Join Ryan on a daredevil stunt with a bike, go-cart, and skateboard in Episode 29.
Join Ryan on a daredevil stunt with a bike, go-cart, and skateboard in Episode 29. Objects can be evaluated individually or as part of a larger object we call a “system.” (1:45) Sketch the force diagrams for each object, (2:25) sum the forces on each object and set that net force equal to mass times acceleration. (6:20)
Question of the Day (9:38) If a 1 kg mass is pushed by a 9 N force into a 2 kg mass along a frictionless surface, how fast will they accelerate?
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Hi and welcome to the APsolute RecAP: Physics 1 Edition. Today’s episode will focus on dynamics as they apply to systems of objects.
Let’s Zoom out:
Unit 2 – Dynamics
Topic - 2.1-2.7
Big Idea – Systems, Field, Force Interactions, Change
As a child, podcast author Ryan used to love riding bikes with friends on his street. They used to get quite creative in ways that they could have fun with their bikes. One time, he tied a rope from his bike to a small pedaled go-cart. The go-cart then had a second bit of rope tied to the rear and connected it to a skateboard. Hopefully you now have the ludicrous and somewhat dangerous image in your head of him as a boy trying to pedal his bike while one hsi friends steers the go-cart that is being dragged behind the bike, and then at the end a third friend with much less ability to steer himself is seated on the skateboard. They all got ready for the ride of a lifetime, and he began to pedal to take out the slack in the rope. He then began to apply as much force to the pedals as he could to begin their journey, and then SNAP! The rope broke. My question for you is, which one?
Let’s Zoom in:
Objects can be evaluated individually or as part of a larger object we call a “system.” In the case of the bike - go-cart - skateboard scenario, we can look at each boy individually and sketch its free-body diagram, or we can look at the entire system and treat it as one more massive object and sketch the free-body diagram. Let’s start by looking at the objects individually first so that you can make sense of how we arrive at the FBD for the system. This would be a good time to grab a pencil and paper, so you can see what I am describing.
Let’s look at the free body diagram for the friend sitting on the skateboard first. It is easiest to look at him first since he has no force pulling him backwards if we assume friction is negligible. There is gravity downward from the Earth, and normal force upwards from the pavement, and then a tension force forwards from the rope, let’s call that tension 1. Gravitational force and the normal force happen to be equal in this case since I can assume there is no vertical acceleration. This leaves tension 1 to be the net force applied to the skateboard sitter, and therefore tension 1 is equal to the mass of the skateboarder (m skate) times his acceleration (a).
Now let’s look at the FBD for the go-cart driver. He too will have a gravitational force down, a normal force up, a tension force from the rope attached to the bike which we will call tension 2, and then because of Newton’s 3rd law he will have to possess a backward tension 1 force that is equal but opposite to the tension 1 we drew on the skateboarder. Remember if the go-cart pulls on the skateboard, then the skateboard MUST pull on the go-cart equally and in the opposite direction. 3rd Law Pair! The gravitational force and normal force are again equal since the go-cart never accelerates vertically. The go-cart accelerates forward in the horizontal direction, so there must be a net force forward after summing the forward tension 2 and the backward tension 1. That means tension 2, MUST be greater than tension 1, so draw it this way. Using the convention of forward being positive, tension 2 minus tension 1 is the sum of the forces and can be set equal to the mass of the go-carter (mass GC) times his acceleration (a) which is the same acceleration as the other individuals since they are all connected via the ropes.
Finally, the FBD for the bike rider must have a gravitational force down, a normal force up, a tension 2 force backward because again… 3rd Law Pair, and finally a forward force we will just call “F”. In reality that forward force is due to static friction between the bike’s tires and the asphalt. Again, the biker is accelerated forward, but not vertically, so normal and gravitational forces are equal, but “F” must be bigger than tension 2 in order for the biker to accelerate forward. And, we come up with the equation “F” minus tension 2 equals the product of the biker’s mass (m bike) and his acceleration (a) which is again the same as the others.
Now, a question like this may say something to the effect of, find the force “F” if the friends are to accelerate at 2 m/s^2 and they each have a mass of 30 kg. Hopefully, you can appreciate that having 3 equations and this many unknowns while solvable, is somewhat challenging given the amount of algebraic gymnastics you’d have to go through. Instead, a problem like this may be better viewed as a system of objects that acts like one larger object with a mass of 90 kg and accelerating at 2 m/s^2. We could sketch a FBD for the system, but what forces should we include? There were so many! Well, some of those forces are not external to the system because they were 3rd law pairs, and therefore will cancel each other out when we look at the whole system. Tension 1 and 2, SEE YA LATER! And, we are then left with an FBD that has “F” forward, normal force up, and gravitational force down. We do need to know that when finding the force of gravity m*g, the mass we use is the 90 kg for the combined system’s mass, but normal force is still equal since there is no vertical acceleration, and then the force “F” is just 90 kg times 2 m/s^2 or 180 N.
If the equation had asked for either tension force, we could have repeated a similar methodology and then applied what we have learned to each of the individual free body diagrams. For example, we can find tension 1 on the skateboarder by multiplying his 30kg mass by his 2 m/s^2 acceleration. Tension 1 must by 60 N. Tension 2 is the worst case scenario since you would have had to complete all of the work mentioned and look at the go-cart FBD to arrive at tension 2: 60 N = 30 kg times 2 m/s^2. Tension 2 must equal 120 N, and VOILA! We now have our answer, the rope between the bike and go-cart would break first assuming the ropes were the same type and knotted sufficiently. Tension 2 is greater than tension 1 because that bit of rope needed to apply force to a larger amount of mass to accelerate it the same 2 m/s*2.
To Recap…
Dynamic systems can be confusing, so hang your hat on a trusted method of solving these types of problems. Sketch the force diagrams for each object, sum the forces on each object and set that net force equal to mass times acceleration. Then, follow the same set of steps for the system as a whole remembering to eliminate internal forces that arise due to 3rd law pairs.
Coming up next on the APsolute RecAP Physics 1 Edition, we start preparing for the exam and point out what is worth memorizing from each unit and tidbits worth storing in your calculator.
Today’s Question of the Day focuses on Dynamic Systems.
Question:
If a 1 kg mass is pushed by a 9 N force into a 2 kg mass along a frictionless surface, how fast will they accelerate?