Sketches are a very efficient way to share ideas. True False
2 answers:
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Answer:
a) 1.16 m/s
b) 1/216000
c) (√15)/6480000
Explanation:
The parameters given are;
Length of boat prototype, lp = 30 m
Speed of boat prototype = 9 m/s
Length of boat model, lm= 0.5 m
a) lm/lp = 0.5/30 = 1/60 = ∝
(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)
vm = 9 × (1/60)^(1/2) = 1.16 m/s
b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;
Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³
Fm/Fp = (1/60)³ = 1/216000
c) The ratio of the model to prototype power pm/p = (Fm/Fp) × (vm/vp) = ∝³×√∝
The ratio of the model to prototype power pm/p = √(1/60) × (1/60)³
pm/p = √(1/60) × (1/60)³ = (√15)/6480000
Answer:
Tension in cable BE= 196.2 N
Reactions A and D both are 73.575 N
Explanation:
The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence
hence
Therefore, tension in the cable,
Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then
Similarly,
Therefore, both reactions at A and D are 73.575 N
