A body weighs 50 N and hangs from a spring with spring constant of 50 N/m. A dashpot is attached to the body. If the body is rai
sed 2 m from its equilibrium position and released from rest, determine the amplitude of vibration if (a) c = 17.7 kg/sec and (b) c = 40 kg/sec. What is the maximum amplitude of vibration in both of these cases?
1 answer:
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Answer:
Given that;
Jello there, see explanstion for step by step solving.
A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.
Explanation:
A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.
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Answer:
M = 281.25 lb*ft
Explanation:
Given
W<em>man</em> = 150 lb
Weight per linear foot of the boat: q = 3 lb/ft
L = 15.00 m
M<em>max</em> = ?
Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):
∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0
⇒ q' = (W + q*L) / L
⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft
⇒ q' = 13 lb/ft (+↑)
The free body diagram of the boat is shown in the pic.
Then, we apply the following equation
q(x) = (13 - 3) = 10 (+↑)
V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)
M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)
The maximum internal bending moment occurs when x = 7.5 ft
then
M(7.5) = 5(7.5)² = 281.25 lb*ft
