George likes to ride his bicycle down steep hills. On this day George finds a steep 60 degree slope to ride down. Determine the force that will pull george and his bike down the hill. George and his bike have a combined mass of 70 kilograms.:confused:

George likes to ride his bicycle down steep hills. On this day George finds a steep 60 degree slope to ride down. Determine the force that will pull george and his bike down the hill. George and his bike have a combined mass of 70 kilograms.:confused:

Lets think about all the forces in play (since friction was not mentioned, I will not mention it as a force).

The inclined plane is positioned 60 degrees, the only forces acting on George is weight and the normal force.

If we reposition out coordinate system so it runs along the inclined plane, rather than the floor, then we have a situation of like triangles, where the normal force is in the y direction, and the force due to weight is 60 degrees from the negative y direction.

We know that the normal force is always equal and opposite of weight, however on an inclined plane the normal forces is only equal to the perpendicular force applied by the weight. Since we know the plane is 60 degrees we can write that

where fn is normal force and fg is the force of weight
F_n=F_g_y=F_gcos(\frac{\pi}{3})

This normal force cancels out the y component of weight, leaving only the x component of weight the only force accelerating George. We can find this force by finding the x component of the weight in regards to our coordinate system.

F_g_x=F_gsin(\frac{\pi}{3})

(remember also that F_g=mg) Hope this helps :o