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
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Answer:
Velocity of ball B after impact is and ball A is
Explanation:
= Initial velocity of ball A
= Initial velocity of ball B = 0
= Final velocity of ball A
= Final velocity of ball B
= Coefficient of restitution = 0.8
From the conservation of momentum along the normal we have
Coefficient of restitution is given by
Adding the above two equations we get
From the conservation of momentum along the plane of contact we have
Velocity of ball B after impact is and ball A is
.
The image is missing, so i have attached it.
Answer:
A) P = 65.11 KN
B) Q = 30 KN
Explanation:
We are given;
The end reaction of the beam; F = 100KN
Coefficient of static friction between two steel surfaces;μ_ss = 0.3
Coefficient of static friction between steel and concrete;μ_sc = 0.6
So, F1 = μ_ss•F =0.3 x 100 = 30 KN
F2 = μ_ss•N_EF = 0.3N_EF
From the screen shot, we see that the angle is 12°
Sum of forces in the Y-direction gives;
F2•sin12 - N_EF•cos12 + 100 = 0
Rearranging gives;
N_EF•cos12 - F2•sin12 = 100
Let's put 0.3N_EF for F2 to give;
N_EF•cos12 - 0.3N_EF•sin12 = 100
Thus;
N_EF(0.9158) - 0.1247 = 100
N_EF(0.9781) = 100 + 0.1247
N_EF = 100.1247/0.9158
N_EF = 109.33 KN
Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN
Wedge will move if;
P = (F1 + F2cos12 + N_EFsin12)
Thus;
P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)
P ≥ 65.11 KN
B) For static equilibrium, Q = F1
Thus, Q = 30 KN
