Answer to Question #100595 in Mechanics | Relativity for Micah Smith

Question

Now the student moves the box up a ramp
(with the same coefficient of friction) inclined
at 10.1
◦ with the horizontal.
b) If the box starts from rest at the bottom
of the ramp and is pulled at an angle of 26.7
◦
with respect to the incline and with the same
173 N force, what is the acceleration up the
ramp?
Answer in units of m/s
2
.

Answer 1

$m=42.4kg$

$a=9.81 m/s2$

Considering the forces perpendicular to the inclined floor 

$mg*cos(10.1)=R+173*sin(26.7)$

solving above equation

$R=331.766 N$

From, F′=μR Where F' is the frictional force

$F′=66.353 N (1)$

Applying $F=ma$ along the inclined floor

$173*cos(26.7)-F'-mg*sin(10.1)=ma$

From (1)

$173*cos(26.7)-66.353-42.4*9.81*sin(10.1)=42.4a$

Solving above equation

$a=0.3598 m/s2$