Calculating Horizontal Force to Prevent Rice Packet Slippage

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To prevent the 2 kg rice packet from slipping, the horizontal force required can be calculated using the coefficient of friction, which is 0.3. The normal force acting on the packet is equal to its weight, approximately 20N (2 kg x 9.8 m/s²). The frictional force can be calculated as 0.3 times the normal force, resulting in a frictional force of 6N. However, to prevent slippage, the applied horizontal force must exceed this frictional force, leading to a required force of 33.3N as indicated in the book. Understanding the relationship between normal force, friction, and applied force is crucial for solving this problem effectively.
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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
f_{k,s} = \mu_{k,s} N
where k is the kinetic friction, and s is static friction. Also note that
\mu_{s} > \mu_{s}
hopes this helps, sincerely, x
 
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