Answer to Question #100626 in Mechanics | Relativity for Jarell Banman

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 188 N at an angle
of 32.5
◦
above the horizontal. The box has a
mass of 25.7 kg, and µk between the box and
the floor is 0.15.
The acceleration of gravity is 9.81 m/s
2
.
Find the acceleration of the box.
Answer in units of m/s

Answer 1

We project all the forces acting on the body on the coordinate axis and write down the laws of motion in projection on these axes:

$N+F \cdot sin{\alpha}-m \cdot g=0$ (1)

$F \cdot cos{\alpha} -F_fr=m \cdot a$ (2)

The sliding friction force is determined by the formula:

$F_fr=\mu\cdot N$ (3)

From expression (2) we express the friction force

$F_fr=F \cdot cos{\alpha} -m \cdot a$

From expression (1) we express the force N

$N=m \cdot g-F \cdot sin{\alpha}$

substitute in (3)

$F \cdot cos{\alpha} -m \cdot a=\mu\cdot (m \cdot g-F \cdot sin{\alpha})$

express acceleration

$a=\frac{F \cdot cos{(\alpha)}-\mu\cdot (m \cdot g-F \cdot sin{(\alpha)})}{m}$

$a=\frac{188 \cdot cos({32.5^0})-0.15\cdot (25.7\cdot 9.81-188 \cdot sin({32.5^0}))}{25.7}=5.288m/s^2$