Rigid Body Statics in 3 Dimensions

In the figure below, an L-shaped bar suspends a cylindrical mass of 400 kg. The bar is suspended by three cables and connected to point O, so that it can rotate freely around the three axes (x, y and z). Determine the reaction forces at point O and the tensile forces on the AC, BD and BE ropes.

Answers:
Ox= 1960 N
Oy= 0
Oz= 6533 N
Tac= 4801 N
Tbe=653 N
Tbd= 2772 N

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

this means that the temperature of both substances will be equal

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the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia