Mechanical Comprehension Study Guide 2 for McGraw-Hill's ASVAB (page 6)

Practice problems for this study guide can be found at:

Mechanical Comprehension Practice Problems for McGraw-Hill's ASVAB

Many Mechanical Comprehension questions have to do with things that move or rotate: gears, pulleys, and other mechanisms.

Systems of Pulleys

You have already learned about pulleys as simple machines used to hoist heavy objects. Pulleys are also used as drive mechanisms. That is, systems of interconnected pulleys are used to transfer power, and often rotational speed, from one shaft to another. Two pulleys (sheaves) that are connected by a belt will run at the same speed if they are the same size. When the sheaves are of different sizes, the smaller one will run faster than the larger one because the smaller one must make more turns to move the belt the same distance.

In most cases on the ASVAB, you will simply be asked which of two or more interconnected pulleys runs fastest or slowest, and you can easily solve these problems by identifying the smallest or largest pulley. However, if you are asked to calculate the speed of a particular pulley in a system of pulleys, you can easily do so by using the following formula:

Speed₁ × diameter₁ = speed₂ × diameter₂

(Note that pulley speed is measured in revolutions per minute, or rpm.)

Example

Pulley 1 measures 9 in. in diameter. Pulley 2 measures 3 in. in diameter. If pulley 1 rotates at 1,200 rpm, how fast will pulley 2 rotate?

Another way to calculate the speed of pulley 2 is to look at the ratio between the two diameters. A ratio of 9:3, or 3:1, will multiply speed × 3. So 1,200 rpm × 3 = 3,600 rpm. Remember that the pulley with the smaller diameter is always the one that rotates faster!

Using these methods, you can calculate the speed of any pulley system as long as you know the diameters of both pulleys and the speed of either pulley.

Example

When pulley A runs at 400 rpm, what will be the speeds of pulleys B, C, and D?

In this system, assume that the linked pulleys (B and C in the example) run at the same rpm, since they are attached to the same shaft. Break the problem down into parts, and calculate them in order:

• Diameter of pulley A/diameter of pulley B = 4/8, so pulley B will run 1/2 as fast as pulley A. 400/2 = 200 rpm
• You already know that pulley C runs at the same speed as pulley B.
• Diameter of pulley C/diameter of pulley D = 4/16 = 1/4, so pulley D will run 1/4 as fast as pulley C.

200/4 = 50 rpm

Systems of Gears

Another way to transfer power between shafts is through systems of gears. The gears in a system typically have different diameters and different numbers of teeth per gear. The teeth of one gear mesh with the teeth of another, and as one gear (the drivinggear) turns, it turns the other gear (the drivengear). When interlocking gears have different numbers of teeth, the gear with fewer teeth will rotate more times in a given period than the gear with more teeth. To see how this works, look at the following example.

Example

Gear A and gear B make up a system of gears. If gear A makes 6 revolutions, how many revolutions will gear B make?

To solve this problem, use the picture and your common sense. Count the teeth on each gear. Gear A has 9 teeth. Gear B has 27 teeth. The ratio of the teeth on the two gears is 9:27 or 1:3. Common sense should tell you that gear A must rotate 3 times to make gear B rotate once. So if gear A rotates 6 times, gear B will rotate twice. Always keep in mind that in this kind of system, the gear with more teeth makes fewer rotations in the same period than the gear with fewer teeth.

Notice, too, that gears change the rotation direction, while pulleys usually do not. To rotate a gear in the same direction as the driving gear, you need a third gear, called an idlergear.

Example

In this system, gear A (the driving gear) is rotating clockwise. Gear B is the idler gear. Gear A makes gear B rotate counterclockwise. Gear B then makes gear C rotate clockwise, the same direction as gear A.

Rotating Wheels and Disks

Another way to drive shafts is to use what is called a pin and slot arrangement. In this arrangement, a pin is attached to a driving shaft, and a slotted disk is attached to a driven shaft. When the driving shaft rotates, the pin enters a slot on the disk and turns the driven shaft.

Example

In this pin and slot arrangement, each time the driving shaft turns one full revolution, the disk on the driven shaft will make 1/4 revolution. How far will the disk rotate when the pin turns three complete revolutions?

3 × 1/4 = 3/4 turn.

You may also be asked simple questions about what happens to points on a wheel when the wheel turns.

Example

How many rotations will point A make when point C makes 5 rotations? Point A will make 5 rotations because Point A and Point C are both fixed on the same wheel.

Example

Which point will travel farthest as the wheel makes 10 rotations? Point B will travel farthest because it is farthest from the center. The distance it travels in each rotation is greater than the distance traveled by the other points.

