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Homework Help: Inclined plane/forces exercises

  1. Jan 28, 2017 #1
    Hello everyone, I have a few problems to solve and questions that need answers. The topic is: basic Force exercises and basic applications of the Newton laws. While the general principles are somewhat clear to me, certain types of exercises confuse me to the point where I simply cannot seem to use logic anymore. It's very frustrating, so please bear with me here, I really need some help.

    I know I'll be writing quite a lot for what may seem like simple problems, but I must make sure that my reasoning is correct. I hope you can see that I wrote a lot so that you would only have to read, and then give brief answers, instead of being the other way around where you have to explain the bulk of it. At least I tried to make it so.

    Problem 1
    Two blocks of known mass m1 (20 kg) and m2 (32 kg) are at direct contact with each other, on a non-incline surface, and they are pushed by a Force F applied on the bigger block (m2). Knowing that the resulting acceleration of the Force is a (1.2 m/s2), determine:
    - The contact forces between the two blocks
    - How much they move after a known time t (2.5s)

    Solution attempt:
    (Note: I know there is a difference in speed and velocity; let me just call it velocity here)
    The second point of the problem seems the easiest. The way I solved it is: knowing that the blocks are in contact with each other, they move the same distance. Using the generic equation for motion, which says:
    x (space) = v0 (starting velocity) * t + 1/2 * a * (t2)

    So the answer to the second point is: x = 0 + 1/2 * 1.2 * (2.52) = 3.75 meters

    The first part of the problem is the one where I have issues with.

    First of all, our teacher wants us to write the forces that act on each block (or rather, the Newton law applied to each block individually). This part right here already drives me crazy. My attempt:

    1. For the first body: [Weight] + [Normal force] + [Contact force between m1 and m2] = m1 * a
    2. For the second body: [Weight] + [Normal force] + [Contact force between m2 and m1] + [Force F] = m2 * a

    The first thing that short-circuits my brain is: when writing the forces that act on each individual body, when we consider the "contact force", which one(s) do we need to actually consider?

    I'll post a picture and three possible answers (only one is correct, but I don't know which one):

    Q1 (question 1): is the scheme correct, at least? Is there anything missing? (Proportionality/vectors starting in the barycenter of the bodies don't matter)

    I called "Action" force the force that m2 exerts on m1, as a result of force F pushing on m2.
    I called "Reaction" force the force that m1 exerts back on m1, because of the third law of Newton that says action-reaction and stuff.

    Q2: assuming it is correct, what's the proper way of writing the forces for each body individually? The three options (A, B, C) that come to mind are:

    A. For each body we only consider the forces exerted ON the body, but not the ones the body exerts on other objects. That's how I wrote it earlier.
    A1. [Weight 1] + [Normal 1] + [Action force] = m1 * a
    A2. [Weight 2] + [Normal 2] + [Reaction force] + [Force F] = m2 * a

    B. Opposite of A, so for each body we only consider the forces exerted BY the body, but not the ones exerted ON the body.
    B1. [Weight 1] + [Normal 1] + [Reaction force] = m1 * a
    B2. [Weight 2] + [Normal 2] + [Action force] + [Force F] = m2 * a

    C. Combined solution, we write all of them for each body.
    C1. [Weight 1] + [Normal 1] + [Reaction force] + [Action force] = m1 * a
    C2. [Weight 2] + [Normal 2] + [Action force] + [Reaction force] + [Force F] = m2 * a

    D. One additional possibility/doubt that floats in my brain is that the Force F should be considered when writing the equation for the first body (m1) too, since it's at direct contact with m2. I really don't think that's the case, but it would then look like this:
    D1. [Weight 1] + [Normal 1] + [Reaction force] + [Action force] + [Force F] = m1 * a
    D2. [Weight 2] + [Normal 2] + [Action force] + [Reaction force] + [Force F] = m2 * a

    Q3: solving the exercise. If I had to choose, I would probably pick option A or C of the above, and if I had to pick one, I would go with C. The way I see it, just like we write Weight + Normal force (and that much I know for sure) for each body, we should likewise also write Action + Reaction force for each body (after all, Weight and Normal are just another action-reaction couple).

