Question involving Newton's Laws and Friction

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SUMMARY

The discussion centers on a physics problem involving Newton's Laws and friction, specifically analyzing a scenario where a worker pulls a 10 kg crate on a rough surface. The coefficients of static and kinetic friction are given as 0.5 and 0.3, respectively. The worker exerts a pulling force of 45N, which is less than the maximum static friction force of 49N calculated using the normal force of 98N. Consequently, the frictional force exerted by the surface equals the pulling force of 45N, preventing the crate from moving.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Knowledge of static and kinetic friction coefficients
  • Ability to calculate normal force using weight (mg)
  • Familiarity with basic force equilibrium concepts
NEXT STEPS
  • Study the implications of static vs. kinetic friction in various scenarios
  • Learn how to calculate frictional forces in different materials
  • Explore advanced applications of Newton's Laws in real-world problems
  • Investigate the effects of angle and surface texture on friction
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the practical applications of Newton's Laws and friction in mechanics.

Chandasouk
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Homework Statement



A worker pulls horizontally on a rope that is attached to a 10 kg crate that is resting on a rough floor. The coefficients of static and kinetic friction are 0.5 and 0.3 respectively. The worker pulls with a force of 45N. What is the frictional force exerted by the surface?

We did this in class and the answer was 45N but I do not understand why that is. Can someone explain?
 
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A worker pulls horizontally on a rope that is attached to a 10 kg crate that is resting on a rough floor. The coefficients of static and kinetic friction are 0.5 and 0.3 respectively. The worker pulls with a force of 45N. What is the frictional force exerted by the surface?

We did this in class and the answer was 45N but I do not understand why that is. Can someone explain?

So vertical direction of the pull does not need to be taken into account, therefore the normal force = weight.

The weight = mg or (10kg)(9.8) = 98 N

so the normal force = 98N

since f = (coefficient) * N

f = .5 * 98 = 49N, so the worker would need to pull with a force GREATER than 49 N in order to start accelerating the box, but since he pulls with only 45N, the force of friction will be equal to the pulling force..

so therefore f = force of pull = 45 N

so f = 45N
 

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