Cams and Cam Followers

Camsare lobes attached to rotating shafts to push separate pieces, called cam followers. Cams are often found in engines, where they push intake and exhaust valves open when the engine turns. For every complete rotation of the camshaft, the cam follower will move away from and then back to its original position. A spring pushes the follower tight to the cam.

Cranks and Pistons

Cranksare used to change motion in a straight line to motion in a circle. You'll find cranks connected to pedals on a bike, and to pistons in a car engine. When a crank makes one complete revolution, the piston must go up and down and return to its original position.

Fluid Dynamics

Fluidsare substances that take the shape of their container. Gases and liquids are both fluids. The behavior of fluids is called fluid dynamics, and it can get rather complicated, but not on the ASVAB. Let's start with air, and move on to hydraulics—the engineering of liquids.

Air Pressure

Air pressure is measured in pounds per square inch. Atmospheric pressure at sea level is 14.7 lb/in2, which is actually quite a bit of pressure. Since it's present all around us, we don't notice it. However, if you create a vacuum inside a weak container, the container will be crushed by all that pressure.

Pneumatics and the Gas LawsSystems that use compressed air to do work are called pneumatic systems. Air is easily compressed, and the calculations are more complicated than they are with liquids, which usually can't be compressed. The larger the driven cylinder, the more air pressure it is exposed to, and the greater the force it can exert.

The "gas laws" apply to air as it is compressed and expanded.

• When a gas is compressed, it gains thermal energy—it warms up. The gas also gains potential energy, which is why compressed air can be used to drive nail guns and pneumatic hammers.
• When a given amount of gas expands, its pressure drops and the gas cools.
• When a gas cools without a change in outside pressure, it loses volume.

What happens when you increase the air pressure outside a balloon? The balloon shrinks until the pressure inside becomes great enough to balance the pressure outside.

Air pressure is also what keeps airplanes aloft. The bulge on the top of an airplane wing increases the speed of air passing over the wing, and that causes a reduction in pressure. Because air pressure does not change below the wing, the result is an unbalanced upward force. This force lifts the airplane.

RefrigeratorsRefrigerators and air conditioners provide interesting applications of the gas laws. A compressor compresses a fluid, called a refrigerant. The refrigerant warms up, as predicted by the gas laws. Then the refrigerant loses heat (but not pressure) in the condenser. The refrigerant is piped into an evaporator, where it goes through a small hole and evaporates under reduced pressure. Expansion causes the temperature to drop, and the cold refrigerant can pick up heat from the surroundings. This is why the evaporator is placed in the area to be cooled. The condenser is placed where it's easy to get rid of excess heat—in the backyard for an air conditioner, or in back for a refrigerator.

Water Pressure

On the ASVAB, water pressure questions often involve flow through pipes. Keep these principles in mind:

1. Total flow through a pipe system must be the same everywhere because water cannot be compressed.
2. When liquid speeds up, pressure falls.
3. When liquid slows down, pressure rises.

In the diagram, the same amount of water is flowing everywhere in the pipe system. For this to be true, water must be flowing faster at point B than at point A. That means that pressure is lower at point B.

Water in a container also exerts pressure on the bottom of the container. The deeper the water, the greater the pressure. To find the amount of water pressure in a tank, calculate the total weight of the water and divide by the area of the base of the tank.

Example

A tank with a base that measures 2 feet × 4 feet holds 1,600 pounds of water. What is the water pressure at the base of the tank?

2 ft × 4 ft = 8 ft²

1,600/8 = 200 lb/ft²

Remember too that 1 ft² = 144 in². To convert pressure between pounds per square inch and pounds per square foot, divide or multiply by 144.

Filling and Emptying Tanks

The Mechanical Comprehension test often includes problems about filling and emptying tanks. Usually the filling and emptying take place at different rates, as in the following examples.

Example

Water is being piped into a tank at the rate of 2 gallons per second. At the same time, it is being piped out of the tank at the rate of 60 gallons per minute. How many gallons will be added in 5 minutes?

Convert the inflow rate so that you are working only with gallons per minute.

Subtract:

120 gal/min inflow – 60 gal/min outflow = 60 gal/min net gain

The net gain in 5 minutes is 5 – 60 = 300 gallons.

Example

A 100-gallon tank contains 10 gallons of water. Water is added through one pipe at the rate of 3 gallons per minute. It is drained away through another pipe at the rate of 2 gallons per minute. How long will it take to fill the tank?

Find the net gain of water per minute:

3 gal/min – 2 gal/min = 1 gal/min

It will take 100 – 10 = 90 gallons to fill the tank. At the rate of 1 gal/min, it will take 90 minutes to fill the tank.