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11Alexandr11 [23.1K]
3 years ago
11

Alarge plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa m1/2. If the plate is expos

ed to a tensile stress of 345 MPa during the service use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.
Physics
1 answer:
Korvikt [17]3 years ago
4 0

Answer:

minimum length of a surface crack is 18.3 mm

Explanation:

Given data

plane strain fracture toughness K = 82.4 MPa m1/2

stress σ = 345 MPa

Y = 1

to find out

the minimum length of a surface crack

solution

we will calculate length by this formula

length = 1/π ( K / σ Y)²

put all value

length = 1/π ( K / σ Y)²

length = 1/π ( 82.4 10^{3/2} / 345× 1)²

length = 18.3 mm

minimum length of a surface crack is 18.3 mm

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A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
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(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

F_f=-\mu mg, the force of friction, with \mu=0.25 being the coefficient of friction, m=30.0 kg being the mass of the crate, and g=9.8 m/s^2. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

where

F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N is the magnitude of the force of friction

d = 4.5 m is the displacement

\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

W=F_f d cos \theta

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cos \theta=0

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

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