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In the case of the wheelbarrow, the output force or load is between the pivot (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.

A wheelbarrow is shown in which the input force F sub I is shown as a vector in vertically upward direction below the handle of wheelbarrow. The weight of the wheelbarrow is downward at the center of gravity. The normal reaction of the ground is acting at the wheel in upward direction. The perpendicular distance between the normal reaction and the input force F sub I is labeled as R sub I and the distance between output force F sub O and normal reaction is labeled as R sub O. In figure b, a man is holding a shovel in his hands. One hand is at one end of the handle and the other hand is holding the shovel at the middle. The center of gravity of the shovel is at its flat end. The weight of the shovel is acting at the center of gravity. The input force is acting at the hand in the middle in upward direction and the end of the shovel is acting as pivot. A free body diagram is also shown at the right side of the figure.
(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 (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.

What is the advantage for the wheelbarrow?

In the wheelbarrow of [link] , 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 45.0 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, F o F i = l i l o 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

F i = F o l o l i . 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

F i = 45.0 kg 9.80 m/s 2 0.075 m 1.02 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 [link] ) gives the following normal force: F i + N = W size 12{F rSub { size 8{1} } +N=W} {} . Therefore, N = ( 45.0 kg ) 9.80 m/s 2 32.4 N = 409 N size 12{N="45" left (9 "." 8 right ) - "32" "." 4="409"" N"} {} . N is the normal force acting on the wheel; by Newton’s third law, the force the wheel exerts on the ground is 409 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 . 6 size 12{ ital "MA"=1 "." "02"/0 "." "075"="13" "." 6} {} .

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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 360º about its pivot, as shown in [link] . 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 r i / r 0 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 1, as it is for the crank, or less than 1, as it is for the simplified car axle driving the wheels, as shown. If the axle’s radius is 2.0 cm size 12{2 "." 0`"cm"} {} and the wheel’s radius is 24.0 cm size 12{"24" "." 0`"cm"} {} , then MA = 2.0 / 24.0 = 0 . 083 size 12{"MA"=1/"12"=0 "." "083"} {} and the axle would have to exert a force of 12,000 N size 12{"12","000"`N} {} on the wheel to enable it to exert a force of 1000 N size 12{"1000"`N} {} on the ground.

In figure a, a crank lever is shown in which a hand is at the handle of the crank lever. The output force F sub O is at the base of the lever and the input force F sub I is at the handle of the lever. The distance between input force and output force is labeled as R sub I. In figure b, a simplified axle of the car is shown. The input force is shown as a vector F sub I on the axle toward right. The output force is shown at the point of contact of the wheel with the ground toward left. The distance between the output force and the pivot point is labeled as R sub O. In figure c, rope over the pulley is shown. The input force is shown as a downward arrow at the left part of rope. The output force is acting on the right part of the rope. The center of the pulley is the pivot point. The distances of the two forces from the pivot are R sub I and R sub O respectively.
(a) A crank is a type of lever that can be rotated 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 1. (c) An ordinary pulley is used to lift a heavy load. The pulley changes the direction of the force T exerted by the cord without changing its magnitude. Hence, this machine has an MA of 1.
Practice Key Terms 1

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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