Lesson Using proportions to solve word problems in Physics

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Using proportions to solve word problems in Physics


A number of problems in Physics can be deduced to proportions and solved using rules for solving proportions.
This lesson explains how to apply proportions to solve problems related to levers.

A Type 1 Lever


A lever is a mechanical tool like rigid beam that is used with an appropriate fulcrum point (supporting point) to get the multiply mechanical force
(output force) at the one end when smaller force (effort) is applied to another end.

In a Type 1 lever (Figure 1), the fulcrum is between the effort and the load. The scheme of the Type 1 lever is shown in Figure 2.


Figure 1. The Type 1 lever


Figure 2. The scheme of the Type 1 lever

People used levers from prehistoric times to move and to raise heavy objects. In the ancient Egypt they used levers to move and raise many-tons stone bricks when built pyramids. The ancient Greek mathematician, physicist and engineer Archimedes was the first who formulated the law of lever in the quantitative form.

If parallel forces F%5B1%5D and F%5B2%5D are applied to both ends of the lever, then the lever is in balance if
F%5B1%5D%2Ad%5B1%5D+=+F%5B2%5D%2Ad%5B2%5D,
where d%5B1%5D and d%5B2%5D are the arms of forces F%5B1%5D and F%5B2%5D respectively.

(The arm of the force is the distance from the force application point to the fulcrum point.
The arms d%5B1%5D and d%5B2%5D are shown in Figure 2.)

You can write the law of lever as the proportion
F%5B2%5D%2FF%5B1%5D+=+d%5B1%5D%2Fd%5B2%5D.
Note that this proportion is not the direct proportionality; it is the inverse proportionality instead.
In the problems below, three terms of proportions are given and one term is unknown. You can apply the proportion properties and rules for solving proportions to calculate the unknown terms.

Problem 1
The arms of a horizontal lever are 0.2 m and 1 m long at opposite sides of the fulcrum.
The shorter arm is loaded with the downward force of 500 Newtons at the end.
What force should be applied at the lever longer arm end to balance the load?

Solution
Let us denote the unknown force as F Newtons.
Write the proportion for the unknown force F using the law of lever:
500%2FF+=+1%2F0.2.
The unknown is the mean term of the proportion.
Apply the rule for solving proportions from the lesson Proportions: the unknown mean of the proportion is equal to the product of extremes divided by the known mean.
You get the unknown force
F+=+500%2A0.2%2F1+=+100 Newtons.

Answer. The downward force of 100 Newtons should be applied at the lever longer arm end to balance the load force of 500 Newtons.

Amazing result, isn't? You apply the effort in five times less than the load is and still keep the balance!
Engineers say that the lever provides the mechanical advantage equal to 5 in this case.

Exercise 1
Solve yourself the same problem with the following data.
The arms of a horizontal lever are 1 ft and 5 ft long at opposite sides of the fulcrum.
The shorter arm loaded with downward force of 200 lbs at the end.
What force (in lbs) should be applied at the lever longer arm end to balance the load?

Problem 2
A man can push down with a force of 160 lbs. He has a 6 ft long crowbar.
The man is going to use the crowbar as a lever to lift the stone. The fulcrum is in 1 foot from the stone.
How heavy a stone could be the man could lift it using crowbar as a lever?

Solution
If a man uses the crowbar as the Type 1 lever, then the shorter arm of the lever is 1 ft, while the longer arm is 5 ft (=6 ft - 1 ft).
Let us denote as W the unknown weight of the stone (in lbs).
To determine the unknown weight W, write the law of lever as the proportion
160%2FW+=+1%2F5.
The unknown is the mean term of the proportion.
Apply the rule for solving proportions from the lesson Proportions: the unknown mean of the proportion is equal to the product of extremes divided by the known mean.
You get the unknown weight
W+=+160%2A5%2F1+=+800 lbs.

Answer. At given conditions, the man can lift the stone of the weight of 800 lbs or less.

Can you imagine lifting such a heavy stone without a lever?

