 2-m cable divided by 2 = 1m, which means that the distance from the pulley to the ceiling is 1m. Hence, the distance from the pulley now, is 1 m from where it was originally.

Since the effort moved twice the distance moved by the load, only ½ the effort is needed to raise the load. Hence if the load weighs 10 kg, an effort of slightly over 5 kg only is needed to move the load.

Simple Machines – The Inclined Plane

This article was originally published in Oct 2008.

As with all machines, if we want the force exerted by the effort to be less than the force exerted by the load, the distance travelled by the effort has to be more than the distance travelled by the load.

Below are 4 illustrations how the inclined plane works.



The more gradual the slope, the greater the distance travelled by the effort compared to the distance travelled by the load. This means that the more gradual the slope, the less effort is needed to move the load.

Examples of applications of the ramp in real life – wheelchair ramps for the disabled, gradual and winding slopes of a road found in mountainous terrain (e.g. roads to Gentling Highlands or Cameron Highlands).



It can be noted that the ‘sharper’ the wedge, the less the effort is needed because for the same distance travelled by the effort, the distance moved by the load is now less.

Examples of applications of the wedge in real life – axe head, blades of knives, metal wedges for prying open flanges of pipes in heavy industries.



Fig 1 - For every 1 turn the screw makes, the screw is driven down by the distance of ‘1 pitch’. Hence, the more gradual the slope of the thread, the smaller the distance it will be between 2 threads, as shown in Fig 2.

Fig 2 - the distance travelled by the load is less than in Fig 1, although in both cases, the effort moves by the same one turn. Since the load moves less in Fig 2, less effort is also needed.

In conclusion, for screws, the more the number of threads there is, OR the smaller the pitch, OR the more gradual the slope of the threads, the less the effort is needed to move the load.



Again, like the screw, the smaller the pitch, the less the ends of the ‘V’ move towards each other, which also means the less the distance the load moves.

Like the screw, for the screw jack, the more the number of threads there is, OR the smaller the pitch, OR the more gradual the slope of the threads, the less effort is needed to move the load.

Take note that for the screw jack, while the mechanism of the inclined plane is used to lift the vehicle off the ground, the handle of the screw jack works on the principle of the wheel and axle. In this case, the handle is the wheel, while the screw of the jack (the part with the threads) is the axle.


Important note to parents and students – The purpose of the above 4 illustrations is to help students understand the mechanics of the inclined plane. Most schools’ practice papers and science textbooks do not use the terms ‘pitch’ or ‘threads’. Those terms are technical terms used in the heavy and light industries.

In trying to make the illustration as simple as possible, I have found that I cannot avoid using those technical terms. The use of the above terms is to assist the student (or parent) to understand the mechanics of screws and screw jacks.

Take note that markers who will be marking your Science PSLE Paper, in all likelihood have been teachers all their lives and may not have worked in industries before, and would probably not be familiar with the terms ‘pitch’ and ‘threads’.

In other words, it is highly advised that you do not use the terms ‘pitch’ and ‘threads’, unless you are able to draw and label the diagrams as accurately as the above.

Parents up in arms again over PSLE Mathematics paper

From Channel News Asia

As requested by some students, here is one possible workout for the solution.

Question

Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim. Jim ate 12 sweets and Ken ate 18 chocolates. The ratio of Jim's sweets to chocolates became 1:7 and the ratio of Ken's sweets to chocolates became 1:4. How many sweets did Ken buy?

Solution

Chocolates ----- C
Sweets ----- S

Jim bought 2 units of chocolates ----- C, C
Ken bought 2 units of sweets ---- S, S

Jim gave 1 unit of chocolates to Ken and Ken gave 1 unit of sweets to Jim.
Jim ----- S, C
Ken ----- S, C

Jim ate 12 sweets and Ken ate 18 chocolates.
Jim ----- S – 12; C
Ken ----- S; C – 18

Jim’s sweet to chocolate ratio 1:7. Therefore for number of sweets to be same as number of chocolates, multiply sweets by 7.
C ----- 7 x (S – 12)
C ----- 7S – 84

Ken’s sweet to chocolate ratio 1:4. Therefore, for number of sweets to be same as number of chocolates, multiply sweets by 4.
4 x S ---- C – 18
4S ---- C – 18

From Jim, (C ----- 7S – 84), replace Ken's C....
(Ken) 4S ----- C - 18
4S ----- (7S – 84) – 18
4S ----- 7S – 102
3S ----- 102
1S ----- 34

Number of sweets Jim had at first
2 units of sweets ----- 2 x 34 = 68

Answer: 68 sweets.

