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)
Then, tension force is cleared within:
If , and , the force needed to pull the crate is:
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:
Which leads to the conclusion that previous force makes pulling the crate easier than pulling horizontally.