Question

A boat and a trailer are being pulled over an undulating road at a velocity v. The contour of the road is such that it can be approximated by a sine wave having a wavelength of l = 10 ft and an amplitude of Y = 0.5 in. The total static deflection of the springs and tires of the trailer due to the weight of the boat and trailer has been measured as 1.5 in. Assume that the damping inherent in the system is viscous and the damping ratio is 0.05. Determine: (a) The speed v at which the amplitude of the boat and the trailer will be a maximum; (b) The value of this maximum amplitude referred to in part (a); (c) The amplitude when the boat and trailer are traveling at the speed of 55 mph

Answer 1

a) v = 25.54 ft/s

b) Xmax = 0.4180

c) Xmax = 0.0048 ft

Step-by-step explanation:

a) determine the speed v at which the amplitude of the boat and the trailer will be a maximum;

to calculate the distance travelled

S = vt

given an expression of wavelength

l = vt

l = 2πv / ω

ω = 2πv / l

equation for counter of the road

y = Y sin ωt

= Y sin ( (2πv / l)t)

next we calculate the angular frequency

ω = √ ( k/m)

= √ ( g / Sst )

= √( 32.2 / ( 1.5/12))

= 16.05 rad/s

Now we calculate the speed v at which amplitude of the boat will be maximum

ω = 2πv / l

16.05 = (2 × π × v ) / 10

160.5 = 2πv

v = 160.5 / 2π

v = 25.54 ft/s

b)

we calculate the maximum amplitude using the following expression;

X/Y = [ √( 1 + ( 2Sr)²)] / [ √(( 1 - r²)² + ( 2Sr)²))]

Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×1)²)] / [ √(( 1 - 1²)² + ( 2×0.05×r)²))]

Xmax / 0.0416 = 1.004987 / 0.1

Xmax 0.1 = 0.0418

Xmax = 0.0418 / 0.1

Xmax = 0.4180

c)

we convert speed of the boat from mph velocity m/s

v = 55 mph × 1.46667ft/s / 1mph

v = 80.66 ft/s

next we calculate angular velocity

ω = 2πv / l

= 2π × 80.66 / 10

= 50.68 rad/s

next is the frequency ratio

r = 50.68 / 16.05

= 3.16

finally we find the amplitude of the boat

X/Y = [ √( 1 + ( 2Sr)²)] / [ √(( 1 - r²)² + ( 2Sr)²))]

Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×3.16)²)] / [ √(( 1 - 3.16²)² + ( 2×0.05×3.16)²))]

Xmax / 0.0416 = √1.0998 / √ ( 80.74 + 0.0998)

Xmax / 0.0416 = 1.0487 / 8.9910

Xmax 8.9910 = 0.0436

Xmax = 0.0436 / 8.9910

Xmax = 0.0048 ft