Question

Kayla, a fitness trainer, develops an exercise that involves pulling a heavy crate across a rough surface. The exercise involves using a rope to pull a crate of mass M = 45.0 kg along a carpeted track. Using her physics expertise, Kayla determines that the coefficients of kinetic and static friction between the crate and the carpet are μk = 0.410 and μs = 0.770, respectively. Next, Kayla has Ramon pull on the rope as hard as he can. If he pulls with a force of F = 325 N along the direction of the rope, and the rope makes the same angle of θ = 23.0 degrees with the floor, what is the acceleration of the box?

Answer 1

Answer:

$a = 3.784\,\frac{m}{s^{2}}$

Explanation:

The Free Body Diagram of the system formed by the crate and the rope and the reference axis are presented below as an attached image. The equations of equilibrium are introduced hereafter:

$\Sigma F_{x} = T\cdot \cos \theta - \mu\cdot N = m\cdot a$

$\Sigma F_{y} = T\cdot \sin \theta + N - m\cdot g = 0$

After some algebraic handling, the system of equations is reduce to a sole expression:

$N = m\cdot g - T\cdot \sin \theta$

$T\cdot \cos \theta - \mu \cdot (m\cdot g - T\cdot \sin \theta) =m\cdot a$

$T\cdot (\cos\theta + \mu \cdot \sin \theta) - \mu \cdot m \cdot g = m\cdot a$

The minimum force to accelerate the crate from rest is:

$T_{min} = \frac{m\cdot \mu_{s}\cdot g}{\cos \theta + \mu_{s} \cdot \sin \theta}$

$T_{min} = \frac{(45\,kg)\cdot (0.770)\cdot (9.807\,\frac{m}{s^{2}} )}{\cos 23^{\textdegree}+(0.770)\cdot \sin 23^{\textdegree}}$

$T_{min} = 278.223\,N$

Since $T > T_{min}$, the crate will experiment an acceleration due to the tension exerted. The acceleration of the crate is:

$a = \frac{T}{m}\cdot (\cos \theta + \mu_{k}\cdot \sin \theta)-\mu_{k}\cdot g$

$a = \frac{325\,N}{45\,kg}\cdot [\cos 23^{\textdegree}+(0.410)\cdot \sin 23^{\textdegree}]-(0.410)\cdot (9.807\,\frac{m}{s^{2}} )$

$a = 3.784\,\frac{m}{s^{2}}$