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

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The discussion centers on a mechanics problem involving a 5.17 kg box on an inclined plane, where the box remains at rest despite unbalanced forces. The user analyzes the forces acting on the box, including gravitational components and friction, and questions why the box does not move when the downward force exceeds static friction. Clarification is provided that the initial conditions indicate the box starts at rest at the bottom of the ramp, without any hidden implications. The user concludes that understanding these initial conditions is essential for solving the problem, emphasizing that they are not extraneous information. The focus remains on applying kinematics and Newton's laws to determine the required force to move the box up the ramp.
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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