Posted by Ed on .
What minimum coefficient of friction between the tires and the road will allow a 3700 kg truck to navigate an unbanked curve of radius 25 m at a speed of 45 km/h?
My answer:
Fnet=ma
Ff=m(v2)/r)
Ff=(3700)(452/25)
Ff=(3700)(2025/25)
Ff=(3700)(81)
Ff=299700
Frictional force
Ff=Fnµ
299700=(3700)(9.8)µ
299700=36260µ
µ= 8.3
Is this right? Thanks for your help.

Physics 
drwls,
I have never heard of a friction coefficient that high. It must be wrong.
45 km/h = 12.5 m/s
You did not make the conversion to m/s. Big error there, since V gets squared.
Ff = Fn*µ = M*g*µ = M V^2/r
µ = V^2/(rg) = 0.64