The crate is moving at constant velocity when the forces acting on it are
balanced.
- The value of the force F required to pull the crate with constant velocity is; [tex]\underline{F = \sqrt{2} \cdot \mu \cdot m\cdot g}[/tex]
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 = [tex]\mathbf{F_N}[/tex]
Each side of the through gives a normal reaction.
The vertical component of the normal reaction on each side of the through
is therefore;
- [tex]F_N[/tex]·j = [tex]\mathbf{F_N}[/tex] × sin(θ)
The sum of the vertical component = [tex]F_N[/tex]·j + [tex]F_N[/tex]·j = 2·[tex]F_N[/tex]·j = 2·[tex]F_N[/tex]×sin(θ)
The sum of the vertical component of the normal reactions = The weight of the crate
Therefore;
2·[tex]F_N[/tex]×sin(θ) = m·g
θ = 45°
Therefore;
2·[tex]F_N[/tex]×sin(45°) = m·g
[tex]\displaystyle sin(45^{\circ}) = \mathbf{\frac{\sqrt{2} }{2}}[/tex]
Therefore;
[tex]\displaystyle 2 \cdot F_N \cdot sin(45^{\circ}) = 2 \cdot F_N \times \frac{\sqrt{2} }{2} = \sqrt{2} \cdot F_N[/tex]
[tex]\displaystyle F_N = \mathbf{ \frac{m \cdot g}{\sqrt{2} }}[/tex]
Which gives;
[tex]\displaystyle Force \ required, \ F = Sum \of \ friction \ forces \ = 2 \times F_N \times \mu = \mathbf{ 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu}[/tex]
[tex]\displaystyle Force \ required, \ F = 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu = \mathbf{ \sqrt{2} \cdot \mu \cdot m \cdot g}[/tex]
- Force required to pull the crate at constant velocity, F = √2·μ·m·g
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