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Force Required to Grip an Object

  1. Dec 9, 2015 #1
    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:


    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.
  2. jcsd
  3. Dec 10, 2015 #2


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  4. Dec 10, 2015 #3


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    It also implies that the greater the coefficient of friction u the greater the force required. Definitely not right.
  5. Dec 10, 2015 #4


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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.


    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..
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