Clarification
Figure 3 to the right illustrates the Problem 2.
Strictly speaking, the lever arms that appear in the law of lever are distances d%5B1%5D and d%5B2%5D in Figure 3. They are lengths of perpendiculars drawn from the force application points to the line through the fulcrum parallel to the force vectors. But in the solution of the Problem 2 above we actually used lengths L%5B1%5D and L%5B2%5D of segments CO and OD instead of d%5B1%5D and d%5B2%5D.
Nevertheless, our solution of the Problem 2 is correct.
The matter is in that, the law of lever uses the ratio d%5B1%5D%2Fd%5B2%5D, in fact.
Note that the ratio d%5B1%5D%2Fd%5B2%5D is equal to the ratio L%5B1%5D%2FL%5B2%5D because triangles OAC and OBD are similar. The later is true because these triangles are right triangles having equal angles AOC and BOD.

Conclusion: when applying the law of lever for a rectilinear lever for a pair of parallel forces, you can use either the ratio of the lever arms or the ratio of the lever segment lengths. The result is the same. For more complex lever shapes use the ratio of lever arms.


Figure 3. Using the lever to lift the stone


Problem 3
Andrew and Bill are playing on seesaw (see Figure 4 to the right). Andrew's weight is 40 pounds, and he is sitting at the distance of 4 ft from the fulcrum.
Bill's weight is 80 pounds. Where Bill should seat to balance the seesaw?

Solution
The idea is to consider the seesaw as a lever and to apply the law of lever to determine the distance from the fulcrum where Bill should seat to balance the seesaw.
Let us denote as x the unknown distance (in ft).
To determine the unknown distance x, write the law of lever as the proportion
40%2F80+=+x%2F4.
The unknown is the mean term of the proportion.
Apply the rule for solving proportions from the lesson Proportions: the unknown mean of the proportion is equal to the product of extremes divided by the known mean.
You get the unknown distance
x+=+40%2A4%2F80+=+2 ft.

Answer. Bill should seat on the opposite side at 2 ft from the fulcrum.


Figure 4. Andrew and Bill
          playing on seesaw


Problem 4
Alice and Barbara are playing on seesaw. Alice’s weight is 30 pounds, and she is sitting at the distance of 4 ft from the fulcrum.
Barbara is sitting at the distance of 3 ft from the fulcrum, and the seesaw is in balance.
Could you determine Barbara’s weight using this data?

Solution
Yes, you can. The idea is to consider the seesaw as a lever and to apply the law of lever to determine the Barbara’s weight.
Let us denote as W the unknown Barbara’s weight (in lbs).
To determine the unknown weight W, write the law of lever as the proportion
30%2FW+=+3%2F4.
The unknown is the mean term of the proportion.
Apply the rule for solving proportions from the lesson Proportions: the unknown mean of the proportion is equal to the product of extremes divided by the known mean.
You get the unknown weight
W+=+30%2A4%2F3+=+40 lbs.

Answer. Barbara’s weight is 40 lbs.

A Type 2 Lever


In a Type 2 Lever (Figure 5), the load is between the fulcrum and the effort. The scheme of the Type 2 lever is shown in Figure 6.


Figure 5. The Type 2 lever


Figure 6. The scheme of the Type 2 lever

Problem 5
A Type 2 horizontal lever is 1 m 20 cm long.
The load of 400 Newtons is applied downward at 30 cm from the fulcrum point.
What force should be applied at the lever end to balance the load?

Solution
Let us denote the unknown force as F Newtons.
Write the proportion for the unknown force F using the law of lever:
400%2FF+=+120%2F30.
The unknown is the mean term of the proportion.
Apply the rule for solving proportions from the lesson Proportions: the unknown mean of the proportion is equal to the product of extremes divided by the known mean.
You get the unknown force
F+=+400%2A30%2F120+=+100 Newtons.

Answer. The upward force of 100 Newtons should be applied at the lever end to balance the downward load of 400 Newtons.

Exercise 2
Solve yourself the same problem with the following data.
The length of a Type 2 horizontal lever is 4 ft.
The load of 80 lbs is applied downward at 1 ft from the fulcrum point.
What force should be applied at the lever end to balance the load?

Note that the same Clarification as above is applicable for the Type 2 levers: when applying the law of lever for a rectilinear Type 2 lever and for a pair of parallel forces, you can use either the ratio of the lever arms or the ratio of lever segment lengths. The result is the same. For more complex lever shapes use the ratio of lever arms.

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