Overview
Opposing-rate problems occur when two or more processes act on the same system but do not all contribute in the same direction. One process may add work while another removes it, or one may increase a quantity while another decreases it.
Typical examples include a pipe filling a tank while a drain empties it, a pump adding water while a leak removes water, or a production system creating units while damaged units are being discarded.
The central principle is:
\[ \boxed{R_{\text{net}} = R_{\text{positive}} - R_{\text{negative}}} \]
Once the net rate is known, the ordinary Work Principle still applies:
\[ \boxed{W = R_{\text{net}}T}. \]
Direction Matters
In combined-rate problems, all workers or processes contribute toward the same result, so their rates are added. In opposing-rate problems, at least one process acts against the desired result.
Suppose a pipe fills
\[ \frac{1}{4} \]
of a tank each hour while a drain removes
\[ \frac{1}{6} \]
of the tank each hour.
The filling rate is positive because it moves the tank toward being full. The draining rate is negative because it moves the tank away from being full.
Therefore,
\[ R_{\text{net}} = \frac{1}{4}-\frac{1}{6}. \]
Choosing Positive and Negative Rates
The signs assigned to rates depend on the goal of the problem.
If the goal is to fill a tank, then:
- filling rates are positive;
- draining or leaking rates are negative.
If the goal were instead to empty the tank, the interpretation could be reversed. What matters is choosing one direction as positive and applying that choice consistently.
A useful general model is
\[ \boxed{R_{\text{net}} = R_1 + R_2 + R_3 + \cdots} \]
where contributing rates are assigned positive values and opposing rates are assigned negative values.
Example 1: Filling a Tank with an Open Drain
A pipe can fill a tank in \(4\) hours. A drain can empty the same tank in \(6\) hours. If the tank begins empty and both are open, how long will it take to fill the tank?
The filling rate is
\[ R_f=\frac{1}{4}. \]
The draining rate is
\[ R_d=\frac{1}{6}. \]
Because the drain opposes filling, subtract its rate:
\[ R_{\text{net}} = \frac{1}{4}-\frac{1}{6}. \]
Use a common denominator:
\[ R_{\text{net}} = \frac{3}{12}-\frac{2}{12} = \frac{1}{12}. \]
The tank fills at a net rate of one-twelfth of the tank per hour.
Use
\[ T=\frac{W}{R}. \]
Since \(W=1\),
\[ T = \frac{1}{\frac{1}{12}} = 12. \]
Therefore,
\[ \boxed{T=12\text{ hours}}. \]
A Reasonableness Check
The filling pipe alone can fill the tank in \(4\) hours. If a drain is open at the same time, the tank must take longer than \(4\) hours to fill.
Therefore the answer should satisfy
\[ T>4. \]
The result
\[ T=12 \]
is consistent with the situation.
A result of \(3\) hours would be impossible because opening the drain cannot make the tank fill faster.
Example 2: A Faster Filling Pipe
A pipe can fill a tank in \(3\) hours, while a leak can empty a full tank in \(12\) hours. How long will the tank take to fill if the leak remains open?
The filling rate is
\[ \frac{1}{3}. \]
The leak rate is
\[ \frac{1}{12}. \]
The net rate is
\[ R_{\text{net}} = \frac{1}{3}-\frac{1}{12}. \]
Rewrite \(\frac{1}{3}\) with denominator \(12\):
\[ R_{\text{net}} = \frac{4}{12}-\frac{1}{12} = \frac{3}{12} = \frac{1}{4}. \]
Therefore,
\[ T = \frac{1}{\frac{1}{4}} = 4. \]
The tank fills in
\[ \boxed{4\text{ hours}}. \]
When the Opposing Rate Is Greater
A problem may produce a net rate of zero or a negative net rate. These results have important physical meanings.
Suppose a pipe fills a tank at
\[ \frac{1}{8} \]
of the tank per hour, but a drain removes
\[ \frac{1}{6} \]
of the tank per hour.
Then
\[ R_{\text{net}} = \frac{1}{8}-\frac{1}{6}. \]
Using a common denominator,
\[ R_{\text{net}} = \frac{3}{24}-\frac{4}{24} = -\frac{1}{24}. \]
The negative rate means the tank is losing water overall. If the tank begins empty, it cannot become full under these conditions.
The equation has not failed. Instead, the sign of the answer describes the direction in which the system is changing.
