Answer to Question #100605 in Mechanics | Relativity for Jeremiah

Question

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

A good sketch is always a good helper:

Write all forces according to Newton's second law:

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

(The answer is correct if the force is 183 N, the coefficient of kinetic friction is 0.28, and the mass of the box is 41.2 kg).