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Force vector mechanic help

  1. Aug 22, 2009 #1

    Mentallic

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    Homework Helper

    1. The problem statement, all variables and given/known data
    http://img41.imageshack.us/img41/6579/mechanics2.jpg [Broken]

    2. Relevant equations
    [tex]w=\frac{d\theta}{dt}[/tex] (1)

    [tex]v=rw[/tex] (2)

    [tex]F=ma[/tex] (3)

    [tex]a=\frac{v^2}{r}[/tex] (4)



    3. The attempt at a solution
    For (i), from (2) and (4) I get [tex]a=rw^2[/tex]

    and I'm thinking [tex]r=\frac{F}{cos\theta}[/tex] but I'm not too sure about it.

    So then I'm left with [tex]cos\theta=mw^2[/tex] for the horizontal component.

    For the vertical, I'm just completely stumped. I tried something, but I know it's probably so wrong that posting it would be to make a mockery out of myself :blushing:

    For (ii) I have [tex]N=mcos\theta(g-rw^2tan\theta)[/tex]

    and since [tex]mcos\theta>0[/tex] since m>0,[tex]cos\theta>0 ,0<\theta<\frac{\pi}{2}[/tex]

    I need to solve for w: [tex]g-rw^2tan\theta>0[/tex]

    Hence, [tex]-\frac{g}{r}cot\theta<w<\frac{g}{r}cot\theta[/tex]

    Is this correct?
     
    Last edited by a moderator: May 4, 2017
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  3. Aug 23, 2009 #2

    zcd

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    Re: Mechanics

    that can't be right, because F is a force vector (dealing with acceleration) and r is a position vector (which can be treated as a scalar here).

    For uniform acceleration to occur, [tex]\vec{\tau}_{net}=\vec{F}_{net}\times\vec{r}=0[/tex], and we already know that [tex]r\neq0[/tex], so [tex]F_{net}[/tex] must equal to 0. Then resolve it into x and y components.
     
  4. Aug 23, 2009 #3

    Mentallic

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    Re: Mechanics

    Sorry where did you get that equation from? I don't understand what
    [tex]\vec{\tau}_{net}[/tex] is.
     
  5. Aug 23, 2009 #4
    Re: Mechanics

    I'm pretty sure that [tex]\tau[/tex]net is the net torque on the particle P. Since the particle P does undergo uniform circular motion there is a torque on particle P, and he was simply using [tex]\tau[/tex]net = Fnet x r = 0 to show you that since (r) isn't 0, Fnet must be equal to 0 in order to fulfill this equation.

    Hope this helps!
     
  6. Aug 24, 2009 #5

    Mentallic

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    Re: Mechanics

    ok thanks for that explanation.

    So we need to resolve the x and y components. How can this be done?
     
  7. Aug 24, 2009 #6

    gabbagabbahey

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    Re: Mechanics

    You seem to be claiming that for a cross product of two vectors to be zero, at least one of the vectors must be zero. That is a false claim.

    As a counter example, consider the vectors [itex]\textbf{u}_1=2\textbf{i}[/itex] and [itex]\textbf{u}_2=3\textbf{i}[/itex]...what is their crossproduct? Are either of these vectors zero?
     
  8. Aug 24, 2009 #7

    gabbagabbahey

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    Re: Mechanics

    Okay, and which direction is this acceleration in? What net force would cause such an acceleration?
     
  9. Aug 24, 2009 #8

    Gib Z

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    Re: Mechanics

    Other posters note, the course Mentallic is doing does not cover cross/dot products, vectors in the "advanced" sense, torque etc . It only allows for elementary calculus to solve the problems.

    Mentallic - The hard part is resolving the forces! For each vector, draw a right angled triangle so that one side is purely horizontal and the other is vertical. This need not be done for the always vertical mg force downwards. Find the lengths of the sides using trig and add up vertical components and horizontal components, taking care to have a negative sign when they are in opposite directions.

    Then take gabbagabbahey's suggestion - Now that we have an expression for the vertical and horizontal components, what can we equate that to? In circular motion, what is the net force, what direction is it in. Hence, what must the components be equal to? Once you form those equations, the rest is relatively easy.
     
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