Newton's Laws Legacy Problem #14 Guided Solution

A good problem solver seldom rushes to their calculator in order to find the answer. Instead, they take the time to get a good visual and conceptual understanding of the situation. They take the time to read the problem and construct a diagram to represent the situation. Often times, like in this unit, it will be a free body diagram that represents the forces acting upon our object. Then they take the time to list what they know and what they are looking for. They may even list this information on their diagram. Finally, they take the time to actually plot out a strategy to get from the given information to the unknown information. Here in this problem, we read about Casey who is in a wagon being pushed by his brother Skyler along the street. We observe that there is a forward force on the wagon applied by Skyler. We would call this an F applied or FF and we would represent it on a diagram, a free body diagram, by a forward or a rightward arrow. There is also a resistance force, probably air resistance and friction, that would be pointing in the opposite direction that the wagon is moving, here to the left. Its value is 24 newtons and I would probably label that on my diagram as F resistance or F fret. Then we observe that Casey and the wagon weigh 304 newtons. That is a down force. You need to know that, that the weight is a force of gravity acting upon the object. So represent it by a down force with an arrow and label it F norm or F grad and put the number 304 newtons next to it. Finally, there is one fourth force not explicitly referenced here, but we know it to be the support force of the street pushing up on the wagon to support the wagon's weight. In cases in which there is no vertical acceleration such as this, that up force or F norm is equal to the down force. Thus, it is also 304 newtons. Now, as you approach problems like this, the first step is to get that diagram and list what you know. List the force values and then list what you are looking for. What you are looking for is net force, mass, and acceleration. We can get to the net force directly off the diagram because we know all individual forces which are acting upon the object. We just need to add them up as vectors. The 304 down and the 304 up add up to zero newtons. The 52 newtons to the right and 24 newtons to the left, they would add up as 28 newtons. So, you have a 28 newton force to the right, or we would just say forward in the direction that Casey and his wagon are moving. So, with that 28 newton forward force, you can now use it to calculate the acceleration. But before you do, you need to know the mass because A equals F net over F. So, you have to ask, how do I get the mass from this information? That is where a good conceptual understanding comes in. You have to understand that the mass and the weight, or the mass and the F graph, are always related to one another. The weight is equal to mg, where g is 9.8 newtons per kilogram, and where the weight is simply the force of gravity acting on the object. So, you take your 304 and plug it into the left side of the equation, weight equals mg, and put your g in and solve for the mass in kilograms. Now that you have the mass in kilograms, it is a straightforward process, A equals F net over F, and you can get your acceleration. Don't forget about the direction. It is always, always, always in the same direction as the net force. According to your diagram you have drawn, that would be to the right. We would probably just say forward, whatever way Casey is pulling.