Force of a spring and static friction

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SUMMARY

The discussion centers on calculating the force of static friction exerted on a backpack resting on a table, with a spring attached. The backpack weighs 54.3N, and the spring has a force constant of 200N/m, stretched by 1.50cm. The correct approach to find the force of static friction is to equate it to the force exerted by the spring, leading to the equation fs = kx. The user initially attempted to derive the coefficient of static friction (mus) incorrectly but later clarified that the force of static friction equals the spring force.

PREREQUISITES
  • Understanding of Hooke's Law (F = kx)
  • Knowledge of static friction and its formula (fs = mus * Normal force)
  • Basic principles of equilibrium in physics
  • Familiarity with force constants and their implications in spring mechanics
NEXT STEPS
  • Study the relationship between force constants and spring behavior in different materials
  • Learn about the implications of static friction in real-world applications
  • Explore advanced topics in dynamics, including forces in equilibrium
  • Investigate the effects of varying spring constants on system stability
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Students in physics, particularly those studying mechanics, as well as educators and anyone interested in understanding the principles of static friction and spring dynamics.

matt@USA
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Homework Statement


A backpack full of books weighing 54.3N rests on a table in a physics laboratory classroom. A spring with a force constant of 200N/m is attached to the backpack and pulled horizontally, as indicated in the figure. If the spring is pulled until it stretches 1.50cm and the pack remains at rest, what is the force of friction exerted on the backpack by the table? Imagine the spring being pulled with a force to the -x, and the force of static friction going in the +x direction.

Homework Equations


I know that the force of the spring is equal to F=kx, the force of static friction is mus*Normal force. I tried to find the mus to solve the equation but I was wrong. I got the sum of the net force in the x direction as fs-F=0, and the sum of the net force in the y direction as N-mg=0.


The Attempt at a Solution


The final equation I used to try to solve for mus was mus=kx/N, with 200N/m being my k, .015m being my x, and 54.3N being my N. Can someone please tell me where, or how I messed up?
 
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matt@USA said:

Homework Statement


A backpack full of books weighing 54.3N rests on a table in a physics laboratory classroom. A spring with a force constant of 200N/m is attached to the backpack and pulled horizontally, as indicated in the figure. If the spring is pulled until it stretches 1.50cm and the pack remains at rest, what is the force of friction exerted on the backpack by the table? Imagine the spring being pulled with a force to the -x, and the force of static friction going in the +x direction.

Homework Equations


I know that the force of the spring is equal to F=kx, the force of static friction is mus*Normal force. I tried to find the mus to solve the equation but I was wrong. I got the sum of the net force in the x direction as fs-F=0, and the sum of the net force in the y direction as N-mg=0.


The Attempt at a Solution


The final equation I used to try to solve for mus was mus=kx/N, with 200N/m being my k, .015m being my x, and 54.3N being my N. Can someone please tell me where, or how I messed up?

How far off was your answer? 200 N/m is a pretty weak spring. 200 kN/m would be a good solid spring.
 
NM, I figured it out. The correct equation was fs=F, and F=kx, so fs=kx!
 

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