The uniform 100-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position with theta = 0.
Determine the initial angular acceleration of the beam and the magnitude of the force FA supported by the pin at A due to the application of the force P = 300 N on the attached cable.
1 answer:
Answer:
ac = 2.86 m / s²
Explanation:
Image can detail the system to determine the force in the FA to understand the system into the applicated force
m = 100 kg , L = 3 m
∑ F = 0 ⇒ Ay - 100 kg + P * cos (45) = 0
Ay = 768.86 N
∑ Mₐ = α * I ₐ
I ₐ = m * L² / 3 ⇒ I ₐ = 100 kg * 4² m / 3
Replacing
P * sin (45) * 3 = α * 100 kg * 4² m / 3
α = 1.193 rad / s²
ac = α *2 ⇒ ac = 1.193 rad / s² * 2
ac = 2.86 m / s²
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Answer:
<em>The depth will be equal to</em> <em>6141.96 m</em>
<em></em>
Explanation:
pressure on the submarine = 62 MPa = 62 x 10^6 Pa
we also know that = ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
= 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine = 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure , we have
62 x 10^6 = 10094.49h
depth h = <em>6141.96 m</em>