Initiation of Review

Hi, I'm Chris (from third period physics) and I will be the scribe for today. I simply titled the post "initiation of review" since all we really covered today was a small review of some sample problems involving forces and free body diagrams. Throughout the class, Mrs. K attempted to detail each of the presented cases (found within a small booklet that was also given out today) sufficiently so that the entire class could comprehend the problems at hand. Though, I'm not so sure everyone in the class was quite as comfortable with these cases as she might have thought. In this post, I will outline each of the cases that were covered in today's class with hopefully proficient detail and explanation that the class might delve into the monstrous test on monday with some apt confidence and security.
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Case 1: Constant Acceleration Motion

"If an applied force of 100 N is opposed by friction μ = 0.15 on an object with a mass of 50 kg, calculate:
a) Ff
b) Fnet
c) a"

Now, this is probably one of the simplest dynamics question that we might encounter on the test on monday.
For part (a), the question asks for us to find Ff. Simple enough, we have been taught that to calculate Ff, the equation would be as follows:

Ff = μFn ; where Ff = friction, μ=coefficient of friction and Fn = normal force.

Now since we know that Fn = -Fg, since the normal force is equal and opposite to the weight of the object, we can then substitute Fg into the equation instead of Fn. We would do so because we know the mass of the object, we can then substitute m*g instead of Fg to determine the weight or normal force (since they are equal in linear dynamics), then we would just multiply it by the coefficient of friction to determine the frictional force. This can be shown algebraically as follows:

Ff = μFg = μmg = (0.15)(50 kg)(-9.81 N/kg) = -73.5 N (negative since it is moving left)

For part (b), we are looking for the net force in the free body diagram, and since we know that Fn and Fg are equal, we do not need to take them into consideration. We can also determine from the free body diagram, and also since there is acceleration occurring, is that the applied force is greater than the frictional force. So in order to determine the net force, we must add up the forces Fa and Ff. Since we know both, this portion becomes quite easy.

Fnet = Fa + Ff = (100 N) + (-73.5N) = 26.5 N (since it's positive, it's moving right)

For part (c), the accceleration can be determined from rearranging the equation Fnet = ma to solve for a:

a = Fnet / m = 26.5 N / 50kg = 0.53 m/s²
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Case Two: Force on Two Masses

"If the applied force is 200 N, m1 = 5 kg and m2 = 3 kg, and μ = 0.1, calculate
a) Fnet
b) a
c) F₂"

Now this question presented tribulations for the class as Mrs. K attempted to present how to approach this problem to the class.
For part (a), to calculate the net force on the objects, all we must do is add Ff and Fa together (since once again, Fn = Fg). Now, we already know Fa, but we then need to determine Ff. Since Ff = μmg, as I outlined earlier, we need to apply that once again but instead of having just one mass, we have two, so we must combine the two masses to get m. By this I mean that in this instant we treat the two masses as one mass by combining them into one. This will give us:

Fnet = Ff + Fa = μ(m₁+ m₂)g + Fa = (0.1)(5 kg + 3 kg)(-9.81 N/kg) + 200N = 192 N

Now for part (b), we are attempting to determine the acceleration of the two masses. All we must do for this is use the net force (since it is the total of all forces in the situation) and divide it by the total mass of the object, once again adding m₁ + m₂. Algebraically:

a = Fnet / (m₁+m₂) = 192 N / 8 kg = 24 m/s²

Part (c) requests that we now determine F₂, this usually presented trouble for some students so I will try my best to outline the process of calculating F₂. In order to calculate this force value, we must see that F₂ is caused by mass 2 and is actually driving m₂ forward, but is pushing back on m₁. This is a direct result of Newton's third law, that every action will have an equal and opposite reaction. We must also realize that this force is a ratio of the overall force being applied, in other words, it is a specific portion of the 200 N being applied to the overall object (m₁ and m₂ together). To determine this so-called ratio, we must divide the mass of m₂ by the total mass, since we're basically looking for how much of the total object is just m₂. Now, here's how the calculations should look like:

F2 = [m₂ / (m₁ + m₂)] F = (3/8) (200 N) = 75 N
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Case Three: Connected Masses

"If m₁ = 2 kg, m₂ = 4 kg, m₃ = 1 kg, and Fa = 10 N, calculate:
a) a
b) T₁
c) T₂"

To determine the acceleration, since there is no friction acting (and Fg = Fn, so they quantitatively cancel each other) Fnet = F. Also, in this case, we must incorporate all three masses into our calculations using m since F is indirectly acting (pulling) the entire connection of the masses together. Now, to calculate acceleration, we can use the formula F = ma rearranged to solve for acceleration:

a = F / (m₁ + m₂ + m₃) = 10N / 7 kg = 1.43 m/s²

To determine the first tension force (acting on m₁ and m₂), we can use the first free body diagram to carry out this calculation, seeing as it is more simple. Now, as we inspect the first diagram, we notice that the only force acting on the object IS T1, meaning that the net force in this case IS T₁. So, in order to calculate T₁, we can use the equation T₁ = m₁a.

