Answer:
i) A force of 187.612 newtons at a direction of 30º above the horizontal to keep the crate moving at constant velocity, ii) The force found in item i) makes pulling easier than pulling horizontally.
Explanation:
i) Let be the crate our system. As first step we draw a free body diagram on the crate, whose outcome is included as attachment below. Tension exerts a force on the crate which moves on the horizontal ground.
By Newton's Laws, a system moving at constant speed must have a net acceleration of zero. The equations of equilibrium are now described:
[tex]\Sigma F _{x} = F \cdot \cos \alpha - \mu_{k}\cdot N = 0[/tex] (Eq. 1)
[tex]\Sigma F_{y} = F\cdot \sin \alpha -m\cdot g +N = 0[/tex] (Eq. 2)
Where:
[tex]F[/tex] - Tension force exerted on crate, measured in newtons.
[tex]\alpha[/tex] - Angle of the tension force, measured in sexagesimal degrees.
[tex]\mu_{k}[/tex] - Coefficient of kinetic friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
[tex]m[/tex] - Mass of the crate, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
We eliminate the normal force in (Eq. 1) by substitution:
(Eq. 2) in (Eq. 1)
[tex]F\cdot \cos \alpha -\mu_{k}\cdot (m\cdot g -F\cdot \sin \alpha) = 0[/tex]
Then, tension force is cleared within:
[tex]F \cdot \cos \alpha -\mu_{k}\cdot m \cdot g + \mu_{k}\cdot F\cdot \sin \alpha = 0[/tex]
[tex]F \cdot (\cos \alpha + \mu_{k}\cdot \sin \alpha) = \mu_{k}\cdot m \cdot g[/tex]
[tex]F = \frac{\mu_{k}\cdot m \cdot g}{\cos \alpha + \mu_{k}\cdot \sin \alpha}[/tex]
If [tex]m\cdot g = 500\,N[/tex], [tex]\mu_{k} = 0.40[/tex] and [tex]\alpha = 30^{\circ}[/tex], the force needed to pull the crate is:
[tex]F = \frac{(0.40)\cdot (500\,N)}{\cos 30^{\circ}+(0.40)\cdot \sin 30^{\circ}}[/tex]
[tex]F = 187.612\,N[/tex]
A force of 187.612 newtons at a direction of 30º above the horizontal to keep the crate moving at constant velocity.
ii) Now, we assume that [tex]\alpha = 0^{\circ}[/tex], the force needed to pull the crate is:
[tex]F = \frac{(0.40)\cdot (500\,N)}{\cos 0^{\circ}+(0.40)\cdot \sin 0^{\circ}}[/tex]
[tex]F = 200\,N[/tex]
Which leads to the conclusion that previous force makes pulling the crate easier than pulling horizontally.
