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sp2606 [1]
3 years ago
5

A belt/pulley system has tight side of 1000N, a slack side of 100N and a wrap angle of 500 degrees. The belt is just on the poin

t of slipping. What is the cofficient of friction?
Engineering
1 answer:
Artemon [7]3 years ago
5 0

Answer:

0.26386

Explanation:

T₁ = Tight side tension = 1000 N

T₂ = Slack side tension = 100 N

θ = Wrap angle = 500° = 500π/180 radians

μ = Coefficient of friction

Belt tension theory

\frac{T_1}{T_2}=e^{\mu\theta}\\\Rightarrow \frac{1000}{100}=e^{\mu\times 500\times \frac{\pi}{180}}\\\Rightarrow 10=e^{\mu\times 500\times \frac{\pi}{180}}

Taking natural logs on both sides

ln10=\mu\times 500\times \frac{\pi}{180}\\\Rightarrow \mu=\frac{ln10}{500\times \frac{\pi}{180}}=0.26386

∴ Coefficient of friction is 0.26386

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The velocity at R/2 (midway between the wall surface and the centerline) is given by (3/4)(Vmax) provided that Vmax is the maximum velocity in the tube.

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So, velocity at any point in the radial direction is

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At the point r = (R/2)

r² = (R²/4)

(r²/R²) = r² ÷ R² = (R²/4) ÷ (R²) = (1/4)

So,

1 - (r²/R²) = 1 - (1/4) = (3/4)

Hence, v at r = (R/2) is given as

v = v(max) × (3/4)

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3 years ago
For a 20 ohm resistor R, the current i = 2 A. What is the voltage V? Submit your answer as a number without units. ​
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Explanation:

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6 0
3 years ago
The collar A, having a mass of 0.75 kg is attached to a spring having a stiffness of k = 200 N/m . When rod BC rotates about the
gladu [14]

Answer:

Speed=1.633 m/s

Force= 20 N

Explanation:

Ideally, v^{2}=\frac {ks^{2}}{m} hence v=s\sqrt {\frac {k}{m}} where v is the speed of collar, m is the mass of collar, k is spring constant and s is the displacement.

In this case, s=100-0=100mm=0.1m since 1 m is equivalent to 1000mm

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3 years ago
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