Force question

You are correct.

 

If you have a body on an incline, then the force of gravity on that body would be F_grav = mg (cos a i + sin a j). But if you compute the magnitude of this vector, you end up with |F_grav| = mg, don't you?

Doesn't this mean that the weight of a body on an incline is the same as on a horizontal surface?

Dot

 

Ah, right. I understand. I misunderstood the original question. Yes, the normal force is less on an incline.

I've been thinking of the inclines in these mechanics problems as splitting the downard force of mg into two components, one parallel to the incline, and one perpendicular to it. So I guess I have been thinking of the incline as a device to "split gravity", although I know that sounds stupid. I haven't really used that term until now.

Could you enlarge on what is really going on? I'd appreciate it.

Dot

 

I understand this. What I don't understand is why this isn't viewed as changing the direction of the objects weight. I've been told I get too hung up on the details, but I do like to understand these things.

Dot