Learning Objectives

Learning Objectives

By the end of this section, you will be able to do the following:

• Describe different simple machines

The information presented in this section supports the following AP® learning objectives and science practices:

• 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4)
• 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4)
• 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3)
• 3.F.1.5 The student is able to calculate torques on a two-dimensional system in static equilibrium by examining a representation or model, such as a diagram or physical construction. (S.P. 1.4, 2.2)

Simple machines are devices that can be used to multiply or augment a force that we apply, often at the expense of a distance through which we apply the force. The word for “machine” comes from the Greek word meaning “to help make things easier.” Levers, gears, pulleys, wedges, and screws are some examples of machines. Energy is still conserved for these devices, because a machine cannot do more work than the energy put into it. However, machines can reduce the input force that is needed to perform the job. The ratio of output to input force magnitudes for any simple machine is called its mechanical advantage (MA).

9.29 $MA=FoFiMA=FoFi size 12{"MA"= { {F rSub { size 8{o} } } over {F rSub { size 8{i} } } } } {}$

One of the simplest machines is the lever, which is a rigid bar pivoted at a fixed place called the fulcrum. Torques are involved in levers, since there is rotation about a pivot point. Distances from the physical pivot of the lever are crucial, and we can obtain a useful expression for the MA in terms of these distances.

Figure 9.24 A nail puller is a lever with a large mechanical advantage. The external forces on the nail puller are represented by solid arrows. The force that the nail puller applies to the nail ($FoFo size 12{F rSub { size 8{o} } } {}$) is not a force on the nail puller. The reaction force the nail exerts back on the puller ($FnFn size 12{F rSub { size 8{n} } } {}$) is an external force and is equal and opposite to $FoFo size 12{F rSub { size 8{o} } } {}$. The perpendicular lever arms of the input and output forces are $lili size 12{l rSub { size 8{i} } } {}$ and $l0l0 size 12{l rSub { size 8{0} } } {}$.

Figure 9.24 shows a lever type that is used as a nail puller. Crowbars, seesaws, and other such levers are all analogous to this one. $FiFi$ is the input force, and $FoFo size 12{F rSub { size 8{o} } } {}$ is the output force. There are three vertical forces acting on the nail puller (the system of interest): $Fi,Fo,Fi,Fo,$ and $NN size 12{N} {}$. $FnFn size 12{F rSub { size 8{n} } } {}$ is the reaction force back on the system, equal and opposite to $FoFo size 12{F rSub { size 8{o} } } {}$. Note that $FoFo size 12{F rSub { size 8{o} } } {}$ is not a force on the system. $NN size 12{N} {}$ is the normal force upon the lever, and its torque is zero since it is exerted at the pivot. The torques due to $FiFi size 12{F rSub { size 8{i} } } {}$ and $FnFn size 12{F rSub { size 8{n} } } {}$ must be equal to each other if the nail is not moving, to satisfy the second condition for equilibrium $net τ = 0 net τ = 0 size 12{ left ("net"τ=0 right )} {}$. In order for the nail to actually move, the torque due to $FiFi size 12{F rSub { size 8{i} } } {}$ must be ever-so-slightly greater than the torque due to $FnFn size 12{F rSub { size 8{i} } } {}$. Hence,

9.30 $liFi=lo,Fo,liFi=lo,Fo, size 12{l rSub { size 8{i} } F rSub { size 8{i} } = l rSub { size 8{o} } F rSub { size 8{o} } } {}$

where $lili size 12{l rSub { size 8{i} } } {}$ and $lolo size 12{l rSub { size 8{o} } } {}$ are the distances from where the input and output forces are applied to the pivot, as shown in the figure. Rearranging the last equation gives

9.31 $FoFi=lilo.FoFi=lilo. size 12{ { {F rSub { size 8{o} } } over {F rSub { size 8{i} } } } = { {l rSub { size 8{i} } } over {l rSub { size 8{o} } } } } {}$

What interests us most here is that the magnitude of the force exerted by the nail puller, $FoFo size 12{F rSub { size 8{o} } } {}$, is much greater than the magnitude of the input force applied to the puller at the other end, $FiFi size 12{F rSub { size 8{i} } } {}$. For the nail puller,

9.32 $MA=FoFi=lilo.MA=FoFi=lilo. size 12{"MA"= { {F rSub { size 8{o} } } over {F rSub { size 8{i} } } } = { {l rSub { size 8{i} } } over {l rSub { size 8{o} } } } } {}$

This equation is true for levers in general. For the nail puller, the MA is certainly greater than one. The longer the handle on the nail puller, the greater the force you can exert with it.

Two other types of levers that differ slightly from the nail puller are a wheelbarrow and a shovel, shown in Figure 9.25. All these lever types are similar in that only three forces are involved—the input force, the output force, and the force on the pivot—and thus their MAs are given by $MA=FoFiMA=FoFi size 12{"MA"= { {F rSub { size 8{o} } } over {F rSub { size 8{i} } } } } {}$ and $MA=d1d2MA=d1d2 size 12{"MA"= { {d rSub { size 8{1} } } over {d rSub { size 8{2} } } } } {}$, with distances being measured relative to the physical pivot. The wheelbarrow and the shovel differ from the nail puller, because both the input and output forces are on the same side of the pivot.

In the case of the wheelbarrow, the output force or load is between the pivot, or the wheel's axle, and the input or applied force. In the case of the shovel, the input force is between the pivot, at the end of the handle, and the load, but the input lever arm is shorter than the output lever arm. In this case, the MA is less than one.

