Calculating frictional force given mass and coefficient?

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SUMMARY

The frictional force for a steel block with a mass of 760.9 g and a static coefficient of friction of 0.15 can be calculated using the formula F = μ * N, where N is the normal force. The normal force is determined by converting the mass to kilograms (0.7609 kg) and multiplying it by the acceleration due to gravity (9.81 m/s²), resulting in a normal force of 7.45 N. Consequently, the maximum static frictional force is calculated as 1.118 Newtons. This value represents the maximum frictional force before sliding occurs, assuming a perfectly horizontal surface and no vertical acceleration.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of the coefficient of friction
  • Ability to convert mass to weight using gravitational acceleration
  • Familiarity with basic physics formulas related to force
NEXT STEPS
  • Study the relationship between static and kinetic friction
  • Learn about the effects of surface area on frictional force
  • Explore the differences between sliding friction and static friction
  • Investigate the impact of vertical acceleration on frictional calculations
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in understanding the principles of friction and force calculations in real-world applications.

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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