Calculating Friction Force on a Resting Block

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Homework Help Overview

The discussion revolves around calculating the friction force on a resting block, given its mass, the coefficient of static friction, and an external force applied horizontally. Participants are examining the conditions under which static or kinetic friction applies.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the completeness of the problem statement and explore the implications of the applied force relative to the static friction threshold. There is discussion about the transition from static to kinetic friction and the necessity of knowing the coefficient of kinetic friction.

Discussion Status

The discussion is active, with participants raising questions about the assumptions in the problem and the conditions for static versus kinetic friction. Some guidance has been offered regarding the relationship between applied force and friction, but no consensus has been reached on the specifics of the kinetic friction coefficient.

Contextual Notes

There are indications of confusion regarding the problem setup, particularly concerning the relationship between the applied force and the static friction force. The need for additional information, such as the coefficient of kinetic friction, is also noted.

physaru86
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A 1.2 kg block is resting on horizontal surface. The coefficient of static friction between block and
surface is 0.5. What is the magnitude and direction of force of friction on block when magnitude of external force acting of block in the horizontal direction is 9.8 N
fs = μs*N, N = mg
upload_2015-3-31_18-52-3.png


m = 1.2 kg, μs = 0.5, g = 9.8
fs = 0.5*1.2*9.8 N = 5.88 N
 
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Is this really the full problem statement, exactly as presented, because something seems wrong about it.

Chet
 
Chestermiller said:
Is this really the full problem statement, exactly as presented, because something seems wrong about it.

Chet
please tell me what is wrong about it?
 
If the applied force is less than the normal force times the coefficient of static friction, then the friction force will equal the applied force and the body will not move. If the applied force is equal to the normal force times the coefficient of static friction, then the friction force will equal the applied force and the body will be on the verge of moving. But, in your example, the applied force is greater than the normal force times the coefficient of static friction, so the body will move, and kinetic friction will prevail; in this case, the coefficient of static friction is irrelevant, as is the normal force times the coefficient of static friction.

Chet
 
S
Chestermiller said:
If the applied force is less than the normal force times the coefficient of static friction, then the friction force will equal the applied force and the body will not move. If the applied force is equal to the normal force times the coefficient of static friction, then the friction force will equal the applied force and the body will be on the verge of moving. But, in your example, the applied force is greater than the normal force times the coefficient of static friction, so the body will move, and kinetic friction will prevail; in this case, the coefficient of static friction is irrelevant, as is the normal force times the coefficient of static friction.

Chet
So can the force of kinetic friction be found? if so how to find
 
physaru86 said:
S

So can the force of kinetic friction be found? if so how to find
To find that, you need to know the coefficient of kinetic friction.

Chet
 
Chestermiller said:
To find that, you need to know the coefficient of kinetic friction.

Chet
Yes it was the same doubt i had. can the coefficient of kinetic friction be found
 
physaru86 said:
Yes it was the same doubt i had. can the coefficient of kinetic friction be found
Sure, it can be measured.

Chet
 
Chestermiller said:
Sure, it can be measured.

Chet
thanks
 
  • #10
Oposed to motion
 

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