07年PHY CE MC 第31條

圖片參考：http://i.imagehost.org/0430/ScreenHunter_16_Apr_17_20_25.gif


http://i.imagehost.org/0430/ScreenHunter_16_Apr_17_20_25.gif

行路隻腳係係向後,所以Frition force acted on foot係向前<<<所以可以向前行
但係因為向前行, a backward friction is acted by the ground on the pedestrian. So, the pedestrian will stop. If the backward friction does not present, he will keep moving fowward constantly.


其實我也不是那麼清楚的:P 試試想象溜冰的情況也許較易明白吧

I think you have got confused. Friction always opposes motion. It could not propel an object forward.

When we walk, friction only prevents out feet from slipping. It is the force from our legs that makes us to move forward. If there is no friction, our legs couldn't exert any force at all because of the slipping of our feet.

Likewise, a car moves forward is due to the force developed by the engine of the car. On a slippery road, the force couldn't act on the car because of slipping of its wheels (i.e the wheels cannot grip firmly onto the road surface), thus the car could not move forward. In case friction is present, the engine could exert a force on the car by acting on the non-slipping wheels.


2010-04-18 00:35:52 補充：
Just imagine if friction did act upward along the slope, the force needed to pull the car up would be reduced when comparing with the case when friction was not present. In other words, a car would travel up a slope even faster if friction was present than if friction was not present....

2010-04-18 00:36:05 補充：
..... This apparently contradicts to common experience where a car needs more power (hence larger force) in going up a slope with rough surface.

我有一個比較易明但比較白痴的方法:
Sub the mass of the car be 10kg
-weight of the car = mg = 10(10) = 100N
-magnitude of friction = 100/2 = 50N

When the car is moving up the incline plane:
-the total downward force = mgsin0 + friction
= 10(10)sin30 + 50
= 100N

Since the car is moving in constant speed, hence:
-the upward force = downward force
= 100N

Velocity of the car is v:
By P=Fv
P=100v
P=10(10)v
P=mgv

希望幫到你=]