You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If
60-kg Dan sits on the left end of the board and 50-kg Tahreen on the right end of the board, where should 54-kg Komila sit to keep the board stable? Ignore the mass of the board and treat each student as point-like objects The positive r-direction is to the right, and the origin is at the center of the chair
Part A Determine the coordinate of the leftmost point on the board where Komila can sit to keep the board stable
Part B Determine the coordinate of the rightmost point on the board where Komila can sit to keep the board stable
Part A Determine the coordinate of the leftmost point on the board where Komila can sit to keep the board stable
Part B Determine the coordinate of the rightmost point on the board where Komila can sit to keep the board stable
1 answer:
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Answer:
The answers are
The average stress = 20000 lb/in²
The maximum tensile stress immediately adjacent to the hole
= 31076.92 lb/in²
Explanation:
To solve the question we have
Weight of tensile load = 30,000 lb
Width of steel bar = 4 in
Thickness of steel bar = 1/2 in
Average Stress = Force/Area
Size of hole drilled = 1.0 in diameter
Available width at cross section where the 1.00 in diameter hole is drilled =
(4 - 1) in = 3 inches
Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)
= 3 in × 1/2 in = 1.5 in²
Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²
the maximum tensile stress immediately adjacent to the hole.
Bending stress = where
0.5*30000/I = 11076.92 lb/in²
Max stress = 31076.92 lb/in²