T1 = m₁a = (2 kg)(1.43 N/kg) = 2.86 N

For T₂, we must use the third free body diagram to calculate it since we can use F as a constituent of the net force to determine T₂. This, in other words, means that we can use the net force (the total of the forces) to determine T₂ since F is known. The vector T₂ would be negative in this situation, since it is moving left, and F would be positive as it would be moving right. This doesn't always have to be the case, but I have designated left negative and right positive, or the force T₂ negative with respect to F. Since the net force is not known, but m₃ and a are both known, we can use those values in place of Fnet since Fnet = m₃a. This gives us:

T₂ --> F - T₂ = m3a --> T2 =-(F - m3a) --> T2 = -[10N - (1 kg)(1.43 N/kg)] = -8.57 N
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Cases Four and Five:Mass on an Incline

Now, for these cases (though we did not go over them in class, but were told to complete them) I will outline how to come about solving such questions. These inhabited a majority of the current unit and is quite likely to be found frequently throughout the upcoming test. Now, the following diagram will be used to describe both scenarios.

The normal force, Fn, will always act perpendicular to the surface. This meaning that if the mass is resting on a surface, the normal force exerted on the mass will form a 90^o with the surface. Cases which involve an object and an incline will always direct gravity downwards, regardless of the slope of the surface, the direction of the normal force, etc. Gravity is always pointed down (as shown in the diagram) in an inclined plane problem. Now, in order to conduct calculations, we must break the force of gravity, Fg, or weight (all the same thing), of the object into two components, a parallel component and a perpendicular component. These are basically just x and y components (or vertical and horizontal components) of the force of gravity, but since Fg is now acting as the hypotenuse pointed directly down, they are named according to their position in the diagram. By this I mean that these components are named relative to the surface, as with in the case of the normal force. These components are illustrated in the freebody diagram depcited above. To calculate these components, we visualize the triangle formed by breaking Fg into it's components as a right triangle, where Fg itself is the hypotenuse. The angle θ will be formed between Fg and F perpendicular, and from there on use basic trigonometric functions (particularly sinθ and cosθ) to determine either F perpendicular or F parallel. And if Fg is not given, but the mass is, then mg can always be used to substitute for Fg.

F parallel plays a specific role in an inclined plane. Fg does not act directly downward on the surface, but in two ways, thus why Fg was dissected and broken down into two components. F perpendicular simply acts equal and opposite to Fn, similar to how gravity does in linear scenarios. But F parallel acts as Fa, driving the object down the incline. The only force to oppose this force (with respect to what we have calculated as of yet) is friction. When friction comes into play, the process might seem more complex and intuitively complicated, though contrary to popular belief this might not be entirely true. As in the case of linear dynamics, Ff opposes Fa (in this case F parallel) and then determines what motion will occur. Here are three important tidbits to remember when regarding net force and Ff vs. Fa:

- If Fa is less than Ff, then the object is not moving.

- If Fa = Ff, then the object is moving with a constant velocity.

- If Fa > Ff, then the object is accelerating.

Continuing with the explanation of frictional force in an inclined plane is imminently complete. As I noted earlier, Ff opposes F parallel, but also, in the case of all situations regarding frictional forces there underlies a specific coefficient of friction (the ratio of friction over the normal force). Using either of these two facts leads to the calculation of the frictional force acting on an object on an inclined plane. To use these two ways, either subtract the F parallel force from the net force to get the frictional force, or manipulate/directly use (depending on for what purpose) the equation Ff = μFn.
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Now, I hope many of you, especially those who were absent from today's class, have acquired more of a formally but yet simplified (to an extent) explanation of some of the main concepts and processes within this unit.

As our class finished covering the above through a somewhat less elaborate volubility of explanation, we were given two sheets entitled "C. Using Concepts" and "Grade 12 Physics: Dynamics Problems" to complete. Though I am not sure whether they are for homework in the sense that we will correct them or hand them in or not, I strongly encourage many to not only attempt and complete the questions, but to shamelessly inquire as necessary. These questions will not only provide more practice and a more definitive grasping of the subject at hand, but also a stronger habitual behaviour to approach problems that we might encounter on the upcoming test. As for the test, it will be on Monday, though some argue (can't remember who else besides Vincent) that the test should be on Tuesday. That is all for my scribe post, I know it was lengthy, albeit an obvious resultant of my instructional endeavor for the rest of the class(es). Hope this post can help anyone at all in this unit, have a great night and don't forget to do homework!

Oh yes, the scribe for tomorrow will be Anthony!