Rosyth Sch 2007 PSLE Math Prelim Q48

A bus was travelling at a constant speed from Town A to Town B. It passed a car travelling at a constant speed of 90 km/h in the opposite direction. 1 and 1/2 hours later, the bus reached Town B but the car was still 25 km away from Town A. If the bus took 4 hours to complete the whole journey, what is the distance between the two towns?

Solution



Answer: 256 km

Rosyth Sch 2007 PSLE Math Prelim Q47

The perimeter of a rectangle to that of a square is in the ratio of 11:6. If the square has an area of 144 square m and the length and breadth of the rectangle are in the ratio of 6:5, find the length and breadth of the rectangle.

Solution



Perimeter
(6 + 6 + 5 + 5) units ----- 88 m
22 units ----- 88 m
1 unit ----- 88m divided by 22 = 4 m

(Length) 6 units ----- 6 x 4 m = 24 m
(Breadth) 5 units ----- 5 x 4 m = 20 m

Answer: Length = 24 m; Breadth = 20 m

Rosyth Sch 2007 PSLE Math Prelim Q46

The total cost of 28 textbooks and workbooks is $784.
3/4 of the books are textbooks and the remaining books are workbooks. A workbook cost half as much as a textbook. Find the difference in the price of a textbook and a workbook.

Solution

Workbook ----- 1 unit
Textbook ------ 2 units


3/4 of the books are textbooks ----- 3/4 x 28 = 21

Workbooks ----- 28 – 21 = 7

21 textbooks + 7 workbooks ---- $784

(21 x 2 units) + 7 units ----- $784

42 units + 7 units ----- $784

49 units ----- $784

1 unit ------ $784 divided 49 = $16

Answer: $16

Rosyth Sch 2007 PSLE Math Prelim Q45

The figure shows a cube with 3 painted parts A, B and C. These painted parts are of the same area and they are touching the midpoints of the sides of the cube. The total area of the painted parts is 54 square cm. Find the volume of the cube.

3 painted parts on 3 faces of the cube ----- 54 square cm

1 painted part on 1 face of the cube
54 square cm divided by 3 = 18 square cm

Area of 1 triangle is 18 square cm divided by 2
= 9 square cm

For area of triangle to be 9 square cm, the base has to be 6 cm and height has to be 3 cm as worked out below
(1/2 x 6 cm x 3 cm = 9 square cm)

1 side of the cube is therefore 6 cm as seen from above diagram.

Volume of cube
(6 x 6 x 6) cubic cm = 216 cubic cm

Answer: 216 cubic cm

Rosyth Sch 2007 PSLE Math Prelim Q44

The figure is not drawn to scale. What fraction of the figure is the total area of the shaded regions A, B, C, D, E, F, G and H?


Solution

The base of the whole figure is (4 + 3 + 6 + 3 + 2) cm = 18 cm

1st row from the top
Fig A has a base of 6 cm.

2nd row
The sum of the bases of Figs B and C is 6 cm.

3rd row
The sum of the bases of Figs D and E is 6 cm.

4th row
The sum of the bases of Figs F and G is 6 cm.

5th row
Fig H has a base of 6 cm.

The average length of the bases of all shaded figures is 6 cm.

The length of the base of the whole figure is 18 cm.

Therefore, the fraction of whole figure that is shaded is
6 cm divided by 18 cm = 1/3

Answer: 1/3

Note that the above method can be used because the vertical columns also correspond to the 6 cm shaded out of the 18 cm total vertical length of the figure.

Rosyth Sch 2007 PSLE Math Prelim Q43

The area of the figure ABCD is 48.5 square cm . Find the length of BE.
(Give your answer to the nearest tenth)


Solution

Area of ACD
1/2 x 18 cm x 10.4 cm = 93.6 square cm

Area of ABC
93.6 square cm - 48.5 square cm = 45.1 square cm

Area of Triangle ABC ----- 1/2 x base x height
45.1 square cm = 1/2 x 18 cm x height

height = (45.1 sq cm x 2 ) divided by18 cm
= 5.01 cm ~ 5.0 cm (to the nearest tenth)

Answer: 5.0 cm