1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vsevolod [243]
3 years ago
10

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

You might be interested in
What happens when a sound wave passes from oil into air? A. the sound wave slows down and bounces back B. the sound wave slows d
Jlenok [28]

Oil is optically denser than water. When sound/light goes from optically denser medium to optically rarer medium, their velocity increase and they moves away for normal.

<h3><u>Appropriate</u><u> </u><u>Answer</u><u>:</u></h3>

The sound wave speeds up and bends

\Large{ \underline{ \boxed{ \pink{ \bf{Option \: (D)}}}}}

As, In optics we learnt that light undergoes refraction when travels from medium of different densities. Similarly, Sound also follows the law of refraction.

  • It is due to the change of speed of water in different mediums, This makes it speed up or down depending upon the medium and their densities.

<u>━━━━━━━━━━━━━━━━━━━━</u>

6 0
3 years ago
A ladder rests against a vertical wall at a point 12 feet from the floor. The angle formed by the ladder and the floor is 63°. C
GenaCL600 [577]

Answer:

length of the ladder is 13.47 feet

base of wall to latter distance 6.10 feet

angle between ladder and the wall is 26.95°

Explanation:

given data

height h  = 12 feet

angle 63°

to find out

length of the ladder ( L) and length of wall to ladder ( A) and angle between  ladder and the wall

solution

we consider here angle between base of wall and floor is right angle

we apply here trigonometry rule that is

sin63 = h/L

put here value

L = 12 / sin63

L = 13.47

so length of the ladder is 13.47 feet

and

we can say

tan 63 = h / A

put here value

A = 12 / tan63

A = 6.10

so base of wall to latter distance 6.10 feet

and

we say here

tanθ = 6.10 / 12

θ = 26.95°

so angle between ladder and the wall is 26.95°

8 0
3 years ago
For a wire has a circular cross section with a radius of 1.23mm.
Mila [183]

Answer:

5.731\times 10^{-5}\ m/s

Decrease

Explanation:

I = Current = 3.7 A

e = Charge of electron = 1.6\times 10^{-19}\ C

n = Conduction electron density in copper = 8.49\times 10^{28}\ electrons/m^3

v_d = Drift velocity of electrons

r = Radius = 1.23 mm

Current is given by

I=neAv_d\\\Rightarrow v_d=\dfrac{I}{neA}\\\Rightarrow v_d=\dfrac{3.7}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times \pi (1.23\times 10^{-3})^2}\\\Rightarrow v_d=5.731\times 10^{-5}\ m/s

The drift speed of the electrons is 5.731\times 10^{-5}\ m/s

v_d=\dfrac{I}{neA}

From the equation we can see the following

v_d\propto \dfrac{1}{n}

So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.

5 0
3 years ago
g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie
polet [3.4K]

Answer: 71.16\ Hz

Explanation:

Given

Capacitance C=100\ \mu F

Resistance R=500\ \Omega

Inductance L=50\ mH

In LCR circuit, current is maximum at resonance frequency i.e.

X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}

Insert the values

\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}

Also, frequency is given by

\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}

\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz

8 0
3 years ago
A locomotive engine pulls a train with a constant force comment on whether its acceleration will increase or decrease if more co
julia-pushkina [17]

Explanation:

the acceleration will be unchanged according to newton second law of motion

7 0
3 years ago
Other questions:
  • If Superman wants to slow down a fast car with speed 20 m/s and mass 1000 kg, how much force in N does he need to apply if he wa
    9·1 answer
  • The distance from one Crest to the next Crest is the blank
    12·2 answers
  • PLZ HELP ME!!!!!!!!! WILL GIVE BRAINLY!!!!!!!! Which photo represents a waxing gibbous moon?
    13·1 answer
  • An electric resistance heater is embedded in a long cylinder of diameter 30 mm. when water with a temperature of 25°c and veloc
    8·1 answer
  • T or F. A virtual image can sometimes be seen on a screen; it just depends on the situation.
    6·1 answer
  • What type of Galaxy is considered the most common?
    12·2 answers
  • PLEASE ASAP ILL GIVE BRAINLIEST.
    14·1 answer
  • A baseball is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stop watch measures the stone's ti
    10·1 answer
  • What is the likely identity of a metal if a sample has a mass of 63.5 g when measured in air and an apparent mass of 60.2 g when
    10·1 answer
  • How long did it take Fernando to drive to
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!