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Valentin [98]
4 years ago
11

*8–52. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the to

p of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P=300 N, the post slips at both B and C simultaneously.

Physics
1 answer:
denis23 [38]4 years ago
8 0

Answer:

μ₁ = 0.1048

μ₂ = 0.1375

Explanation:

Using  static equation can find in both point the moment and the forces so:

∑ M = F *d  , ∑ F = 0

∑ M A = 0

N₁ * 3 - 200 * 9.81 * 1.5 = 0

N₁ = 981  

∑ M y = 0

N₂ + 300 * ³/₅ - 981 - 20 * 9.81 = 0

N₂ = 997.2 N

∑ M C = 0

F₁ * 1.75  - 300 * ⁴/₅  * 0.75 = 0

F₁ = 102.86

∑ M B = 0

300 * ⁴/₅ * 1 - F₂ * 1.75 = 0

F₂ = 137.14 N

The Force F1 and F2 related the coefficients of static friction

F₁ = μ₁ * N₁   ⇒  102.86 N = μ₁ * 981 ⇒ μ₁ = 0.1048

F₂= μ₂ * N₂  ⇒  137.14 N = μ₂ * 997  ⇒ μ₂ = 0.1375

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3 years ago
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5 0
3 years ago
A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing
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Given:

Shaft Power, P = 7.46 kW = 7460 W

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7 0
3 years ago
Read 2 more answers
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IgorC [24]

ANSWER:

F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.

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F(h) is Horizontal Force = 200 N

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Since the velocity is constant so acceleration is zero; a=0

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F(h) = F(f)

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g is gravitational acceleration = 9.8 m/s^2

200 = mg*u

Since weight increases by 42% and friction coefficient decreases by 19%

New weight = 1+0.42 = 1.42 = (1.42*m*g)

New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u

F(h) = (0.81μ) (1.42 m g)

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3 years ago
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Answer:

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