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:
(Eq. 1)
(Eq. 2)
Where:
- Tension force exerted on crate, measured in newtons.
- Angle of the tension force, measured in sexagesimal degrees.
- Coefficient of kinetic friction, dimensionless.
- Normal force, measured in newtons.
- Mass of the crate, measured in kilograms.
- Gravitational acceleration, measured in meters per square second.
We eliminate the normal force in (Eq. 1) by substitution:
(Eq. 2) in (Eq. 1)
![F\cdot \cos \alpha -\mu_{k}\cdot (m\cdot g -F\cdot \sin \alpha) = 0](https://tex.z-dn.net/?f=F%5Ccdot%20%5Ccos%20%5Calpha%20-%5Cmu_%7Bk%7D%5Ccdot%20%28m%5Ccdot%20g%20-F%5Ccdot%20%5Csin%20%5Calpha%29%20%3D%200)
Then, tension force is cleared within:
![F \cdot \cos \alpha -\mu_{k}\cdot m \cdot g + \mu_{k}\cdot F\cdot \sin \alpha = 0](https://tex.z-dn.net/?f=F%20%5Ccdot%20%5Ccos%20%5Calpha%20-%5Cmu_%7Bk%7D%5Ccdot%20m%20%5Ccdot%20g%20%2B%20%5Cmu_%7Bk%7D%5Ccdot%20F%5Ccdot%20%5Csin%20%5Calpha%20%3D%200)
![F \cdot (\cos \alpha + \mu_{k}\cdot \sin \alpha) = \mu_{k}\cdot m \cdot g](https://tex.z-dn.net/?f=F%20%5Ccdot%20%28%5Ccos%20%5Calpha%20%2B%20%5Cmu_%7Bk%7D%5Ccdot%20%5Csin%20%5Calpha%29%20%3D%20%5Cmu_%7Bk%7D%5Ccdot%20m%20%5Ccdot%20g)
![F = \frac{\mu_{k}\cdot m \cdot g}{\cos \alpha + \mu_{k}\cdot \sin \alpha}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%5Cmu_%7Bk%7D%5Ccdot%20m%20%5Ccdot%20g%7D%7B%5Ccos%20%5Calpha%20%2B%20%5Cmu_%7Bk%7D%5Ccdot%20%5Csin%20%5Calpha%7D)
If
,
and
, the force needed to pull the crate is:
![F = \frac{(0.40)\cdot (500\,N)}{\cos 30^{\circ}+(0.40)\cdot \sin 30^{\circ}}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%280.40%29%5Ccdot%20%28500%5C%2CN%29%7D%7B%5Ccos%2030%5E%7B%5Ccirc%7D%2B%280.40%29%5Ccdot%20%5Csin%2030%5E%7B%5Ccirc%7D%7D)
![F = 187.612\,N](https://tex.z-dn.net/?f=F%20%3D%20187.612%5C%2CN)
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
, the force needed to pull the crate is:
![F = \frac{(0.40)\cdot (500\,N)}{\cos 0^{\circ}+(0.40)\cdot \sin 0^{\circ}}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%280.40%29%5Ccdot%20%28500%5C%2CN%29%7D%7B%5Ccos%200%5E%7B%5Ccirc%7D%2B%280.40%29%5Ccdot%20%5Csin%200%5E%7B%5Ccirc%7D%7D)
![F = 200\,N](https://tex.z-dn.net/?f=F%20%3D%20200%5C%2CN)
Which leads to the conclusion that previous force makes pulling the crate easier than pulling horizontally.