The crate is moving at constant velocity when the forces acting on it are
balanced.
- The value of the force <em>F</em> required to pull the crate with constant velocity is;
Reasons:
Mass of the crate = m
Cross section of through = Right angled
Orientation (inclination) of the through to horizontal = 45°
Coefficient of kinetic friction = μ
Required:
The value of the required to pull the crate along the through at constant
velocity.
Solution:
When the through is moving at constant velocity, we have;
Friction force acting on crate = Force pulling the crate
Friction force = Normal reaction × Coefficient of kinetic friction
Normal reaction on an inclined plane =
Each side of the through gives a normal reaction.
The vertical component of the normal reaction on each side of the through
is therefore;
·j =
× sin(θ)
The sum of the vertical component = ·j +
·j = 2·
·j = 2·
×sin(θ)
The sum of the vertical component of the normal reactions = The weight of the crate
Therefore;
2·×sin(θ) = m·g
θ = 45°
Therefore;
2·×sin(45°) = m·g
Therefore;
Which gives;
- Force required to pull the crate at constant velocity, <u>F = √2·μ·m·g</u>
Learn more about force of friction here:
