Unit 2 Example Problems: Forces AP Physics B

Welcome to this little tutorial about the Unit 2 forces example problems. The focus of this tutorial, and of these problems, is to familiarize you with the following ideas… - Weight - Normal force -How to apply Newton’s 2 nd law, and equilibrium ideas, to a variety of situations with multiple forces acting.

Intro : The format of this tutorial will be to first introduce some important ideas that will be helpful on a variety of problems, and then to work through each example problem. So if all you’re looking for is help on a particular problem, skip ahead to that problem.

Background : Weight is defined as the magnitude of the force of gravity acting on an object. On earth, we always calculate weight in the following manner: F g = mg =(mass)(9.8m/s 2 )

Background : Normal force is defined as a contact force exerted by one object on another, in a direction perpendicular to the surface of contact. On our basic problems, F N will typically be an upward force acting on an object that is resting on a table or the ground or some other flat horizontal surface.

Background : Newton’s 2 nd law is by far the most important formula we’ll be using this chapter. It can be stated mathematically in the following manner: ΣF = ma ΣF = net force acting on an object m = mass of the object a = acceleration experienced by the object, when the net force acts on it

Background : ΣF = ma This formula can be easy to use, if two of the three variables are given to you. Things get a little more complex, though, as soon as multiple forces have an effect on the object’s acceleration. Notice that ΣF doesn’t just mean force. It means net, or total, force.

Background : ΣF = ma So if multiple forces are involved, they all become part of the net force, ΣF, which is the vector sum of all forces on the object. Though that last statement sounds difficult, it really doesn’t have to be. The standard way to think about ΣF is to think of forward forces (helping the accel.) minus resistive forces (hindering the accel.)

Background : ΣF = ma With the previous thinking, on multiple-force problems, Newton’s 2 nd law really becomes… Forward – Resistive = ma

Example #1a : Calculating weight is easy. We always find it in the same way. F g = mg = (5)(9.8) = 49N Normal force can be found with the 2 nd law ideas we just talked about… Forward – Resistive = ma F N - F g = ma F N - 49 = (5)(0) → F N = 49N

Example #1a : Another simpler way to think about the normal force part of the problem is with the idea of equilibrium. Equilibrium is simply a state of balanced forces, characterized by zero acceleration. On this problem, it is easy to see that the rock is in equilibrium, since it’s not even moving. Therefore, forces must be balanced, so… F N = 49N

Example #1b : Nothing changes about weight of the rock. F g = mg = (5)(9.8) = 49N Normal force… Forward – Resistive = ma F N - F g = ma F N - 49 = (5)(2.3) → F N = 60.5N

Example #2 : This one’s quite similar to the last one… Forward – Resistive = ma F Tension - F g = ma 22 - (m)(9.8) = (m)(2.5) m = 1.79kg

Example #3 : For connected objects, the important thing is to realize that the situation can be viewed in several different ways, depending on which way is easiest.

Example #3 : One way is to view the entire system of objects as one thing, all experiencing one acceleration, influenced by forces that are external to the system. This way is great for simplifying the problem. This way doesn’t work at all if the problem deals specifically with force that are internal to the system. (Like the tension in a cord connecting two blocks, for instance.)

Example #3 : The other way to view the problem is to look at just one object at a time. This way tends to be more complicated, since internal forces come into play. But this way is necessary if those internal forces are what you’re looking for on the problem.

Example #3a : Since this is just asking for acceleration of the whole system, and not asking anything about internal forces, view the entire system as one object. Forward – Resistive = ma = (58)(a) a =.862m/s 2 Notice that the mass is 58kg, because we chose to think of the system as one big object.

Example #3b : Now we’re interested in a force that’s internal to the system, so we have to look at smaller parts of the system. Let’s choose just to look at the front 30kg block. Forward – Resistive = ma 50 - F Tension = (30)(.862) F Tension = 24.14N

Example #3b : One point about the previous method is that it’s only really helpful if you already know the acceleration of the object. (Luckily, it will always be the same as the acceleration of the system, for connected objects.)

Example #3b : Another point is that you can arrive at the right answer, even if you choose to look at a different portion of the system. For example, this problem can be solved by thinking of the last 2 connected blocks as a 28kg system. Forward – Resistive = ma F Tension - 0 = (28)(.862) F Tension = 24.14N

Example #3b : One last point about the previous problem is that it doesn’t really make sense to break apart the 2 nd and 3 rd objects to look at either one individually. This is because there’s another cord between them, and therefore internal tension forces that we don’t care about.

Example #4 : This problem is just a twist on the last idea. The only thing that’s weird is that you have to think of ‘forward’ and ‘resistive’ in reference to the total motion of the system, rather than as ‘up’ or ‘down’.

Example #4 : Forward – Resistive = ma F g of the 8kg - F g of the 5kg = ma (8)(9.8) - (5)(9.8) = 13a F Tension = 2.26m/s 2 Notice that the mass is 13kg, because we chose to think of the system as one big object.

Example #5 : This problem really isn’t too difficult for vector-experts like you all. If we had to categorize it, we’d call it an equilibrium problem in two dimensions. All this means as the object is sitting still, so it’s not accelerating, so it’s in a state of equilibrium. And if it’s in equilibrium, then forces are balanced.

Example #5 : And if forces are balanced, then we can choose to think of those balanced forces in terms of their components. Then we would say that the x-direction forces are balanced, and that the y-direction forces are also balanced.

Example #5 : Let’s begin by thinking about the x-direction forces (because there’s only two of them to consider). The object isn’t moving sideways, so the x- component of Tension 1 must equal the x- component of Tension 2. T 1x = T 2x T 1 cos24 = 250cos52 T 1 = 168.5N

Example #5 : For the 2 nd part, which has to do with y- direction forces (like weight), let’s think about the y-components of the forces. The object isn’t moving vertically, so any upward forces must be balanced by downward forces.

Example #5 : T 1y + T 2y = F g 168.5sin sin52 = F g F g = 265.5N

Example #5 : One related note that might help you on a problem or two on your HW: If an object is supported by just one cable instead of two, you still think as if there are two tensions acting on the object. But since the two tensions are really in the same cable, they must be equal in strength.

Example #6 : This one’s just an interesting twist on force vector components and Newton’s 2 nd law. Begin by thinking about the following diagram of the yo-yo and forces on it… Fg Fg FT FT Notice that the tension force can be broken into components.

Example #6 : When you draw in the components, you notice that they can be thought of as accomplishing two different things… The y-component is balancing F g. Fg Fg FT FT The x-component is responsible for causing the yo-yo to accelerate forward.

Example #6 : Now the math… In the y-direction, the yo-yo is in equilibrium. Fg Fg FT FT F Ty = F g F T cos34 = (0.032)(9.8) F T = 0.378N

Example #6 : In the x-direction, the yo-yo is accelerating. Fg Fg FT FT F Tx = ma (0.378)(sin34) = (0.032)(a) a = 6.61m/s 2

Example #6 : One related note that might help you on a problem or two on your HW: On the problem we just worked, we really would not have needed to know the mass of the yo-yo. Even though we used it in the problem, if we would have just written ‘m’ where we used it, we would have noticed that eventually an ‘m’ on top and an ‘m’ on bottom of a division would cancel.

The End : Good luck on the assignment. I hope this tutorial made some sense. Feel free to revisit any parts of it that you need to as you work the problems over the weekend. See you Monday!