A student pushes a 5 kg box initially at rest along a flat, rough table with a rightward force of 10 N. The frictional force act
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Answer:
43.41 mm
Explanation:
Given:
thickness of sheet, t = 6 mm
Force exerted by punch, F = 45 KN
Average shearing stress, T = 55 MPa
From average shearing stress T = Force F / Area A
Hence area = force/stress =45000/ 55 =
From area = pi*diameter*thickness
diameter = area/(pi* thickness)
= 818.18/(3.142*6)
= 43.41 mm