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Acceleration on an inclined plane

  1. Apr 4, 2016 #1
    • Member advised to use the formatting template for questions posted to the homework forums
    I have a problem with conceptualising what happens to an object as it travels down an incline slope. Online resources say that as the gradient of the slope increases the acceleration increases, but isn't this incorrect in a sense.

    Shouldn't it be said that as the incline of a plane increases the component force in the direction of motion increases but the resultant of the components of acceleration stays the same, i.e. 9.81 m/s2? OR am i losing my mind and gravity isn't constant anymore...

    Because I am trying to investigate the difference in acceleration of a two bodies of different masses at rest on an incline plane. To my understanding, these both have the same components of acceleration in the x-y axes and the same resultant acceleration. Once they're 'released' then their velocities will definitely be different, but their acceleration should be identical given no external factors such as friction/drag?

    Thanks!
     
  2. jcsd
  3. Apr 4, 2016 #2

    BvU

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    Hello Sirsh,

    You have me confused somewhat: are you looking at straight inclines and studying the effect of steeper / gentler or are you looking at curved inclines where the slope increases/decreases during the motion ?
    This too is confusing: if accelerations are the same and both start with velocity 0, then how can the velocites be different ?

    Can you perhaps post a few drawings for a few cases ?
     
    Last edited: Apr 5, 2016
  4. Apr 4, 2016 #3
    Sorry for the confusion, let me clarify some more.

    I have two masses, A and B. A is 10kg and B is 5kg. They both are modeled separately on an incline plane which is 10 degrees from the horizontal.

    I've been told to compare the initial acceleration of these two masses from rest.

    I have calculated the acceleration in both the x and y axes using Newton's 2nd Law, as the mass cancels out when you do this, you end up with the same values for acceleration in both directions and hence the same resultant. As it's dependent upon the angle of incline only.

    So how could you 'compare' these two masses?
     
  5. Apr 4, 2016 #4

    BvU

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    Seems to me the two masses behave identically.
    Galileo found that out dropping things from the Pisa tower long ago. Big stone and small stone arrive at the same time when let go at the same time.
    Ambiguous question :smile: If you want to know which of the two is heavier, this kind of experiment doesn't help you, as you calculated correctly.

    So far all this is without any friction we assume, right ?
     
  6. Apr 4, 2016 #5
    Thank you, that's what I thought as well.

    There is no explicit friction stated in the problem, but there is drag due to wind. However, it is modeled as a function of velocity, and at rest this would equate to zero. Thus, leaving us with the situation of two stationary blocks.

    However, I need to find the maximum velocity of the blocks. I tried to do a force balance and solve for the velocity in the drag equation but had an issue.

    if drag is equal to 0.5*v2, for mass A of 10kg, and a incline of 20°. I have this force balance in the x-axis:

    Fx=max
    mg*sinθ-0.5v2 = max
    If I rearrange this for v, I get:
    mg*sinθ-max=0.5v2
    v2 = (mg*sinθ-max)/0.5
    v = √((mg*sinθ-max)/0.5)

    However, if I use the acceleration in the x direction from the initial at rest conditions, then this is zero. So how would I go about figuring out the maximum acceleration?
     
    Last edited: Apr 4, 2016
  7. Apr 4, 2016 #6

    BvU

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    If the force is zero when the velocity is zero, the wind isn't blowing and we usually call that air resistance :smile:.
    At least: initally stationary. But I do expect you let them start sliding down at t = 0 ?
    Is a differential equation ([edit] corrected, see below)$${d^2x\over dt^2} + {1\over 2m} {dx\over dt} - mg\sin\theta = 0 $$with inital conditions ## {dx\over dt}_{t=0} = 0## and ##x_{t=0} = x_0## do you know how to solve such equations ?

    Quantitatively you can see the maximum acceleration is when ##v^2 = 0##, so at t=0. After a while,a terminal velocity is reached , for which ##a_x## = 0.
     
    Last edited: Apr 5, 2016
  8. Apr 4, 2016 #7
    I'm not 100% sure how to solve that differential equation, my first thoughts are that I can cancel the first order term out and then rearrange the equation so that d2x/dt2 = mgsinθ and then maybe integrate it once so then it's in the form of dx/dt which is the equivalent to velocity?

    Ah! I was assuming that when ax = 0, then it'll be at zero velocity, but its actually at maximum velocity because it cannot accelerate any further, but is not decelerating!
     
  9. Apr 5, 2016 #8

    BvU

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    Yes, so solving ##a_x=0## from your ##mg\sin\theta-ma_x=0.5v^2## (which then no longer is a differential equation but a simple equation that in fact you solved already) gives you the maximum velocity (aka terminal velocity).

    ----
    I really have to correct myself with this differential equation o:)o:)o:) ; it should have been $$
    {d^2x\over dt^2} + {1\over 2m} \left ({dx\over dt}\right )^2 - g\sin\theta = 0
    $$and I am surprised neither you or anyone else jumped in to point that out !!​
    ----

    As to this full equation for the falling velocity, you can simply change to a 'new' variable, ##v = {dx\over dt}## to get $$
    {dv\over dt} + {1\over 2m} v^2 - mg\sin\theta = 0
    $$ with initial condition ##v_0 = 0 ##.
     
  10. Apr 5, 2016 #9
    I was actually going to note that it wasn't squared! But didn't post until you had replied.

    So for my mass A which is 10kg going down a 20 degree slope this is my interpretation of how to find it's terminal velocity (acceleration is zero in the axis of motion).

