A 1000 kg car carrying two 100 kg football players travels over abumpy \"washboa

Question

A 1000 kg car carrying two 100 kg football players travels over abumpy "washboard" road with the bumps spaced 3.0 m apart. Thedriver finds that the car bounces with a max amplitude when hedrives at a speed of 5.0 m/s. The car then stops and picks up threemore 100 kg passengers.

By how much does the car body sag on its suspension when thesethree additional passengers get in?


My attempt:
I tried solving this by figuring out frequency as v /. With frequency, I calculated (2pi*f) whichis equal to K/m. Figuring out K, I found out two separatelengths using L = mg / K and solved for L.

I'm not sure if this is the correct way to solve this problem (itseems a little too simple since this is considered a challengeproblem). Any help would be much appreciated. Thank you.

Explanation / Answer

Given : Mass ( M ) = 1000 kg m = 100 kg    x = 3 m    v = 5 m/s    T = x / v = 3m / 5m/s = 0.6s The external frequency due to the bumps matches thecar's natural frequency fnet = Fo = 1/ T = 1/2k/m1      k = ( 1200kg ) ( 213.14 ) ( 1.667Hz)2         = 131600 N/m     m1 = M +2 m = 1200 kg when car at rest Fnet = 0     y = mg /k    Initial compression when m1 = 1200kg     yi = 0.0894 m m2 = M +5 m = 1500 kg    yf = 0.1117 m      The car sage on it suspension is =yf - yi = ------- m Solve it I hope it helps you      

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