Zero Net Rate
If the positive and negative rates are equal, then the net rate is zero.
For example,
\[ R_{\text{net}} = \frac{1}{5}-\frac{1}{5} = 0. \]
In this case, the amount of work does not change.
For a tank, water would enter at exactly the same rate that it leaves. The water level would remain constant.
Because
\[ W=RT, \]
a net rate of zero produces no change in work regardless of how much time passes.
Example 3: Two Filling Pipes and One Drain
Pipe A can fill a tank in \(6\) hours, Pipe B can fill it in \(8\) hours, and a drain can empty the full tank in \(12\) hours. If all three are open, how long will it take to fill the tank?
The rates are
\[ R_A=\frac{1}{6}, \qquad R_B=\frac{1}{8}, \qquad R_D=-\frac{1}{12}. \]
The net rate is
\[ R_{\text{net}} = \frac{1}{6} + \frac{1}{8} - \frac{1}{12}. \]
The least common denominator is \(24\):
\[ R_{\text{net}} = \frac{4}{24} + \frac{3}{24} - \frac{2}{24} = \frac{5}{24}. \]
The completion time is
\[ T = \frac{1}{\frac{5}{24}} = \frac{24}{5}. \]
Therefore,
\[ \boxed{T=4.8\text{ hours}}. \]
This is \(4\) hours and \(48\) minutes.
Finding Partial Work with Opposing Rates
Opposing-rate problems do not always ask for the time required to complete an entire job. Sometimes the goal is to determine how much work has been accomplished after a specified time.
Suppose a pump fills a tank at a rate of
\[ \frac{1}{5} \]
of the tank per hour while a leak removes
\[ \frac{1}{20} \]
of the tank per hour.
The net rate is
\[ R_{\text{net}} = \frac{1}{5}-\frac{1}{20} = \frac{4}{20}-\frac{1}{20} = \frac{3}{20}. \]
If both operate for \(4\) hours, then
\[ W = \frac{3}{20}(4) = \frac{12}{20} = \frac{3}{5}. \]
Therefore,
\[ \boxed{\frac{3}{5}} \]
of the tank is filled after \(4\) hours.
Example 4: Finding an Unknown Leak Rate
A pipe can fill a tank in \(4\) hours. With a leak present, the tank requires \(6\) hours to fill. How long would the leak take to empty a full tank by itself?
The filling rate is
\[ R_f=\frac{1}{4}. \]
The net filling rate is
\[ R_{\text{net}}=\frac{1}{6}. \]
Let \(R_L\) represent the leak rate. Then
\[ \frac{1}{4}-R_L=\frac{1}{6}. \]
Solve for \(R_L\):
\[ R_L = \frac{1}{4}-\frac{1}{6} = \frac{3}{12}-\frac{2}{12} = \frac{1}{12}. \]
The leak removes one-twelfth of the tank per hour. Therefore, the leak alone would empty the tank in
\[ \boxed{12\text{ hours}}. \]
The General Opposing-Rate Model
For several processes acting simultaneously, assign a sign to each rate according to whether it contributes toward or opposes the chosen goal.
The net rate is
\[ R_{\text{net}} = R_1+R_2+R_3+\cdots. \]
Then apply
\[ W=R_{\text{net}}T. \]
For one complete job,
\[ 1 = T(R_1+R_2+R_3+\cdots). \]
This form can represent multiple filling pipes, several drains, production and waste, or any other system involving simultaneous positive and negative rates.
Method for Solving Opposing-Rate Problems
- Identify the goal. Decide what represents positive progress, such as filling a tank or completing a job.
- Represent the complete job by \(1\).
- Convert each completion time into a rate. Use \(R=\frac{1}{t}\).
- Assign signs to the rates. Rates that move toward the goal are positive; rates that move away from it are negative.
- Find the net rate. Add the signed rates.
- Apply the Work Principle. Use \(W=R_{\text{net}}T\).
- Solve for the unknown quantity.
- Interpret the sign and magnitude of the result. Confirm that the answer is physically and mathematically reasonable.
Common Errors
- Adding an opposing rate instead of subtracting it. A drain or leak works against filling and must be represented by a negative rate when filling is the goal.
- Subtracting completion times. As with combined-rate problems, completion times must first be converted into rates.
- Ignoring the meaning of a negative net rate. A negative result may correctly indicate that the system is moving opposite the chosen goal.
