Question

A delivery company has just unloaded a 500 N crate full of home exercise equipment in your level driveway. Suppose you try to move the crate by pulling upward on the rope at an angle of 30∘ above the horizontal. How hard do you have to pull to keep the crate moving with constant velocity? Is this easier or harder than pulling horizontally? For this crate the coefficient of kinetic friction, μk, is 0.40.

Answer 1

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:

$\Sigma F _{x} = F \cdot \cos \alpha - \mu_{k}\cdot N = 0$ (Eq. 1)

$\Sigma F_{y} = F\cdot \sin \alpha -m\cdot g +N = 0$ (Eq. 2)

Where:

$F$ - Tension force exerted on crate, measured in newtons.

$\alpha$ - Angle of the tension force, measured in sexagesimal degrees.

$\mu_{k}$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$m$ - Mass of the crate, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

We eliminate the normal force in (Eq. 1) by substitution:

(Eq. 2) in (Eq. 1)

$F\cdot \cos \alpha -\mu_{k}\cdot (m\cdot g -F\cdot \sin \alpha) = 0$

Then, tension force is cleared within:

$F \cdot \cos \alpha -\mu_{k}\cdot m \cdot g + \mu_{k}\cdot F\cdot \sin \alpha = 0$

$F \cdot (\cos \alpha + \mu_{k}\cdot \sin \alpha) = \mu_{k}\cdot m \cdot g$

$F = \frac{\mu_{k}\cdot m \cdot g}{\cos \alpha + \mu_{k}\cdot \sin \alpha}$

If $m\cdot g = 500\,N$, $\mu_{k} = 0.40$ and $\alpha = 30^{\circ}$, the force needed to pull the crate is:

$F = \frac{(0.40)\cdot (500\,N)}{\cos 30^{\circ}+(0.40)\cdot \sin 30^{\circ}}$

$F = 187.612\,N$

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 $\alpha = 0^{\circ}$, the force needed to pull the crate is:

$F = \frac{(0.40)\cdot (500\,N)}{\cos 0^{\circ}+(0.40)\cdot \sin 0^{\circ}}$

$F = 200\,N$

Which leads to the conclusion that previous force makes pulling the crate easier than pulling horizontally.