Suppose you were planning to move a house by transporting it on a large trailer. The house has an estimated weight of 45.6 tons \((1\) ton \(=2000 \mathrm{lb})\). The trailer is expected to weigh 8.3 tons. Each wheel of the trailer will have tires inflated to a gauge pressure of 85 psi (which is actually 85 psi above atmospheric pressure). If the area of contact between a tire and the pavement can be no larger than \(100.0 \mathrm{in}^{2}(10.0\) in. \(\times 10.0\) in. \(),\) what is the minimum number of wheels the trailer must have? (Remember, tires are mounted in multiples of two on a trailer. Assume that atmospheric pressure is 14.7 psi.)

Understanding unit conversion is a fundamental skill in both physics and chemistry, as it allows for the comparison and computation of various measurements in a consistent manner. In the problem at hand, we are presented with weights in tons, but to perform precise calculations, we need to convert these to pounds. To do this, the given value in tons is multiplied by the conversion factor of 2000 pounds per ton. For instance, the house weighing 45.6 tons is equivalent to \(45.6 \times 2000 = 91,200\) pounds.

During unit conversion, it is essential to pay attention to the units you start with and the units you need to end up with, ensuring that your conversion factor is accurate. This step-by-step approach breaks down potentially complex problems into manageable parts, making them easier to solve.

Unit conversion not only makes calculations consistent but also reduces the likelihood of errors in your final result. It is a critical step before proceeding to more intricate calculations like those needed for pressure and weight distribution.

Pressure calculations are integral to various fields of science, especially in chemistry where the behavior of gases under different pressures is a common study. In our exercise, the pressure exerted by the tires is a central focus. Pressure is defined as force per unit area and in this case, the force is the weight that the tire needs to support.

The problem specifies a gauge pressure, which is the pressure above atmospheric pressure. For accuracy, we must add atmospheric pressure, 14.7 psi, to our gauge pressure of 85 psi to account for the total pressure exerted by each tire. This is mathematically represented by \(85 \text{ psi (gauge)} + 14.7 \text{ psi (atmospheric)} = 99.7 \text{ psi (total)}\).

Many students are initially puzzled by the different measures of pressure, but it's crucial to remember that gauge pressure plus atmospheric pressure equals the absolute pressure, which is what affects the tire's ability to support weight. Issues with pressure calculations often arise from not considering all relevant pressures or misunderstanding the units used. Ensuring clarity on these points is vital for accurate problem-solving.

Weight distribution is a term often associated with physics and engineering, referring to how weight is spread across an object or system. For the trailer scenario, weight distribution involves how the total weight is supported by the number of wheels. One might overlook that the weight isn't simply supported by one wheel but dispersed among all wheels.

To determine if the weight distribution is adequate, we calculate the weight each tire can support by multiplying the total pressure by the contact area: \(99.7 \text{ psi} \times 100.0 \text{ in}^2 = 9,970 \text{ pounds/tire}\). This step is critical – ensure the contact area is in square inches, as the pressure is in psi (pounds per square inch). By performing this calculation, you identify the maximum weight each tire can support without exceeding the contact area limit.

Problems often arise when learners assume an even distribution without accounting for real-world variables, such as uneven road surfaces. In practice, these factors must be considered to ensure safe and effective weight distribution across all tires.

Problem-solving in chemistry involves analytical and methodical approaches to resolve complex queries, often requiring a blend of theoretical knowledge and practical application. In the context of our textbook exercise, the process involves sequential steps that build upon each other to arrive at a solution. It starts with unit conversion, moves to total weight calculation, incorporates pressure, and ends with the calculation of the number of wheels needed.

Approaching problems methodically, like in this exercise, instills a logical progression from known to unknown and ensures all relevant factors are considered. Additionally, rounding up to the nearest even number because tires are mounted in pairs is an example of applying practical knowledge to a theoretical problem. This not only develops problem-solving skills but also demonstrates the application of chemistry principles to everyday scenarios. Remember, piecing together each part of the puzzle with accuracy and attention to detail is the crux of problem-solving in chemistry.