A torque T 5 3 kN ? m is applied to the solid bronze cylinder shown. Determine (a) the maximum shearing stress, (b) the shethe 1
1 answer:
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vₐ = 0.771 ft/s
vb = -1.54 ft/s
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<u>Explanation:</u>
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Block A:
F = ma
Nₐ = - 60 cos60° = 0
Nₐ = 30 lb
Fₐ = 0.1 X 30 = 3lb
Block B:
F = ma
Nb = - 40 cos30° = 0
Nb = 34.64lb
Fb = 0.1 X 34.64 = 3.464lb
T1 + ∑U = T2
(0 + 0) + 60 sin 60° |Δsₐ| - 40 sin 30° |Δsb| - 3|Δsₐ| - 3.464 |Δsb|
= 1/2 (60/32.2) vₐ² + 1/2 (40/32.2) vb²
and
2vₐ = - vb
On solving, we get
vₐ = 0.771 ft/s
vb = -1.54 ft/s
