Calculating frictional force given mass and coefficient?

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Discussion Overview

The discussion revolves around calculating the frictional force exerted by a steel block given its mass and the coefficient of static friction. Participants explore the necessary components for this calculation, including the role of the contact area and the distinction between mass and force.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks how to find the frictional force given the mass of a steel block and the static coefficient of friction.
  • Another participant suggests that the area of contact between the block and the surface is relevant to the calculation.
  • Some participants express that the provided information is insufficient to determine the frictional force without considering additional factors.
  • There is a discussion about the units of the coefficient of friction, with some participants affirming that it is a scalar ratio of frictional force to the normal force.
  • One participant calculates the force required to support the block and derives a maximum static friction value based on that force.
  • Another participant confirms the calculation of the maximum static friction but notes that it does not represent the actual static friction unless certain conditions are met.
  • Participants discuss the formula for frictional force, indicating that it involves the coefficient of friction and the normal force, while also distinguishing between static and sliding friction.

Areas of Agreement / Disagreement

Participants generally agree on the formula for calculating frictional force but express differing views on the necessity of additional information, such as contact area, and the implications of the calculations presented. The discussion remains unresolved regarding the completeness of the information needed for a definitive answer.

Contextual Notes

There are limitations regarding the assumptions made about the conditions under which the frictional force is calculated, such as the orientation of the surface and the absence of vertical acceleration.

radaballer
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The mass of a steel block is 760.9 g, the static coefficient of friction is 0.15, how do i find the frictional force in Newtons?
 
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The area of the block in contact with the surface needs to come into the picture.
 
radaballer said:
The mass of a steel block is 760.9 g, the static coefficient of friction is 0.15, how do i find the frictional force in Newtons?
You can't, from this information.
anorlunda said:
The area of the block in contact with the surface needs to come into the picture.
How would that help?
 
0.00032258 meters^2

anorlunda said:
The area of the block in contact with the surface needs to come into the picture.

A.T. said:
You can't, from this information.
How would that help?
 
OK, and the unite of the coefficient 0.15 are what?
 
anorlunda said:
OK, and the unite of the coefficient 0.15 are what?
anorlunda said:
OK, and the unite of the coefficient 0.15 are what?
I thought Coeff of friction was the ratio of Frictional force to force pushing the bodies together, it is scalar, right?
 
radaballer said:
I thought Coeff of friction was the ratio of Frictional force to force pushing the bodies together, it is scalar, right?

I beg your pardon, you are correct.
 
anorlunda said:
I beg your pardon, you are correct.
So multiply 0.15 by mass in kg?
 
radaballer said:
So multiply 0.15 by mass in kg?
The kilogram is not a unit of force. It is a unit of mass. Given an objects mass and the local acceleration of gravity, you can determine how much force is required to support it, however.
 
  • #10
jbriggs444 said:
The kilogram is not a unit of force. It is a unit of mass. Given an objects mass and the local acceleration of gravity, you can determine how much force is required to support it, however.
Ok, I got 7.45 for the force required to support it, and 0.15 x 7.45 is 1.118 Newtons. Look good?
 
  • #11
Yes. 0.15 times 7.45 Newtons is 1.118 Newtons.
 
  • #12
radaballer said:
Ok, I got 7.45 for the force required to support it, and 0.15 x 7.45 is 1.118 Newtons. Look good?
Note that this is not the actual static friction, just the maximal value it can reach given a normal force of 7.45N.
 
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  • #13
Usually, F=mu*Wt frictional force equals some constant (coefficient of friction) times normal (perpendicular) force. Although there are differences between sliding friction and static friction, their correlation is close to 1. That doesn't mean they are equal.
 
  • #14
Sliding friction can often be approximated by F=mu*N. You know mu, you know N (normal force). Sliding friction is almost constant for slow sliding (<<<<<<<<<<c).
 
  • #15
max static friction calculated as 1.118 Newtons is correct provided there is no vertical acceleration of the surface, and it is perfectly horizontal.
 

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