1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Forces to hold an object

  1. Nov 24, 2011 #1
    This is a problem I came up with out of curiosity but can't solve.. The image shows a situation similar to a thumb an index finger grasping a marble. Probably obvious to figure out, but I can't, so I'll ask. What are the minimum force values F1 and F2 necessary to keep the marble from falling if mass is known?

    Attached Files:

  2. jcsd
  3. Nov 24, 2011 #2
    Start by drawing a correct free body diagram.:wink:
    edit: Actually you still need a friction coefficient in order to solve it.
  4. Nov 24, 2011 #3
    Found a link saying the coefficient of friction is .05

    So if the mass is M, the force of gravity on it would be mg, the force of friction must equal this so it would require a force of mg/.05 so for a marble weight .003kg, it would be .588N
  5. Nov 25, 2011 #4
    Force of friction is [itex] \mu N [/itex] where [itex] N [/itex] is the normal force. Since we also know that this needs to be [itex] \geq Mg [/itex] you can solve :)
    As a side note I've included two frictional forces since there's no reason to assume one side is preferable to another and I know from experience there is no net torque, however if friction is independent of surface area would the force be distributed between the two contact points? In other words would each frictional force be half what I've written? It seems like it should be but how could each side "know" that the other side had some frictional force? Then again I guess it might have to since the same argument must be wrong when applied to any single atom "knowing" about the friction on any other atom with a block sliding down an incline.
    I'll change the value back but leave this here as kind of an interest thing I guess. :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook