# Angle of friction force of a weighted ball on a surface

If a block had constant density and laid on a slope with angle $$\theta$$ then the angle of friction and the horizontal would be $$\theta$$. If there was a ball with constant density then the friction force would also pass through the contact point and have angle $$\theta$$ with the horizontal.

However, if there was a weighted part of the ball then I learned the friction force would still pass through the contact point, but it would not have an angle $$\theta$$ with the horizontal.

What causes this and what is the angle? I read it’s the angle of line from the weighted section to the center and the vertical line, why is this so?

• What type of friction are you considering? If you are taking into account rolling friction via deformation of surfaces, then certainly the friction force does not need to be parallel to the incline. – Aaron Stevens Sep 17 at 16:15
• It's been my understanding that friction is defined as the component of the contact force which is parallel to the surface. (The other component is the normal force.) Each of these may vary from point to point. – R.W. Bird Sep 17 at 18:03