Why Is the Box Not Moving Despite Unbalanced Forces on an Inclined Plane?

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Homework Help Overview

The problem involves a mass at rest on an inclined plane, with a focus on understanding the forces acting on the box and the conditions required to move it up the ramp. The scenario includes parameters such as mass, ramp length, incline angle, and coefficients of friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the forces acting on the box, particularly the static friction and the weight components. Questions arise about the interpretation of the initial conditions and whether the box's state of rest is a pre-condition for the problem.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup. Some guidance has been offered regarding the relevance of initial conditions, but no consensus has been reached on the necessity of certain information for solving the problem.

Contextual Notes

There is a focus on the conditions at the bottom of the slope and the role of static friction in determining the box's motion. Participants are questioning the assumptions related to the forces and the box's state of rest.

BOAS
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Hi,

i'm stuck on a mechanics problem where the initial conditions are a mass at rest on an inclined plane, but the forces are seemingly unbalanced.

1. Homework Statement

A 5.17 kg box sits at rest at the bottom of a ramp that is 8.56 m long and that is inclined
at 40.0◦ above the horizontal. The coefficient of kinetic friction is 0.40, and the coefficient
of static friction is 0.50.
What constant force F, applied parallel to the surface of the ramp, is required to push the
box to the top of the ramp in a time of 4.22 s ?

Homework Equations

The Attempt at a Solution



I have drawn a freebody diagram with the weight of the box acting straight down. perpendicular to the slope, we have the component $$mg \cos(\theta)$$ and the normal force being $$-mg \cos(\theta)$$. parallel to the slope, a component of the weight acts down the the slope $$mg \sin(\theta)$$ and the force of static friction opposes this motion.

Static friction $$f_{s} \leq \mu_{s} n$$ where $$n = -mg \cos(\theta) = -19.43 N$$

This is the max force that can be applied parallel to the slope without the box beginning to move, but the component of weight acting down the slope is greater than this, so why is the box not moving?

Was I supposed to consider "at the bottom of the slope" to mean the normal force = weight?
 
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Maybe it is just being held there before the force parallel to incline is applied. It is just a pre-condition to start the problem.
 
It seems to me that I don't need this information to solve the problem.

The applied force is parallel to the slope, so kinetic friction remains constant. Using the equations of kinematics, I can determine the acceleration required to cover the distance in that time. Using Newton's second law, I can determine what the net force must be, and therefore the applied force is equal to the difference between the kinetic friction and resultant force.
 
BOAS said:
It seems to me that I don't need this information to solve the problem.
You need to know that it starts at the bottom and that it starts from rest. No hidden meaning beyond that.
 
Doc Al said:
You need to know that it starts at the bottom and that it starts from rest. No hidden meaning beyond that.
Thanks
 

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