Question

A student moves a box of books down the hall
by pulling on a rope attached to the box. The
student pulls with a force of 174 N at an angle
of 35.0 above the horizontal. The box has a
mass of 25.1 kg, and μk between the box and
the floor is 0.25.
The acceleration of gravity is 9.81 m/s2 .
Find the acceleration of the box.
Answer in units of m/s2

Answer 1

Make a summary of all of the forces acting on the box. I chose the applied force to be up and right, but of course it could have been to the up and left as well.
\[\begin{array}{c|c|c|c}
&\text{name}&\text{direction}&\text{magnitude} \\ \hline
\vec F_g & \text{gravitational force} & \text{downward} & mg \\
\vec F_n & \text{normal force} & \text{upward} & (??)\\
\vec F_a & \text{applied force} & \text{up & right, angle }\theta & \text{given}\\
\vec F_f & \text{friction} & \text{leftward} & \mu_k F_n
\end{array}\]Let us now apply Newton's first and second laws to the net forces in the \(x\) (horizontal) and \(y\) (vertical) directions. Since the acceleration in the \(y\) direction is zero, it must be that all corresponding forces cancel (and thus add to zero). Upwards and rightwards are treated as the positive directions.\[\sum F_x = F_a \cos \theta - \mu_k F_n=ma_x\]\[\sum F_y = F_n+F_a\sin \theta -mg = 0 \text{ (no vertical acceleration)}\]The second equation allows us to find \(F_n=mg-F_a \sin \theta\). We will substitute this value into the first equation. It is now easy to solve for \(a_x\). Plug in given values to get numerical result.\[F_a \cos \theta - \mu_k \underbrace{(mg-F_a \sin\theta)}_{F_n}=ma_x\]\[\boxed{\displaystyle a_x = \frac{F_a}{m}(\cos\theta - \mu_k \sin\theta)-\mu_kg}\]