A bicyclist is driving up a hill which is tilted at 8 degrees. How big is the force caused by gravity that "drags" the bicyclist down the hill?
Answer is: 110N
2006-09-20 09:42:27 · 3 answers · asked by - 1 in Science & Mathematics ➔ Physics
And I'm of course asking because I need to know how this is calculated.
2006-09-20 09:49:39 · update #1
This is a problem involving vectors, so I have to assume that you are studying vectors in class. What you are doing is "resolving" the force of gravity into two components. One of those components tends to hold the bicyclist against the road surface on the hill (i.e., it is perpendicular to the road). The other component is parallel to the road surface and is pulling the bicyclist back down the hill. The sum of these two force vectors is a "resultant" vector that is equal to the force of gravity (i.e., equal to the weight of the biker plus his bike). That resultant vector is directed vertically downward (because gravity acts vertically downward).
Now, in order to calculate the magnitude of the force pulling the bicyclist down the hill, you have to multiply the weight of the bike and biker by a trigonometric factor. That factor is the sin of 8 degrees, and it equals approximately .1392.
Your question states that the answer is 110 newtons; so we know that:
Weight x .1392 = 110N
This equation can be solved to determine the weight.
Hope that helps.
2006-09-20 10:04:03 · answer #1 · answered by actuator 5 · 0⤊ 0⤋
It is sin(8 degrees) times the weight of the bicyclist.
2006-09-20 09:51:52 · answer #2 · answered by campbelp2002 7 · 0⤊ 0⤋
If the answer is 110 N then what are you asking?
2006-09-20 09:45:54 · answer #3 · answered by aryeh_cls 2 · 0⤊ 2⤋