Calculating Horizontal Force to Prevent Rice Packet Slippage

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The discussion centers on calculating the horizontal force required to prevent a 2 kg packet of rice from slipping, given a coefficient of friction of 0.3. The correct horizontal force to prevent slippage is 33.3 N, as derived from the equation f_{k,s} = \mu_{k,s} N, where N is the normal force. The confusion arises from the misapplication of the normal force and frictional force in the calculations. The normal force in this scenario is 20 N, leading to the conclusion that the applied force must equal the frictional force to maintain equilibrium.

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A shopper picks up a 2 kg packet of rice with the thumb and index finger of one hand.The coefficient of friction between her finger and the wrapping is 0.3.What horizontal force must she exert to prevent the packet from slipping?

What I have done for this question
I don't know whether the friction force is (0.3x20).Is it?But then if the force wanted is horizontal,can I still use the normal force of 20N?The answer given in the book is 33.3N,but from what I have done,where force applied is equal to the friction force,it is only 6N.

Thanks for helping.
 
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Try drawing a picture, and show all the forces on the bag, sum the forces in each direction. Not the non-slip condition would apply if the force applied exactly cancels the force of friction. The normal force comes in with the introduction of the frictional force. Usually, in magnitude
[tex]f_{k,s} = \mu_{k,s} N[/tex]
where k is the kinetic friction, and s is static friction. Also note that
[tex]\mu_{s} > \mu_{s}[/tex]
hopes this helps, sincerely, x
 

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