Analyzing Tension Forces with friction

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SUMMARY

The discussion centers on calculating the tension in a string connecting a 9.00-kg hanging object and a 5.00-kg block on a table, with a coefficient of kinetic friction of 0.200. The correct approach involves applying Newton's second law (F = ma) and recognizing that the tension (T) must account for both the frictional force and the acceleration of the system. The calculations yield a tension of approximately 37.8 N when using an acceleration of 2 m/s², confirming the importance of accurately determining the forces acting on each block and ensuring consistent acceleration across the system.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of free body diagrams (FBD)
  • Familiarity with kinetic friction and its coefficient
  • Basic algebra for solving equations
NEXT STEPS
  • Study how to draw and interpret free body diagrams for multi-body systems
  • Learn about the effects of friction in mechanical systems
  • Explore advanced applications of Newton's laws in dynamics
  • Investigate the role of tension in various pulley systems
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Students studying physics, particularly those focusing on mechanics, as well as educators looking to enhance their teaching of tension and friction concepts in dynamic systems.

Mindstein
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Homework Statement


A 9.00-kg hanging object is connected, by a light, inextensible cord over a light, frictionless pulley, to a 5.00-kg block that is sliding on a flat table. Taking the coefficient of kinetic friction as .200, find the tension in the string


Homework Equations


F = ma


The Attempt at a Solution


Fk = ma
a = (9.8 N friction force)/(5.00 kg) = 1.96 which is approximately 2
ΣFx= T - Fk = ma
T = Fk + ma
T = (9.8N) + total mass 14N(1.96 m/s^2)
T = 37.24 or if you use 2 for the acceleration you get 37.8 which is the right answer

Did I do this right?

Please help me to understand why.
 
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Mindstein said:

Homework Statement


A 9.00-kg hanging object is connected, by a light, inextensible cord over a light, frictionless pulley, to a 5.00-kg block that is sliding on a flat table. Taking the coefficient of kinetic friction as .200, find the tension in the string


Homework Equations


F = ma


The Attempt at a Solution


Fk = ma
a = (9.8 N friction force)/(5.00 kg) = 1.96 which is approximately 2
I am not sure why you say that Fk = ma. Try Fk=uN.
ΣFx= T - Fk = ma
yes, looking at all forces acting in the x direction on the block sitting on the table, this equaton is correct, as it applies to the 5kg block.
T = Fk + ma
T = (9.8N) + total mass 14N(1.96 m/s^2)
you must make the Fk correction. You are right now just looking at the block on the table, where m = 5, not 14, and 'a' is unknown.
T = 37.24 or if you use 2 for the acceleration you get 37.8 which is the right answer

Did I do this right?

Please help me to understand why.
In problems of these types, or any types using Newton's laws, you MUST draw free body diagrams. Isolate each block and note all the forces acting on it (weight, friction, normal force, applied forces, etc., in the x and y directions). Try also drawing a FBD of the hanging object. What are the forces acting on it? Then apply Newton's laws to each block. Note that since the blocks are connected by an inextensible cord over a light frictionless pulley, that the magnitude of the accelerations of each block must be the same, and the rope tension acting on each block must be the same.
 

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