    However, if that's the case, then the first body should have no motion at all, since (C1) Weight 1 and Normal 1 null each other, and Reaction + Action also null each other. So I would exclude C.

    That's why I thought that option D might be valid. However, since Action = Reaction (for Newton's third), then option D would then become Force F = m1*a and Force F = m2*a, which shouldn't be possible since m1 and m2 are different and a is the same. So I exclude D as well.

    So, going by exclusion, it could either be A or B (unless there is a fifth option E which I'm missing).

    I really can't figure which one is the correct one, and solutions for the exercise were not given. The more I think about it, the less I think logically, so I figured I'd ask here before having a nervous breakdown.

    Anyway, with "option A" the solution would become:

    Action = m1*a
    Reaction + F = m2*a

    What bugs me here is that we don't even need a system (mathematical system, I mean) to solve the problem, because the first equation is already sufficient: knowing m1, and knowing a, Action = 20*1.2 = 24.

    Knowing the Action = Reaction (Newton's third), Action should also be 24 (+24 or -24 depending on the chosen positive direction).

    With "option B" the solution would become:

    Reaction = m1*a
    Action + F = m2*a

    I am aware that the numbers would be the same in the end, but I need to know which one is technically and formally correct.

    If I had to guess, I would say that option A is the correct one. The reason is that m1 (scalar) * a (vector) should give a vector (Action or Reaction) with the same direction as a, and since Action is the vector that has that direction, while Reaction has the opposite direction, then Action = m1*a should be the correct one.

    Q4: this is an additional question/"interpretation" of the problem that I have. Since the blocks are at direct contact, I thought we could also consider them as a single mass m = m1+m2.

    If that were the case, then this unified block "m" would have an equation of forces like this:
    [Weight] + [Normal] + [Force F] = m*a

    Where Weight = Weight 1 + Weight 2; Normal = Normal 1 + Normal 2; m = m1+m2

    Is this correct?

    If that were the case, we could proceed by saying:
    Force F = m*a
    Force F = 62.4

    At this point we could go back to the previous "options", option A and option B, and verify the numbers. With option A:

    Action = m1*a
    Reaction + F = m2*a

    I'll consider positive the vectors pointing to the left (Force F, Action, and a), negative the others (Reaction).

    Action = m1*a = 20*1.2 = 24
    Reaction = -Action = -24
    F = 62.4
    m2*a = 32*1.2 = 38.4

    Therefore, Reaction + F = m2*a becomes: -24 + 62.4 = 32*1.2, which would actually be correct.

    Problem 1 Questions Recap:
    - Q0: is the answer to the second question correct?
    The answer was: x = 0 + 1/2 * 1.2 * (2.52) = 3.75 meters

    - Q1: is the scheme of forces correct? (see image)

    - Q2: is option A the correct representation of the forces for each individual body (block) ? Option A was:
    A1. [Weight 1] + [Normal 1] + [Action force] = m1 * a
    A2. [Weight 2] + [Normal 2] + [Reaction force] + [Force F] = m2 * a

    - Q3: if option A is correct, why is it that for the forces on the Y axis we write both "active" (Weight) and "passive" (Normal) forces, whereas for the X axis we write only the "passive" forces?
    My interpretation, "if I had to guess", is that they are all actually "passive" forces, and we always only write "passive forces". In other words, the Weight force is "passive" because it's a force external to the body, that acts on the body by pushing it down (hence "passive", from the body's perspective); likewise, the Normal force is "passive" because it's a force that "comes from the surface" and acts on the body by pushing it up. Likewise, the force F and force Reaction are obviously external, so they are also "passive" from the body's point of view.

    So, am I to understand that when we write the forces for a body, in a general situation, we only write the forces that act ON the body, but never the forces that the body exerts on other objects? Is this interpretation correct?

    Problem 2
    This one is simpler, I promise.

    On an incline plane of known angle α (30°) lies a block of mass m1 (0.23 grams), connected through a pulley without friction to a second block of mass m2 (0.18 grams). Determine:
    - The acceleration of the block
    - The Tension force
    - The Normal force of the plane

    Solution attempt:
    The simplest of the requests is actually the one that troubles me the most. The Normal force of the plane should be:
    Normal = -Weight1_Y
    (Weight1_Y = Y-axis component of the weight force of the first block)
    Therefore, Normal = Weight 1 * cos (30°), so Normal = 0.23*9.81*0.87 = 1.96

    As simple as it should be, after frying my brain with Problem 1 I now have doubts that there might be more to it, such as some sort of influence on the Normal force by the second object in the system.
    So, in other words, I'm asking if the Normal force in this case is still just -Weight1_Y, or if there's something else to consider.

    Also, note that the problem asks the "Normal force of the plane", so if there is any other "Normal force", other than the plane's reaction to the Weight force (Y component only) of the body that lies on it, please let me know.

    Like in Problem 1, I have to write the forces for each body individually.

    1. [Tension] + [Weight 1] + [Normal] = m1*a
    2. [Tension] + [Weight 2] = m2*a

    Choosing +X towards the motion (assuming block m1 is sliding down), and +Y upwards (assuming block m2 is going up), this would become:

    1. -T + W1_X + W1_Y - Normal = m1*a
    2. +T - W2 = m2*a

    The #1 becomes (W1_Y - Normal = 0): T = W1_X - m1*a
    The #2 becomes: T = W2 + m2*a
    Solving the system, we have W1_X - m1*a = W2 + m2*a, so:
    a = (W1_X-W2) / (m1+m2)
    a = (0.23*9.81*sin (30°) - 0.18*9.81) / (0.23-0.18)
    a = -1.55

    Since acceleration is negative, it means that the motion happens in the direction opposite the chosen axis system. Therefore, I conclude that the block m1 doesn't actually slide down the incline plane, but actually goes up the plane with an acceleration of +1.55 m/s2

    Normally I wouldn't have a problem with this conclusion, but I just want to make sure.

    Anyhow, the Tension is then calculated from the same system, for instance the first equation, so:
    T = 0.23*9.81*sin (30°) - 0.23*(-1.55)
    T = +1.48
    Same number results from T = W2+m2*a (0.18*9.81+0.18*(-1.55))

    Problem 3
    This one should also be simple, but I have a stupid doubt that needs to disappear.

    A block of mass m1 (8.5 kg) lies on a flat (non-incline) plane, with friction coefficient μ (0.2). It is pulled by a force F (32 N) that has makes an angle α (22°) with the plane. Determine:
    - Acceleration
    - After how much time t the body will have a velocity v = 1.2 m/s2

    The doubt I have is in regards to what component of the force F we should consider. Here's a drawing:

    I know there is also a Weight force pointing down, and a Normal force pointing up; I omitted them, as they shouldn't matter.

    First, I need to write the forces on the body, and separate those on the X axis from those on the Y axis. I'll choose +Y going up, and +X going right.

    X. [Friction] + [Force F_X] = m*a
    Y. [Weight] + [Normal] + [Force F_Y] = 0 (since there is no vertical movement)

    So here's the two stupid questions that I have.

    Q1: Friction force = [coefficient] * [weight force_Y] ? I would have to say so, but for some reason in my mind now I have a doubt that it might be:
    Friction force = [coefficient] * [weight force] * [cos (α)]

    In other words, I don't know if I should take into consideration the fact that the force is "inclined" or not. My guess would be not...but my brain refuses to cooperate here, so please just give me the correct answer since I'll never get it on my own.

    So, it's either A or B:
    A. Friction force = 0.2*8.5*9.81 = 16.7
    B. Friction force = 0.2*8.5*9.81* cos (22°) = 15.5

    Q2: assuming we know the correct Friction force, now we need to find the acceleration. Obviously I'll be using the equation on the X axis, but I need Force F_X, which is the X-axis component of Force F. Correct?

    If so, then I would say that Force F_X = Force F * cos (α)
    That's because of this triangle here:
    This sounds obvious and straight-forward to me, but I need confirmation please.

    So at the end it would be:
    a = (-Friction + Force F_X) / (m)
    a = (-16.7 + 32*cos (22°) ) / (8.5)
    a = (-15.5 + 32*cos (22°) ) / (8.5)

    I know this was a long post, so thanks to anyone that will read and reply.
  2. jcsd
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