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don't get it........?
1. A 1200kg car rounds a curve of radius 70m banked at 12 degrees at 25 m/s. is there a (static) friction force? if so, how much and in what direction?
I think it's asking if a friction force is needed to contribute to the centripetal force or if a component of the normal force is enough.
1 Answer
- TBTLv 71 decade agoFavorite Answer
force // to slope due to weight
= 1200 * g * sin 12
= 2447.54 N
centripetal force
= m v^2 /r
= 1200 * 25^2 / 70
= 10714.29 N
net force // to slope
= 2447.54 - 10714.29 cos 12
= - 8032.1 N
in the cetripetal force direction to be countered by friction force.
Yes, the friction force is 8032.1N -direction down the slope --answer
Source(s): my brain (Prof TBT)