    [itex] F_x = ma_x [/itex]
    [itex] F_x = mgsin(\theta)-0.5v^2 [/itex]
    [itex] mgsin(\theta)-0.5v^2 = ma_x [/itex]
    [itex] v = \sqrt(mgsin(\theta)-ma_x/0.5) [/itex]
    As ax is zero at terminal velocity, then:
    [itex] v = \sqrt(mgsin(\theta)/0.5)[/itex]
    Therefore, for the conditions of this situation:
    [itex] v = \sqrt(10*9.81*sin(20°)/0.5) = 8.19 m/s [/itex]
     
  11. Apr 6, 2016 #10

    BvU

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    That's what I get, too.
     
  12. Apr 6, 2016 #11
    Awesome, thanks for your help it's really appreciated.

    If it's not a hassle, would you be able to explain how I could figure out the terminal velocity of the block if it was travelling up the incline in addition it has 100 W of energy assisting it.

    From my understanding the acceleration components are the same as for when it was going down the incline, but because I have assigned my positive x axis as down the ramp, my motion will be going in the negative direction.

    Does this seem the correct way to model the velocity up the incline? I'm assuming that the 100W will assist it 'up' the incline so it'll be in the same direction as the motion.

    [itex] F_x = ma_x [/itex]
    [itex] F_x = 0.5v^2-mgsin(\theta)-100 [/itex]
    [itex] 0.5v^2-mgsin(\theta)-100 = ma_x [/itex]
    [itex] v = \sqrt((ma_x+mgsin(\theta)+100)/0.5) [/itex]
    As ax is zero at terminal velocity, then:
    [itex] v = \sqrt((mgsin(\theta)+100)/0.5)[/itex]
    Therefore, for the conditions of this situation:
    [itex] v = \sqrt((10*9.81*sin(20°)+100)/0.5) = 188.87 m/s [/itex][/QUOTE]

    This seems like a crazy amount of increase in speed for the block?

    Thanks again.
     
  13. Apr 6, 2016 #12

    BvU

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    100 W can not be in this equation: it has the dimension of power, not force.
    So we'll have to think of something else !
     
  14. Apr 6, 2016 #13
    Wow, that totally slipped my mind, the units. Obviously I need the equivalent of 100 W in newtons to use it in the equation that I wrote above.

    Power is the amount of work done in a time period, however, in this case there is no time period stated. P = W/t = F*d/t = F*v if I rearrange this to be dimensionally correct I get F = P/v. My initial thoughts are that I may be able to put into my equation + 100/v, but then the equation becomes a quadratic I think and then I may be able to solve for the velocity at ax = 0?
     
  15. Apr 6, 2016 #14

    BvU

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    Looks good to me. It becomes third order, actually, so you may need a solver (or you can iterate).
    And: If you go up the slope, the air resistance points the other way !
     
  16. Apr 6, 2016 #15
    [itex] F_x = ma_x [/itex]
    [itex] F_x = 0.5v^2-mgsin(\theta)-(100/v) [/itex]
    [itex] 0.5v^2-mgsin(\theta)-(100/v) = ma_x [/itex]
    As ax is zero at terminal velocity, then:

    [itex] 0.5v^2-(100/v) =10*9.81*sin(20°)[/itex]
    Solving this with the quadratic equation leaves me with the solutions v = -5.60, -3.79 and 9.40 m/s. I'm 99% sure that it should be 9.40 m/s but because my positive direction is stated to be down the ramp, it makes me think that the velocity is down the ramp even though the block is supposed to be travelling up the ramp. Otherwise, maybe my direction assumption is wrong?

    Thanks again.
     
  17. Apr 6, 2016 #16

    BvU

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    Posts crossed. You are searching in the right direction though!
    But I can't follow "solving with the quadratic"equation ..." ?

    If you want to make things easier conceptually: start without air resistance. easy to get the velocity, and with resistance it can only be smaller...
     
  18. Apr 6, 2016 #17
    Okay, modeling without air resistance, I'm going to take up the ramp as positive in this case and down the ramp as negative.
    [itex] F_x = ma_x [/itex]
    [itex] F_x = 100/v - mgsin(\theta) [/itex]
    [itex] 100/v - mgsin(\theta) = ma_x [/itex]
    At terminal velocity ax = 0.
    [itex] 100/v = mgsin(\theta) [/itex]
    [itex] v = 100/mgsin(\theta) = 100/(10*9.81*sin(20°) = 2.98 m/s[/itex] (up the ramp)
     
  19. Apr 6, 2016 #18
    This is my interpretation of the system:
    Screen_Shot_2016_04_06_at_6_19_33_pm.png
     
  20. Apr 6, 2016 #19

    BvU

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    Beautiful picture :smile:, and yes, I get 2.98 m/s as well.
     
  21. Apr 6, 2016 #20
    So that was w/o air resistance, this is with air resistance:

    [itex] F_x = ma_x [/itex]
    [itex] F_x = 100/v - mgsin(\theta)-0.5v^2 [/itex]
    [itex] 100/v - mgsin(\theta)-0.5v^2 = ma_x [/itex]
    At terminal velocity ax = 0.
    [itex] 100/v - 0.5v^2 = mgsin(\theta) [/itex]
    [itex] 100/v - 0.5v^2 = 10*9.81*sin(20°) [/itex]
    Using a solver to solve this equation, yields:
    v = 2.69 m/s and like you said should be smaller than without the resistance which was 2.98 m/s!
     
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