- Assuming every problem can reach completion. If the opposing rate equals or exceeds the positive rate, the desired job may never be completed.
- Forgetting the reasonableness check. A leak should make filling take longer, not shorter.
Practice Problems
- A pipe can fill a tank in \(5\) hours, while a drain can empty it in \(20\) hours. How long will the tank take to fill if both are open?
- A pump can fill a reservoir in \(8\) hours. A leak can empty the full reservoir in \(24\) hours. What is the net filling rate?
- A pipe fills a tank in \(6\) hours, while a drain empties it in \(9\) hours. What fraction of the tank is filled after \(4\) hours if both remain open?
- Two pipes can fill a tank individually in \(8\) hours and \(12\) hours. A drain can empty it in \(24\) hours. How long will the tank take to fill if all three are open?
- A filling pipe alone fills a tank in \(3\) hours. With a leak present, the tank takes \(4\) hours to fill. How long would the leak take to empty the tank by itself?
- A pipe fills a tank at \(\frac{1}{10}\) tank per hour while a drain empties it at \(\frac{1}{8}\) tank per hour. What is the net rate, and can an empty tank become full under these conditions?
- A pump adds \(\frac{1}{6}\) of a tank per hour while two leaks remove \(\frac{1}{24}\) and \(\frac{1}{12}\) of the tank per hour. What is the net rate?
- A tank is initially half full. A pipe fills at \(\frac{1}{8}\) tank per hour while a drain removes \(\frac{1}{20}\) tank per hour. How long will it take for the tank to become full?
Practice Answers
- \[ R = \frac{1}{5}-\frac{1}{20} = \frac{3}{20}. \] Therefore, \[ T = \frac{1}{\frac{3}{20}} = \boxed{\frac{20}{3}\text{ hours}} \] or approximately \(6.67\) hours.
- \[ R = \frac{1}{8}-\frac{1}{24} = \frac{3}{24}-\frac{1}{24} = \boxed{\frac{1}{12}} \] of the reservoir per hour.
- \[ R = \frac{1}{6}-\frac{1}{9} = \frac{3}{18}-\frac{2}{18} = \frac{1}{18}. \] After \(4\) hours, \[ W = \frac{1}{18}(4) = \boxed{\frac{2}{9}}. \]
- \[ R = \frac{1}{8} + \frac{1}{12} - \frac{1}{24} = \frac{3}{24}+\frac{2}{24}-\frac{1}{24} = \frac{1}{6}. \] Therefore, \[ \boxed{6\text{ hours}}. \]
- \[ R_L = \frac{1}{3}-\frac{1}{4} = \frac{1}{12}. \] Therefore, \[ \boxed{12\text{ hours}}. \]
- \[ R = \frac{1}{10}-\frac{1}{8} = \frac{4}{40}-\frac{5}{40} = \boxed{-\frac{1}{40}}. \] The tank is losing water overall, so an empty tank cannot become full under these conditions.
- \[ R = \frac{1}{6} - \frac{1}{24} - \frac{1}{12} = \frac{4}{24}-\frac{1}{24}-\frac{2}{24} = \boxed{\frac{1}{24}}. \]
- The remaining work is \[ 1-\frac{1}{2}=\frac{1}{2}. \] The net rate is \[ \frac{1}{8}-\frac{1}{20} = \frac{5}{40}-\frac{2}{40} = \frac{3}{40}. \] Therefore, \[ T = \frac{\frac{1}{2}}{\frac{3}{40}} = \frac{1}{2}\times\frac{40}{3} = \boxed{\frac{20}{3}\text{ hours}}. \]
Summary
Opposing-rate problems involve processes that act in different directions. The rates must therefore be assigned positive or negative signs according to the goal of the problem.
The fundamental relationship is
\[ \boxed{R_{\text{net}}=R_{\text{positive}}-R_{\text{negative}}}. \]
More generally,
\[ \boxed{R_{\text{net}}=R_1+R_2+R_3+\cdots}, \]
where each rate carries the appropriate sign.
Once the net rate is found, use
\[ \boxed{W=R_{\text{net}}T}. \]
A positive net rate means progress is being made toward the chosen goal, a zero net rate means no overall change occurs, and a negative net rate means the system is moving away from the goal.
The next class of problems introduces another complication: workers or processes that do not begin working at the same time.
Continue
Continue to 4. Staggered Work Problems
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