Homework Help: Physics

Posted by Ed on Friday, November 19, 2010 at 11:32pm.

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=Fnľ
299700=(3700)(9.8)ľ
299700=36260ľ
ľ= 8.3

Is this right? Thanks for your help.

Physics - drwls, Saturday, November 20, 2010 at 1:02am
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*ľ = M*g*ľ = M V^2/r

ľ = V^2/(rg) = 0.64

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