blakester
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The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.30 with the floor. If the train is initially moving at a speed of 43 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?
I use weird variables I've noticed browsing these forums but here we go
fg = friction with ground
N=normal force
ok so from my free body diagram i got Force of x and y is:
x---- Fs = ma
y----N - mg = ma (which a is 0 since there is not vertical acceleration)
working it down I got N = mg, and using Fg=UN and combining the equations i get
ma=U*mg
divide by m on each side and plug in what i know i get
a=ug
a=.3 * 9.81
a= 2.943
Then using my constant acceleration equations i have Vf^2 - Vo^2=2ad
(0)^2 - (43)^2 = 2(2.943)d
d=314.135
This is the wrong answer apparently though according to the website and I can't for the life of me see what i did wrong. Any help is appreciated thanks! My midterms tomorrow so wish me luck too heh
I use weird variables I've noticed browsing these forums but here we go
fg = friction with ground
N=normal force
ok so from my free body diagram i got Force of x and y is:
x---- Fs = ma
y----N - mg = ma (which a is 0 since there is not vertical acceleration)
working it down I got N = mg, and using Fg=UN and combining the equations i get
ma=U*mg
divide by m on each side and plug in what i know i get
a=ug
a=.3 * 9.81
a= 2.943
Then using my constant acceleration equations i have Vf^2 - Vo^2=2ad
(0)^2 - (43)^2 = 2(2.943)d
d=314.135
This is the wrong answer apparently though according to the website and I can't for the life of me see what i did wrong. Any help is appreciated thanks! My midterms tomorrow so wish me luck too heh