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spayn [35]
3 years ago
13

From the top of a vertical cliff 80m high, the angles of depression of 2 buoys lying due west of the cliff are 23° and 15° respe

ctively. How far are the buoys apart?​
Engineering
1 answer:
Umnica [9.8K]3 years ago
6 0

Answer:

  110 m

Explanation:

The geometry of the problem can be modeled by a right triangle. The height of the cliff forms one leg, and the distance to a buoy forms the other leg. The angle of depression is opposite the height of the cliff. An appropriate trig relation is ...

  Tan = Opposite/Adjacent

Solving for the Adjacent side (the distance to the buoy), we find ...

  Adjacent = Opposite/Tan

  distance to buoy 1 = (80 m)/tan(23°) ≈ 188.468 m

  distance to buoy 2 = (80 m)/tan(15°) ≈ 298.564 m

Then the distance between the buoys is ...

  298.564 -188.468 m ≈ 110 m

The buoys are about 110 meters apart.

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a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8
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b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam  along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we  determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = \frac{MC}{I}  =  \frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

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