Answer to Question #100610 in Mechanics | Relativity for Jeremiah

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 183 N at an angle
of 27.6
◦
above the horizontal. The box has a
mass of 41.2 kg, and µk between the box and
the floor is 0.28.Now the student moves the box up a ramp
(with the same coefficient of friction) inclined
at 10.6
◦ with the horizontal.
b) If the box starts from rest at the bottom
of the ramp and is pulled at an angle of 27.6
◦
with respect to the incline and with the same
183 N force, what is the acceleration up the
ramp?
Answer in units of m/s^2

Answer 1

b) Sketch what happens:

Now simply and correctly applying Newton's second law calculate the acceleration. Point x-axis to the right along the incline and y-axle at 90 degrees to the slope upward:

$$Ox: F\text{ cos}27.6^\circ-\mu_kN-mg\text{ sin}10.6^\circ=ma,\\ Oy: F\text{ sin}27.6^\circ-mg\text{ cos}10.6^\circ+N=0.$$

Hence:

$$Ox: F\text{ cos}27.6^\circ-\mu_kN-mg\text{ sin}10.6^\circ=ma,\\ Oy: F\text{ sin}27.6^\circ-mg\text{ cos}10.6^\circ+N=0.\\ \space\\ \text{Therefore:}\\ \space\\ N=mg\text{ cos}10.6^\circ-F\text{ sin}27.6^\circ,\\ a=\frac{F}{m}\text{ cos}27.6^\circ-\mu_k\frac{N}{m}-g\text{ sin}10.6^\circ=\\ \space\\ =\frac{F}{m}(\text{cos}27.6^\circ+\text{ sin}27.6^\circ)-\\ \space\\-g(\mu_k\text{ cos}10.6^\circ-\text{ sin}10.6^\circ)=\\ =5.10\text{ m/s}^2.$$