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3 years ago

Rigid Body Statics in 3 Dimensions

Physics

1 answer:

slamgirl [31]3 years ago

7 0

Explanation:

Draw a free body diagram of the bar.

There are 3 reaction forces at O in the x, y, and z direction (Ox, Oy, and Oz).

There is a tension force Tac at A in the direction of the rope. There are also tension forces Tbd and Tbe at B in the direction of the ropes.

Finally, there is a weight force mg pulling down halfway between A and B, where m = 400 kg.

There are 6 unknown variables, so we'll need 6 equations to solve. Summing the forces in the x, y, and z direction will give us 3 equations. Summing the torques about the x, y, and z axes will give us 3 more equations.

First, let's find the components of the tension forces.

Tbe is purely in the z direction.

Tbd has components in the y and z directions. The length of Tbd is √8.

(Tbd)y = 2/√8 Tbd

(Tbd)z = 2/√8 Tbd

Tac has components in the x, y, and z directions. The length of Tac is √6.

(Tac)x = 1/√6 Tac

(Tac)y = 1/√6 Tac

(Tac)z = 2/√6 Tac

Sum of the forces in the +x direction:

∑F = ma

Ox − (Tac)x = 0

Ox − 1/√6 Tac = 0

Sum of the forces in the +y direction:

∑F = ma

Oy + (Tac)y + (Tbd)y − mg = 0

Oy + 1/√6 Tac + 2/√8 Tbd − mg = 0

Sum of the forces in the +z direction:

∑F = ma

Oz − (Tac)z − (Tbd)z − Tbe = 0

Oz − 2/√6 Tac − 2/√8 Tbd − Tbe = 0

Sum of the torques counterclockwise about the x-axis:

∑τ = Iα

mg (2 m) − (Tac)y (2 m) − (Tbd)y (2 m) = 0

mg − (Tac)y − (Tbd)y = 0

mg − 1/√6 Tac − 2/√8 Tbd = 0

Sum of the torques counterclockwise about the y-axis:

∑τ = Iα

-(Tac)x (2 m) + (Tbd)z (1.5 m) + Tbe (1.5 m) = 0

-4 (Tac)x + 3 (Tbd)z + 3 Tbe = 0

-4/√6 Tac + 6/√8 Tbd + 3 Tbe = 0

Sum of the torques counterclockwise about the z-axis:

∑τ = Iα

-mg (0.75 m) + (Tbd)y (1.5 m) = 0

-mg + 2 (Tbd)y = 0

-mg + 4/√8 Tbd = 0

As you can see, by summing the torques about axes passing through O, we were able to write 3 equations independent of those reaction forces. We can solve these equations for the tension forces, then go back and find the reaction forces.

-mg + 4/√8 Tbd = 0

4/√8 Tbd = mg

Tbd = √8 mg / 4

Tbd = √8 (400 kg) (9.8 m/s²) / 4

Tbd = 2772 N

mg − 1/√6 Tac − 2/√8 Tbd = 0

1/√6 Tac = mg − 2/√8 Tbd

Tac = √6 (mg − 2/√8 Tbd)

Tac = √6 ((400 kg) (9.8 m/s²) − 2/√8 (2772 N))

Tac = 4801 N

-4/√6 Tac + 6/√8 Tbd + 3 Tbe = 0

3 Tbe = 4/√6 Tac − 6/√8 Tbd

Tbe = (4/√6 Tac − 6/√8 Tbd) / 3

Tbe = (4/√6 (4801 N) − 6/√8 (2772 N)) / 3

Tbe = 653 N

Now, using our sum of forces equations to find the reactions:

Ox − 1/√6 Tac = 0

Ox = 1/√6 Tac

Ox = 1/√6 (4801 N)

Ox = 1960 N

Oy + 1/√6 Tac + 2/√8 Tbd − mg = 0

Oy = mg − 1/√6 Tac − 2/√8 Tbd

Oy = (400 kg) (9.8 m/s²) − 1/√6 (4801 N) − 2/√8 (2772 N)

Oy = 0 N

Oz − 2/√6 Tac − 2/√8 Tbd − Tbe = 0

Oz = 2/√6 Tac + 2/√8 Tbd + Tbe

Oz = 2/√6 (4801 N) + 2/√8 (2772 N) + 653 N

Oz = 6533 N

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<h3>5) Gravitational force between Earth and Moon</h3>

In this case we have:

$F_{earth-moon}$ is the gravitational force between Earth and Moon

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$m=7.46(10)^{23} kg$ is the mass of the Moon

$R=3.86(10)^{8} m$ is the distance between Earth and Moon

Solving:

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<h3>6) Gravitational force between Jupiter and Venus</h3>

Assuming for a moment that the planets are perfectly aligned and all are in the same orbital period, we can make a rough estimation of the distance between Jupiter and Venus, knowing the distance of each to the Sun:

distance between Sun and Jupiter - distance between Sun and Venus=distance between Jupiter and Venus= $R_{jupiter-venus}$ (4)

$R_{jupiter-venus}=778.3(10)^{9} m - 108(10)^{9} m=6.703(10)^{11} m$ (5)

Using this value in the Law of Universal Gravitation equation:

$F_{jupiter-venus}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.90(10)^{27} kg)(4.87(10)^{24} kg)}{(6.703(10)^{11} m)^{2}}$ (6)

$F_{jupiter-venus}=1.37(10)^{18} N$ (7)

<h3>7) Gravitational force between Saturn and Mars</h3>

Using the same assumption we made in the prior question:

distance between Sun and Saturn - distance between Sun and Mars=distance between Saturn and Mars= $R_{saturn-mars}$ (8)

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Using this value in the Law of Universal Gravitation equation:

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Firstly, we need to find $F_{earth-pluto}$ and then find $F_{venus-pluto}$ in order to find the difference.

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$m=1.46(10)^{22} kg$ is the mass of Pluto

$R_{earth-pluto}=5.7504(10)^{12} m$ is the distance between Earth and Pluto

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<u></u>

<u>For $F_{venus-pluto}$ :</u>

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$m=1.46(10)^{22} kg$ is the mass of Pluto

$R_{venus-pluto}=5.792(10)^{12} m$ is the distance between Venus and Pluto

$F_{venus-pluto}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(4.87(10)^{24} kg)1.46(10)^{22} kg)}{(5.792(10)^{12} m)^{2}}$ (14)

$F_{venus-pluto}=1.33(10)^{11} N$ (15) Force between Venus and Pluto

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$F_{earth-pluto}-F_{venus-pluto}=1.759(10)^{11} N-1.33(10)^{11} N$

Finally:

$F_{earth-pluto}-F_{venus-pluto}=4.29(10)^{10} N$ (16)

Hence:

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