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3 years ago

15

The vector position of an object is given by r what is the torque acting on the object about the origin when a force f = (−12.5i

hat) n acts on it? express your answer using unit vector notation.

1 answer:

attashe74 [19]3 years ago

7 0

Let the vector position of the object in the (x-y) plane be
$\vec{r} = x \hat{i} + y \hat{j}$

The applied force is
$\vec{f} = -12.5 \hat{i}
$

By definition, the applied torque is
$\vec{T} = \vec{r} \times \vec{f} = (x\hat{i} + y\hat{j}) \times (-12.5y \hat{i}) = 12.5\hat{k}$

Answer: $12.5y \, \hat{k}$

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Answer:

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3 years ago

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saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

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We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

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$x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2$

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and this force produces acceleration of

$a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .$

With this acceleration, wooden block would reach at the foot of ramp in.

$x(t) = 12m = 16.2m/s^2*t^2$

$t = 1.217s$

and final velocity will be

$v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.$

this velocity of wooden box just before going up the ramp.

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and deceleration along the ramp is

$a = \frac{F_{s} }{m}$

Where $F_{s}$ force of friction along the inclined plane.

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is a component of g along normal of the inclined plane.

$F_{s} = 0.25*5kg*9.2089m/s^2$

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$a = \frac{11.5112N}{5kg} = 2.3022m/s^2$

And with this deceleration time needed to get wooded block to stop is.

$v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0$

$t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s$

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$x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m$

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Explanation:

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