Find all forces & the coeff. static friction of a pulley at equilibrium

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SUMMARY

The discussion focuses on calculating the forces and static friction of a pulley system in equilibrium involving two blocks: block A (0.500 kg) on a flat surface and block B (0.200 kg) hanging over a pulley. The gravitational forces acting on the blocks are calculated as 4.905 N for block A and 1.962 N for block B. The user initially miscalculates the acceleration as 2.8 m/s², but correctly identifies that the system is at equilibrium, meaning the acceleration is 0 m/s². The key takeaway is that static friction is directly proportional to the normal force, and the user is advised to draw free body diagrams (FBDs) to analyze the forces effectively.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of free body diagrams (FBDs)
  • Familiarity with static friction and normal force concepts
  • Basic calculations involving gravitational force (Fg = mg)
NEXT STEPS
  • Learn how to draw and analyze free body diagrams (FBDs) for mechanics problems
  • Study the relationship between static friction and normal force in detail
  • Explore the equations of motion for systems in equilibrium
  • Investigate tension calculations in pulley systems
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of pulley systems and static friction analysis.

minadd
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Homework Statement
It is given that that block a (.5kg) is on a flat surface at rest and block b (.2kg) is hanging over a table attached to a pulley. I tried to find friction but im stuck. And I still have to find tension for block a and b. I've had experience will inclines but forgot how to do flat surfaces...
Relevant Equations
Fg = mg
g= 9.81ms^2
(Sigma)F = ma
Ff = (mu)Fn
Given:
(block a) m = .500kg
(block b) m = .200kg
that's all...
Fg (block a) = (.5)(9.81) = 4.905N = Fn
Fg (block b) = (.2)(9.81) = 1.962N

i think acceleration might go like...
a = (1.962)/(.5 + .2) = 2.8m/s^2 but isn't it supposed to be at equilibrium? So 0m/s^2.
How am I supposed to find friction without acceleration?

the photo attached is the question.
 

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minadd said:
Homework Statement: It is given that that block a (.5kg) is on a flat surface at rest and block b (.2kg) is hanging over a table attached to a pulley. I tried to find friction but I am stuck. And I still have to find tension for block a and b. I've had experience will inclines but forgot how to do flat surfaces...
Homework Equations: Fg = mg
g= 9.81ms^2
(Sigma)F = ma
Ff = (mu)Fn

Given:
(block a) m = .500kg
(block b) m = .200kg
thats all...
Fg (block a) = (.5)(9.81) = 4.905N = Fn
Fg (block b) = (.2)(9.81) = 1.962N

i think acceleration might go like...
a = (1.962)/(.5 + .2) = 2.8m/s^2 but isn't it supposed to be at equilibrium? So 0m/s^2.
How am I supposed to find friction without acceleration?

the photo attached is the question.
You are trying to jump to the end without doing the work of setting up the problem. The first task to draw the FBDs, the same as every other mechanics problem. Once you have all of the forces on each mass drawn and labeled, you can write the force balance equation for each mass and solve the equations.
One thing that might help: static friction is a force that holds an object still by preventing it from sliding along a surface. It is directly proportional to the normal force of the surface on the object.
 
tnich said:
static friction is a force that holds an object still by preventing it from sliding along a surface. It is directly proportional to the normal force of the surface on the object.
Small correction: the maximum static friction force is directly proportional to the normal force of the surface on the object. The actual force is whatever force, up to that limit, that is required to prevent sliding.
 
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