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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?

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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.

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  • $\begingroup$ Thanks for the feedback. By "forward of the "CM line,"" do you mean further down the incline than the CM? $\endgroup$ Commented Apr 13, 2023 at 2:06
  • $\begingroup$ On a side note: In this case with a stationary object on an incline, since the torques of the normal and friction forces cancel out, then the magnitude of the friction force times its distance from the CM would equal the magnitude of the normal force times its distance from the CM. Then using how the friction force equals the coefficient of friction times the normal force, you get the normal forces to cancel and have that the distance of the normal force from the CM equals the distance of the friction force from the CM times the coefficient of friction. $\endgroup$ Commented Apr 13, 2023 at 2:15
  • $\begingroup$ Yes, for the case I discussed. Also, consider the case where the normal force acts for a person leaning while standing on a horizontal surface. $\endgroup$
    – John Darby
    Commented Apr 13, 2023 at 2:17
  • $\begingroup$ I would say it is at the front of the person's feet if they are leaning forward, right? The more they lean forward, assuming they keep their body straight, the more force they feel closer to their toes since putting more of their mass forward would mean they press down more (or get pressed up on) closer to the front of where they are in contact with the ground. $\endgroup$ Commented Apr 13, 2023 at 2:28
  • $\begingroup$ I agree with this. $\endgroup$
    – John Darby
    Commented Apr 13, 2023 at 12:00

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