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Maru [420]
3 years ago
15

A tool chest has 950 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.

The surface has a coefficient of friction of 0.3. Determine the value of the horizontal force P necessary to cause motion of the chest to the right, and determine if the motion is sliding or tipping. The value of P is N. The motion is .
Engineering
1 answer:
Mazyrski [523]3 years ago
6 0

Answer:

P > 142.5 N  (→)

the motion sliding

Explanation:

Given

W = 959 N

μs = 0.3

If we apply

∑ Fy = 0 (+↑)

Ay + By = W

If  Ay = By

2*By = W

By = W / 2

By = 950 N / 2

By = 475 N (↑)

Then  we can get F (the force of friction) as follows

F = μs*N = μs*By

F = 0.3*475 N

F = 142.5 N (←)

we can apply

P - F  > 0

P  > 142.5 N (→)

the motion sliding

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Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

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Given that

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Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

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