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dezoksy [38]
2 years ago
9

vertical gate in an irrigation canal holds back 12.2 m of water. Find the average force on the gate if its width is 3.60 m. Repo

rt your answer with proper units and 3 sig figs.
Engineering
1 answer:
DanielleElmas [232]2 years ago
5 0

Answer:

The right solution is "2625 kN".

Explanation:

According to the question,

The average pressure will be:

= density\times g\times \frac{h}{2}

By putting values, we get

= 1000\times 9.8\times \frac{12.2}{2}

= 1000\times 9.8\times 6.1

= 59780

hence,

The average force will be:

= Pressure\times Area

= 59780\times 3.6\times 12.2

= 2625537 \ N

Or,

= 2625 \ kN

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In an orthogonal cutting operation, the tool has a rake angle = 12°. The chip thickness before the cut = 0.32 mm and the cut yie
Snezhnost [94]

Answer:

The shear plane angle and shear strain are 28.21° and 2.155 respectively.

Explanation:

(a)

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.  

Given:  

Rake angle is 12°.  

Chip thickness before cut is 0.32 mm.

Chip thickness is 0.65 mm.  

Calculation:  

Step1  

Chip reduction ratio is calculated as follows:  

r=\frac{t}{t_{c}}

r=\frac{0.32}{0.65}

r = 0.4923

Step2  

Shear angle is calculated as follows:  

tan\phi=\frac{rcos\alpha}{1-rsin\alpha}

Here, \phi is shear plane angle, r is chip reduction ratio and \alpha is rake angle.  

Substitute all the values in the above equation as follows:  

tan\phi=\frac{rcos\alpha}{1-rsin\alpha}

tan\phi=\frac{0.4923cos12^{\circ}}{1-0.4923sin12^{\circ}}

tan\phi=\frac{0.48155}{0.8976}

\phi=28.21^{\circ}

Thus, the shear plane angle is 28.21°.

(b)

Step3

Shears train is calculated as follows:

\gamma=cot\phi+tan(\phi-\alpha)

\gamma=cot28.21^{\circ}+tan(28.21^{\circ}-12^{\circ})\gamma = 2.155.

Thus, the shear strain rate is 2.155.

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