# How does normal force affect torque?

SO, I understand that F(c,h) and F are considered a couple so the net torque they produce about any point is F*(perpendicular distance) so the torque of this couple is F(2R-h) in the clockwise direction. However, according to the answer key above the other forces contributing the net torque of the system are mg and normal force. So they took the corner of the step as the axis of rotation; thus the net torque of the couple is still F(2R-H), BUT the torque by Mg about the corner should be Mg(R) and with a little bit of geometry, the torque by the normal force about the corner should be (normal force)[square root of (2Rh)] FINALLY, the net torque about the corner should be torque = Mg(R) - F(2R-H) - (normal force)[square root of (2Rh)]. Why does my equation not match with theirs? did I do something wrong? or was there a concept applied that I do not know of. I acknowledge that Mg is not equal the normal force in this scenario since mg is counteracted by both the normal force + F(c,v)

• Why do you think the torque by gravity should be Mg(R)? Feb 8, 2017 at 3:52
• Because it is taken about the corner of the table, the distance from this point to the center of gravity is R. Feb 8, 2017 at 3:54
• R is the total distance, but it is not the perpendicular distance. Feb 8, 2017 at 3:56
• why would we use the perpendicular distance? Mg is not coupled with any other forces, since the the net force in the y direction is 0 whilst the forces that exist in the vertical direction are Mg, fn, and fcv; thus, we see that Mg is not equal to fn suggesting that they are not a couple Feb 8, 2017 at 3:58
• hyperphysics.phy-astr.gsu.edu/hbase/torq2.html#tc Feb 8, 2017 at 4:03

Torque due to gravity is not mg(R) Torque is defined as $r \times F$. You must either resolve $r$ and multiply the component perpendicular to $F$ with $F$ or resolve $F$ and multiply the part perpendicular to $r$ with $r$ to get its magnitude.