Kinetic and static frictional forces

AI Thread Summary
The discussion focuses on determining the horizontal force needed to initiate movement of a block with a mass of 755N, given static friction of 0.800 and kinetic friction of 0.600. The relationship between the coefficient of friction and the weight of the block is highlighted, emphasizing that the force required to overcome static friction is calculated using the static friction coefficient multiplied by the normal force. The normal force in this case equals the weight of the block, which is 755N. Therefore, the horizontal force necessary to start moving the block is 0.800 times 755N. Understanding these frictional forces is crucial for calculating the required force to initiate motion.
Leestyn
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static and frictional forces applied to a block. what is the horizontal force required to start the box from moving. known is kinetic friction - .600 and static friction .800 the block has a mass of 755N
 
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Leestyn said:
static and frictional forces applied to a block. what is the horizontal force required to start the box from moving. known is kinetic friction - .600 and static friction .800 the block has a mass of 755N

Welcome to PF.

How is the coefficient of friction related to the weight?
 

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