The friction force required to keep a car from skidding on a curve is given by [MATH]\vec{F} _s(t) = ma_{N} \vec{N} (t)[/MATH]. Find the friction force needed to keep a car of mass [MATH]m = 100[/MATH] (slugs) from skidding if [MATH]\vec{r} (t) = \ <100 \cos \pi t, 100 \sin \pi t>[/MATH].
Friction Force
- Thread starter nasi112
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this question seems small
But
it is very Big
let us solve it by finding piece by piece
[MATH]|\vec{a} (t)|^2 = {a_T}^2 + {a_N}^2[/MATH]
where [MATH]\vec{a} (t)[/MATH] is the acceleration vector, [MATH]a_T[/MATH] is the tangential acceleration, and [MATH]a_N[/MATH] is the normal acceleration
then
[MATH]a_N = \sqrt{|\vec{a} (t)|^2 - {a_T}^2}[/MATH]
[MATH]\vec{v} (t) = \frac{d \vec{r} (t)}{dt} = \ <-100\pi \sin \pi t, 100\pi \cos \pi t>[/MATH]
[MATH]|\vec{v} (t)| = 100\pi[/MATH]
[MATH]\vec{a} (t) = \frac{\vec{v} (t)}{dt} = \ <-100{\pi}^2 \cos \pi t, -100{\pi}^2 \sin \pi t>[/MATH]
[MATH]|\vec{a} (t)| = 100{\pi}^2[/MATH]
[MATH]a_T = \frac{d \ |\vec{v} (t)|}{dt} = 0[/MATH]
then
[MATH]a_N = \sqrt{(100{\pi}^2)^2 - {0}^2} = 100{\pi}^2[/MATH]
[MATH]\vec{N} (t) = \frac{\frac{d \ \vec{T} (t)}{dt}}{\left |\frac{d \ \vec{T} (t)}{dt}\right |}[/MATH]
where [MATH]\vec{N} (t)[/MATH] is the unit normal vector and [MATH] \vec{T} (t)[/MATH] is the unit tangent vector
[MATH]\vec{T} (t) = \frac{\vec{v} (t)}{|\vec{v} (t)|} = \frac{<-100\pi \sin \pi t, 100\pi \cos \pi t>}{100\pi} = \ <-\sin \pi t, \cos \pi t>[/MATH]
[MATH]\frac{d \ \vec{T} (t)}{dt} = \ <-\pi \cos \pi t, -\pi \sin \pi t>[/MATH]
then
[MATH]\vec{N} (t) = \frac{<-\pi \cos \pi t, -\pi \sin \pi t>}{\pi} = \ <-\cos \pi t, -\sin \pi t>[/MATH]
And finally
[MATH]\vec{F} (t) = m \cdot a_N \cdot \vec{N} (t) = 100 \cdot 100{\pi}^2 \cdot <-\cos \pi t, -\sin \pi t> \ = \ <-10000{\pi}^2\cos \pi t, -10000{\pi}^2\sin \pi t>[/MATH] lbf
But
it is very Big
let us solve it by finding piece by piece
[MATH]|\vec{a} (t)|^2 = {a_T}^2 + {a_N}^2[/MATH]
where [MATH]\vec{a} (t)[/MATH] is the acceleration vector, [MATH]a_T[/MATH] is the tangential acceleration, and [MATH]a_N[/MATH] is the normal acceleration
then
[MATH]a_N = \sqrt{|\vec{a} (t)|^2 - {a_T}^2}[/MATH]
[MATH]\vec{v} (t) = \frac{d \vec{r} (t)}{dt} = \ <-100\pi \sin \pi t, 100\pi \cos \pi t>[/MATH]
[MATH]|\vec{v} (t)| = 100\pi[/MATH]
[MATH]\vec{a} (t) = \frac{\vec{v} (t)}{dt} = \ <-100{\pi}^2 \cos \pi t, -100{\pi}^2 \sin \pi t>[/MATH]
[MATH]|\vec{a} (t)| = 100{\pi}^2[/MATH]
[MATH]a_T = \frac{d \ |\vec{v} (t)|}{dt} = 0[/MATH]
then
[MATH]a_N = \sqrt{(100{\pi}^2)^2 - {0}^2} = 100{\pi}^2[/MATH]
[MATH]\vec{N} (t) = \frac{\frac{d \ \vec{T} (t)}{dt}}{\left |\frac{d \ \vec{T} (t)}{dt}\right |}[/MATH]
where [MATH]\vec{N} (t)[/MATH] is the unit normal vector and [MATH] \vec{T} (t)[/MATH] is the unit tangent vector
[MATH]\vec{T} (t) = \frac{\vec{v} (t)}{|\vec{v} (t)|} = \frac{<-100\pi \sin \pi t, 100\pi \cos \pi t>}{100\pi} = \ <-\sin \pi t, \cos \pi t>[/MATH]
[MATH]\frac{d \ \vec{T} (t)}{dt} = \ <-\pi \cos \pi t, -\pi \sin \pi t>[/MATH]
then
[MATH]\vec{N} (t) = \frac{<-\pi \cos \pi t, -\pi \sin \pi t>}{\pi} = \ <-\cos \pi t, -\sin \pi t>[/MATH]
And finally
[MATH]\vec{F} (t) = m \cdot a_N \cdot \vec{N} (t) = 100 \cdot 100{\pi}^2 \cdot <-\cos \pi t, -\sin \pi t> \ = \ <-10000{\pi}^2\cos \pi t, -10000{\pi}^2\sin \pi t>[/MATH] lbf
Steven G
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The vector was defined. Do you understand what [MATH]\vec{A} (t)[/MATH] means?
it is a definition
the principal unit normal vector [MATH]\vec{N} (t)[/MATH] is a unit vector having the same direction as [MATH] \frac{d \ \vec{T} (t)}{dt} [/MATH] and is defined by [MATH]\vec{N} (t) = \frac{\frac{d \ \vec{T} (t)}{dt}}{\left|\frac{d \ \vec{T} (t)}{dt}\right|}[/MATH] provided that [MATH]\frac{d \ \vec{T} (t)}{dt} \neq 0[/MATH].
you can find this definition in any Calculus book
D
Deleted member 4993
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Work through (re-do) response #3 line by line - with a pencil and paper. You'll see that it has been answered.