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3 years ago

Need help solving this question.

Physics

1 answer:

MatroZZZ [7]3 years ago

7 0

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

$-15+T-W=0\\T-W=15$

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

$(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]$

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Answer:

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Explanation:

From the question we are told that

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Taking a wild guess here, but it sounds like you're asked to find |A| and |B| given the magnitude of the resultant A + B, i.e. |A + B| = 50 N, and the sum of the individual magnitudes, |A| + |B| = 70 N.

Recall that the dot product of a vector with itself is equal to the square of that vector's magnitude,

A • A = |A|²

Then

|A + B|² = (A + B) • (A + B)

|A + B|² = (A • A) + 2 (A • B) + (B • B)

|A + B|² = |A|² + 2 (A • B) + |B|²

Since A and B are perpendicular to one another, their dot product is

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Substitute |B| = 70 N - |A| and solve for |A| :

2500 N² = |A|² + (70 N - |A|)²

2500 N² = |A|² + 4900 N² - (140 N) |A| + |A|²

2 |A|² - (140 N) |A| + 2400 N² = 0

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|A|² - (70 N) |A| = - 1200 N²

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|A| = 35 N ± 5 N

so that |A| = 30 N or |A| = 40 N. If we fix |A| to be one of these, then |B| will have the other value.

So, the magnitudes are |A| = 30 N and |B| = 40 N.

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