Block on an incline and frictional force

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Homework Help Overview

The problem involves a block resting on a 30-degree incline, with a focus on determining the horizontal force required to initiate movement up the incline, considering static friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the applied horizontal force and its components, questioning the original poster's force diagram and its accuracy.

Discussion Status

Participants are actively engaging in clarifying the relationship between the applied force and the frictional force, with some guidance offered regarding the direction of forces and their contributions to the normal force.

Contextual Notes

There is a noted confusion regarding the representation of forces in the force diagram, particularly how the horizontal force affects the normal force and frictional force on the incline.

Carrie
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Homework Statement


A 4.00-kg block rests on a 30.0 degree incline. If the coefficient of static friction between the block and the incline is 0.700, with what magnitude force must a horizontal force act on the block to start it moving up the incline?

Homework Equations


F=ma

Ffriction=μs*Fnormal

The Attempt at a Solution


I attached the force body diagram I did.

Fy
Fn - mgsin(30) = 0
Fn = mgsin(30)[/B]


Fx

Ffr - FA = 0
Ffr = FA
μ*N = FA
mgsin(30) + .7*mgcos(30)) = FA
(4)(9.8)sin(30)+.7*(4)(9.8)cos(30) = FA
FA = 43.36 N

However, the answer is 84.1 N. I'm pretty rusty with this stuff. Thank you!
Snapshot.jpg
 
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Carrie said:
a horizontal force
Consider what does it means to be "horizontal" here.
 
Hi Carrie, Welcome to Physics Forums.

Note that the problem stated that the applied force is horizontal. You've drawn it as parallel to the face of the slope. This will make a difference :smile:

Edit: D'oh! sometimes=1 beat me to it!
 
Hello and thank you!

I'm sorry, I realize what you mean by me drawing the force diagram incorrectly, but I'm still confused as to what that changes.
 
Carrie said:
Hello and thank you!

I'm sorry, I realize what you mean by me drawing the force diagram incorrectly, but I'm still confused as to what that changes.
Some part of the applied force pushes the block upslope, but another component will act to press the block into the slope. What other force will be affected by this?
 
The frictional force, right?
 
Carrie said:
The frictional force, right?
Yup.
 
I thought that's what I represented when I said:

Carrie said:
Ffr - FA = 0
 
That's not a valid equation. FA (which is horizontal) and the friction force (which acts along the slope) do not act along the same direction.

The force due to friction is obtained from the normal force and the friction coefficient. The normal force is made up of two contributions: one due to a component of the block's gravitational weight, the other due to a component of the force FA.

So FA increases the total normal force, and hence increases the friction too.
 
  • #10
Thank you!:smile:
 

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