Figure 9.25 (a) In the case of the wheelbarrow, the output force or load is between the pivot and the input force. The pivot is the wheel's axle. Here, the output force is greater than the input force. Thus, a wheelbarrow enables you to lift much heavier loads than you could with your body alone. (b) In the case of the shovel, the input force is between the pivot and the load, but the input lever arm is shorter than the output lever arm. The pivot is at the handle held by the right hand. Here, the output force, which is supporting the shovel's load, is less than the input force from the hand nearest the load, because the input is exerted closer to the pivot than is the output.

Example 9.3What is the Advantage for the Wheelbarrow?

In the wheelbarrow of Figure 9.25, the load has a perpendicular lever arm of 7.50 cm, while the hands have a perpendicular lever arm of 1.02 m. (a) What upward force must you exert to support the wheelbarrow and its load if their combined mass is 45 kg? (b) What force does the wheelbarrow exert on the ground?

Strategy

Here, we use the concept of mechanical advantage.

Solution

(a) In this case, $FoFi=liloFoFi=lilo size 12{ { {F rSub { size 8{o} } } over {F rSub { size 8{i} } } } = { {d rSub { size 8{1} } } over {d rSub { size 8{2} } } } } {}$ becomes

9.33 $Fi=Fololi.Fi=Fololi. size 12{F rSub { size 8{i} } =F rSub { size 8{o} } { {d rSub { size 8{2} } } over {d rSub { size 8{1} } } } } {}$

Adding values into this equation yields

9.34 $Fi=45.0 kg9.80 m/s20.075 m1.02 m=32.4 N.Fi=45.0 kg9.80 m/s20.075 m1.02 m=32.4 N. size 12{F rSub { size 8{i} } = left ("45"" kg" right ) left (9 "." 8" m/s" rSup { size 8{2} } right ) { { left (0 "." "075"" m" right )} over {1 "." "02"" m"} } ="32" "." 4" N"} {}$

The free-body diagram (see Figure 9.25) gives the following normal force: $Fi+N=WFi+N=W size 12{F rSub { size 8{1} } +N=W} {}$. Therefore, $N=(45.0 kg)9.80 m/s2−32.4 N=409 NN=(45.0 kg)9.80 m/s2−32.4 N=409 N size 12{N="45" left (9 "." 8 right ) - "32" "." 4="409"" N"} {}$. $NN$ is the normal force acting on the wheel; by Newton's third law, the force the wheel exerts on the ground is $409 N409 N size 12{"409"N} {}$.

Discussion

An even longer handle would reduce the force needed to lift the load. The MA here is $MA=1.02/0.0750=13.6MA=1.02/0.0750=13.6 size 12{ ital "MA"=1 "." "02"/0 "." "075"="13" "." 6} {}$.

Another very simple machine is the inclined plane. Pushing a cart up a plane is easier than lifting the same cart straight up to the top using a ladder, because the applied force is less. However, the work done in both cases, assuming the work done by friction is negligible, is the same. Inclined lanes or ramps were probably used during the construction of the Egyptian pyramids to move large blocks of stone to the top.

A crank is a lever that can be rotated $360º360º$ about its pivot, as shown in Figure 9.26. Such a machine may not look like a lever, but the physics of its actions remain the same. The MA for a crank is simply the ratio of the radii $ri/r0ri/r0 size 12{r rSub { size 8{i} } /r rSub { size 8{0} } } {}$. Wheels and gears have this simple expression for their MAs too. The MA can be greater than one, as it is for the crank, or less than one, as it is for the simplified car axle driving the wheels, as shown. If the axle's radius is $2 cm2 cm size 12{2 "." 0"cm"} {}$ and the wheel's radius is $24 cm24 cm size 12{"24" "." 0"cm"} {}$, then $MA=2/24=0.083MA=2/24=0.083 size 12{"MA"=1/"12"=0 "." "083"} {}$, and the axle would have to exert a force of $12,000 N12,000 N size 12{"12","000"N} {}$ on the wheel to enable it to exert a force of $1,000 N1,000 N size 12{"1000"N} {}$ on the ground.

Figure 9.26 (a) A crank is a type of lever that can be rotated $360º 360º$ about its pivot. Cranks are usually designed to have a large MA. (b) A simplified automobile axle drives a wheel, which has a much larger diameter than the axle. The MA is less than one. (c) An ordinary pulley is used to lift a heavy load. The pulley changes the direction of the force $TT$ exerted by the cord without changing its magnitude. Hence, this machine has an MA of one.

An ordinary pulley has an MA of one; it only changes the direction of the force and not its magnitude. Combinations of pulleys, such as those illustrated in Figure 9.27, are used to multiply force. If the pulleys are friction-free, then the force output is approximately an integral multiple of the tension in the cable. The number of cables pulling directly upward on the system of interest, as illustrated in the figures given below, is approximately the MA of the pulley system. Since each attachment applies an external force in approximately the same direction as the others, they add up, producing a total force that is nearly an integral multiple of the input force $TT$.

Figure 9.27 (a) The combination of pulleys is used to multiply force. The force is an integral multiple of tension if the pulleys are frictionless. This pulley system has two cables attached to its load, thus applying a force of approximately $2T2T$. This machine has $MA≈2MA≈2 size 12{ ital "MA" approx 2} {}$. (b) Three pulleys are used to lift a load in such a way that the mechanical advantage is about three. Effectively, there are three cables attached to the load. (c) This pulley system applies a force of $4T4T$, so that it has $MA≈4MA≈4 size 12{ ital "MA" approx 4} {}$. Effectively, four cables are pulling on the system of interest.