# Is the normal force greater below the center of mass than above?

Say there's an object that becomes electrically charged and then stuck on a wall using the electrostatic attraction between the object and the wall. For the object to stay in place, there is a friction force acting from where the wall and object meet going upwards, equal to the force of gravity acting from the center of mass of the object going downwards. I am thinking how these forces are not aligned would create rotational motion (clockwise if the wall is on the left). To counteract this, the object must have more than one horizontal line of contact on the wall (so it can't be a sphere or a cylinder oriented with the curved side in contact with the wall in a horizontal line). The electrostatic force acts equally across all points of contact, so the "main" direction of the force would be in line with the center of mass of the object and start from where the wall and object meet and go into the wall. As a result, would there have to be more normal force exerted below the center of mass on the wall than above the center of mass? This would serve to counteract the unalignment of the vertical forces, causing an opposite rotational force with a different unalignment. The normal force doesn't have to be uniformly distributed, so that is the one with a "different" position.

Force diagrams I see in class show forces as one vector in a line, but forces like weight, the electrostatic force, and normal force are spread out, right?

• This may help - Toppling of a cylinder on a block Commented Apr 12, 2023 at 21:03
• Yea that doesn't really help. I'll try adding a diagram later. Commented Apr 12, 2023 at 22:36

You are correct. The normal force does not always "line up" with the center of mass (CM). Here is an example. Consider a block at rest on an inclined surface. Friction acts up the block, gravity acts down the block, and the normal force is normal to the incline-block interface. For equilibrium the net torque must be zero as well as the net force. Consider the net torque about the center of mass (CM). With respect to the CM, the force of gravity provides no torque, the force of friction provides torque $$fd$$ where $$f$$ is the force of friction and $$d$$ is the distance from the CM to the interface. The net normal force must act forward of the "CM line" to provide torque to counter-balance that from friction.