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eduard
3 years ago
14

A link in a mechanism is to be subjected to a tensile force that varies from 3500 to 500 N in a cyclical fashion as the mechanis

m runs. It has been decided to use SAE 1040 cold-drawn steel. Complete the design of the link, specifying a suitable cross-sectional shape and dimensions.
Physics
1 answer:
shtirl [24]3 years ago
5 0

Answer:

Design explained below

Explanation:

We are designing a link which is able to sustain 3500 N force because it will automatically sustain any lower force.

 

The tensile strength of AISI 1040 steel is : 585 MPa

Shear Strength = 338.15 MPa from data book

Now, 585 * 10^6 = 3500 / [Area]

(from pressure = force / area)

So, area = 5.98*10^-6= pi / 4 * d^2

So, diameter = 2.76 mm

Also, 338.15 *10^6 = 3500 / (pi * d * L)

So, L = 37.5 mm

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A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a
grigory [225]

Answer:

a)  fr = 266.92 N,   fy = 1300 N,  b)    μ = 0.36

Explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

             -w x - W x₂ + R y = 0         (1)

usemso trigonometry to find distances

            cos 60.08 = x / 7.5

            x = 7.5 cos 60.08

            x = 3.74 m

fireman

           cos 60.08 = x₂ / 4

           x2 = 4 cos 60

           x2 = 2 m

wall support

           sin 60.08 = y / 15

           y = 15 are 60.08

           y = 13 m

we substitute in equation 1

           R y = w x + W x2

            R = (w x + W x2) / y

            R = (500 3.74 +800 2) / 13

            R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

           R -fr = 0

           R = fr

           fr = 266.92 N

Y Axis  

           Fy - w-W = 0

           fy = 500 + 800

           fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

          cos 60.08 = x₃ / 9.00

          x₃ = 9 cos 60

          x₃ = 5.28 m

from equation 1

          R = (w x + W x₃) / y

          R = 500 3.74 + 800 5.28) / 13

          R = 468.769 N

we saw that

          fr = R = 468.769

The expression for the friction force is

          fr = μ N

in this case the normal is the ratio to pesos

        N = Fy

       N = 1300 N

        μ N = fr

        μ = fr / N

        μ = 468,769 / 1300

         μ = 0.36

7 0
3 years ago
In what factors sound can be heard​
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Explanation:

Loudness of sound is a measure of response of sound to our ear. Loudness of sound is not simply the energy reaching the human ear, but it also tells about the sensitivity of human ear detecting this energy. Loudness of sound is measured in decibel (dB). As energy reaching the ear depends on square of amplitude, loudness of sound depends on various factors namely,

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Question #14
Katena32 [7]

The decrease in energy in the hydrogen molecule is what allows its formation on Earth, but in stars the great energy of the explosion has a kinetic energy so great that electrons cannot bind to another atom, which is why hydrogen has a single atom.

The hydrogen molecule is a form that two hydrogen atoms share their electrons decreasing the total energy of the molecule, this bond has a covalent and hydrogen bonding characteristic.

In a stellar explosion, the energy released increases the energy of the hydrogen atom, for which we have two possibilities:

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