Mechanics - find the coefficient of friction

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SUMMARY

The discussion focuses on calculating the coefficient of friction for a block on an inclined plane, where a horizontal force of 2 N prevents a 1 kg block from sliding down. The incline angle is given as arcsin(7/25), and the gravitational acceleration is 9.8 m/s². The initial calculation yielded a coefficient of friction (μ) of 0.0876, while the correct value is 0.0827. Additionally, the acceleration of the block when the force is removed is calculated to be 1.97 m/s².

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  • Understanding of Newton's laws of motion
  • Knowledge of forces acting on inclined planes
  • Familiarity with trigonometric functions and their application in physics
  • Ability to resolve forces into components
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  • Explore the application of Newton's second law in two dimensions
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mohdakram
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Homework Statement


A horizontal force of 2 N is just sufficient to prevent a block of mass 1 kg from sliding down a rough plane inclined at arcsin [tex]\frac{7}{25}[/tex] to the horizontal. Find the coefficient of friction between the block and the plane and the acceleration with which the block will move when the force is removed.

g = 9.8

Homework Equations


F = [tex]\mu[/tex]R

The Attempt at a Solution


I didn't try the second part, but this is the first part.
Ncos[tex]\theta[/tex]+[tex]\mu[/tex]R=mgsin[tex]\theta[/tex]
I replace N and [tex]\theta[/tex] with the values given and R with mgcos[tex]\theta[/tex] and solve for [tex]\mu[/tex].

I get [tex]\mu=0.0876[/tex]

The answer at the back of the book is 0.0827 for coefficient.
1.97 ms^-2 for acceleration
 
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mohdakram said:

Homework Statement


A horizontal force of 2 N is just sufficient to prevent a block of mass 1 kg from sliding down a rough plane inclined at arcsin [tex]\frac{7}{25}[/tex] to the horizontal. Find the coefficient of friction between the block and the plane and the acceleration with which the block will move when the force is removed.

g = 9.8


Homework Equations


F = [tex]\mu[/tex]R


The Attempt at a Solution


I didn't try the second part, but this is the first part.
Ncos[tex]\theta[/tex]+[tex]\mu[/tex]R=mgsin[tex]\theta[/tex]
I replace N and [tex]\theta[/tex] with the values given and R with mgcos[tex]\theta[/tex] and solve for [tex]\mu[/tex].

I get [tex]\mu=0.0876[/tex]

The answer at the back of the book is 0.0827 for coefficient.
1.97 ms^-2 for acceleration
You don't have all the forces listed, and your geometry/trig and sum of force component equations are off. When you do inclined plane problems, first identify all forces acting, both the gravity force and all the contact forces. Then, before applying Newton's laws, let the x-axis be parallel to the incline , and let the y-axis be perpendicular to the incline. Now tilt your head, break forces into their x and y components, and solve using Newton's laws in both the x direction and y direction.
And welcome to PF!
 
Thank you PhantomJay for the help. I forgot to include the vertical component of the horizontal force when calculating R, which affected my final answer.
 

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