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Hello, I have some questions regarding the image I have attached. As you can likely tell, the image is my (un-artistic) representation of a force being applied to both sides of a gear, in an effort to increase the effort force required to move it. I've found that the gear needs to have 1010.7 N of frictional force to reach the desired goal of applying an upper bound of 20.214 Nm of torque to the gear (hence, it is acting 0.02m from the center of the gear). When calculating the normal force applied to the gear in this situation, would I be correct in saying that the kinetic co-efficient of friction is doubled, such that the normal force, $F_N$ is given by:

$F_F = 2 \times \mu_{k} \times F_N$

Where I can just re-arrange for $F_N$? Or, does the does the normal force change because it is acting from both sides of the gear, in which case i'd assume that it is halved.

Any help is much appreciated.

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    $\begingroup$ There iare two $F_N$s applied in the oppsite direction. They combine to be zero, Therefore, you cannot and should not combine two normal forces. Instead, the friction forces generated from these two normal force are in parallel directions, These two frictional force can be added together. $\endgroup$
    – ytlu
    Commented Jul 10, 2021 at 12:31

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The short answer

The Kinetic Coefficient of Friction is just a property of a pair of materials (e.g. (steel and steel)) which models the frictional force between any two surfaces made of those materials. The frictional force increases as a result of the increase of the total gear-to-pad normal force, whereas the coefficient of friction remains constant.

The Long Answer

Consider individually the friction $F_f$ caused by the contact on each side of the gear, and double it to get the overall frictional force opposing the motion of the gear. (You can double the frictional force of one side to get the total because the situation is perfectly symmetrical down the middle).

Clearly the resultant force on each braking pad and the gear itself is 0 in the horizontal direction.

As there is a force $F_N$ incident on the braking pad, and it is in equilibrium, there must be another force of magnitude $F_N$ incident on the braking pad from the gear itself. By Newton's 3rd Law, the force incident on the gear itself from the braking pad must also be of magnitude $F_N$ (the magnitude of the normal force between them is $F_N$). So the frictional force caused by one braking pad is $\mu_kF_N$.

Of course, the gear is in equilibrium in the horizontal axis as well; this is because an equal normal reaction force is incident upon it by the other braking pad, using the same logic as before. Thus there is a second frictional force of magnitude $\mu_kF_N$ caused by the contact on the other side.

Thus, the overall frictional force is $2\mu_kF_N$ not because the coefficient of friction has doubled, but because the total normal reaction force incident on the gear has doubled and all of that additional normal reaction force is between the same materials.

EDIT Make sure you pay attention to ytlu's comment as well, the answer of $2\mu_kF_N$ is arrived at by first calculating the frictional forces, realising that they act in the same direction, and then adding them together - you should not add the normal forces together before you calculate the total frictional force, because while this may sometimes arrive at the right answer (as it does here), it usually won't, because that's not how the model of frictional force used here operates.

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