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artcher [175]
2 years ago
9

Analyze the example of this band saw wheel and axle. The diameter of the wheel is 14 inches. The diameter of the axle that drive

s the wheel is 3/4 inch. The actual force needed to cut through a one-inch-thick softwood board is 1.75 pounds. Consider the efficiency of this band saw to be 22%.
Questions: Calculate ideal mechanical advantage when the effort force is applied to the axle.

Questions: considering the efficiency calculate the actual mechanical advantage of the wheel and axle.

Questions: If you used the same wheel and axle in a different way and applied effort force to the wheel to drive the axle, what is the ideal mechanical advantage of the wheel and axle?
Engineering
1 answer:
Kazeer [188]2 years ago
5 0

The answer for the ideal mechanical advantage and actual mechanical advantage for the different scenarios are;

A) Ideal Mechanical Advantage = 18.67

B) Actual Mechanical Advantage = 4.1067

We are given;

Input distance; The diameter of the wheel; d_w = 14 inches

Output distance; The diameter of the axle that drives the wheel; d_a = 3/4 inches

The force needed to cut a one-inch-thick softwood board; F = 1.75 pounds

The efficiency of the band saw; η = 22% = 0.22

A) Formula for Mechanical advantage is;

M.A = Force output/Force input = (Input distance)/(Output distance)

Thus;

Ideal mechanical advantage = 14/(3/4)

Ideal mechanical advantage = 18.67

B) Now, we are given that efficiency of the band saw is η = 22% = 0.22.

Thus using the mechanical advantage formula above;

Actual mechanical advantage = 0.22 × Expected output

Actual mechanical advantage = 0.22 × 18.67

Actual mechanical advantage ≈ 4.1067

Read more about Mechanical Advantage at; brainly.com/question/18345299

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Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t
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Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

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A = 45

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a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

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Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

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P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

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Tan 2a = 30/(45+60)/2

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p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

b) Shear stress in plane

Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

Average = (45-(-60))/2 = -7.5 MPa

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