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2 years ago

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A barbellbell is loaded with two 20 kg plates on its right side and two 20kg plates on its left side. The barbell is 2.2m long,

and its unloaded mass is 20kg. The two 20kg plates on the right side are locked in place at 35cm and 40cm from the right end of the bar. The two 20kg plates on the left side have slipped one iss 30 cm from the left end of the bar and the other is 20 cm from the left end of th bar. where is the center of gravity of this bar?

1 answer:

Aleks [24]2 years ago

7 0

Answer:

<h2>Center of gravity of the system of all masses is 105 cm from the left end of the rod</h2>

Explanation:

Let the position of Left end of the rod is our reference

So here we will have

$m_1 = 20 kg$

$x_1 = 20 cm$

$m_2 = 20 kg$

$x_2 = 30 cm$

$m_3 = 20 kg$

$x_3 = 110 cm$

$m_4 = 20 kg$

$x_4 = (220 - 40) = 180 cm$

$m_5 = 20 kg$

$x_5 = (220 - 35) = 185 cm$

so we will have

$x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4 + m_5x_5}{m_1 + m_2 + m_3 + m_4 + m_5}$

$x_{cm} = \frac{20(20 + 30 + 110 + 180 + 185)}{20 + 20 + 20 + 20 + 20}$

$x_{cm} = 105 cm$

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An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

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Answer:

The total distance traveled is 28 inches.
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Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector $\hat{i}$ pointing in the west direction, the ant start at position

$\vec{r}_0 = 16 \ inch \ \hat{i}$

Then, moves to

$\vec{r}_1 = 29 \ inch \ \hat{i}$

so, the distance traveled here is

$d_1 = |\vec{r}_1 - \vec{r}_0 | = | 29 \ inch \ \hat{i} - 16 \ inch \ \hat{i} |$

$d_1 = | 13 \ inch \ \hat{i} |$

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after this, the ant travels to

$\vec{r}_2 = 14 \ inch \ \hat{i}$

so, the distance traveled here is

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2 years ago

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Answer:

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When we add vectors, vectors that point in the same direction add up and vectors that point in opposite directions are subtracted.

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