Force Required to Grip an Object

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After using this equation I've come to a few questions and concerns about the physics of it. The only article I could find over the amount of force necessary for a robot hand to hold an object in the air produced the following equation:

fb789dbfded277f07d91b487330c4f05.png


where
65dea38223e26b71a0419a7c21d41f70.png
is the force required,
77316bd1d5862dd8a074a4378a5b4b45.png
is the coeffecient of friction,
b0dc5035f0bee3e371161fbcb77491b4.png
is the number of fingers on the gripper and
45ed2756f8ecb4c8179e77153386f4a6.png
is the weight of the object held.

It seems counter-intuitive to me. If you increase the amount of grippers, more force needs to be applied? The higher the coefficient the more force needs to be applied? What...

Using this formula I came to the conclusion that with a
77316bd1d5862dd8a074a4378a5b4b45.png
of 0.8 between a human hand and a metal object of 43 kg, with 5 fingers, it would take 1,687N to hold it in midair. I need to know if I'm doing it right or if these numbers are totally out of the ballpark.
 
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This situation is similar to the "object wedges between two walls" problem.

The friction at each wall is f where

f = uF

F is the force squeezing the object.

The object doesn't slide down so if W is the weight of the object..

W - 2f = 0

"2" because there are two walls where friction force f acts.

So

W = 2uF

or the required force F is given by..

F = W/2u

Now if the number of walls (aka fingers) or u increases the force F reduces.

Found an image..
2014911-134956948-9786-Capture.PNG