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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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A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
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Answer:

a) V = 0.354

b)  G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

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Using Area, A = π/4 · d²

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E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

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Also final diameter d(f) = 19.9837 mm

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Poisson said that V = Е(elasticity)/Е(long)

= -  <u>( 19.9837 - 20 /20)</u>

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= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

=  68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

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