- #1
BOAS
- 546
- 19
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 ?
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?
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?
