A 3.00-- kg rifle fires a 0.00500 m/s. Which force is greater in magnitude: (i) the force that the kg bullet at a speed of 300 r
1 answer:
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Answer:
Explanation:
In the x direction the force will be
½(-w₀)L/2 = -¼w₀L
acting ⅔(L/2) = L/3 below the x axis.
In the y direction the force will be
½(-w₀)L + ½w₀L/2 = -¼w₀L
the magnitude of the resultant will be
F = w₀L √((-¼)² + (-¼)²) = w₀L√⅛
in the direction
θ = arctan(-¼w₀L / -¼w₀L) = 225°
to find the distance, we balance moments
(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]
(√⅛)[d] = ½ [⅔L] + ¼ [⅔L/2] - ¼ [L - ⅓L/2]
(√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]
(√⅛)[d] = ⅓L + ⅟₁₂L - ¼L + ⅟₂₄L
(√⅛)[d] = 5L/24
d = 5L/24 / (√⅛)
d = 5√⅛L/3
Answer:
life (N) of the specimen is 117000 cycles
Explanation:
given data
ultimate strength Su = 120 kpsi
stress amplitude σa = 70 kpsi
solution
we first calculate the endurance limit of specimen Se i.e
Se = 0.5× Su .............1
Se = 0.5 × 120
Se = 60 kpsi
and we know strength of friction f = 0.82
and we take endurance limit Se is = 60 kpsi
so here coefficient value (a) will be
a = ......................1
put here value and we get
a =
a = 161.4 kpsi
so coefficient value (b) will be
b =
b =
b = −0.0716
so here number of cycle N will be
N =
put here value and we get
N =
N = 117000
so life (N) of the specimen is 117000 cycles