- #1

chudd88

- 23

- 1

Intuitively, I feel like the horizontal force should be mg*sin(@)*cos(@). My reasoning is that the mg*sin(@) force is just any other force at that point, so I can multiply is by cos(@) to get the component that is parallel to a horizontal line. But from what I've been able to find, the actual horizontal component should be mg*tan(@).

Could someone give me an explanation for this, or point me in the right direction? I'm having trouble searching for an answer because I'm not really sure what to call this horizontal force (thus the somewhat confused title of this post.)

Thanks.