physics tension problem?

Question

5 (part 1 of 3)
A block is at rest on the incline.The coefficients of static and ki-
netic friction are ¹s = 0:49 and ¹k = 0:42, respectively.
The acceleration of gravity is 9:8 m/s2. The mass is 31 kg.
The angle of the incline is 23 degrees. What is the frictional force acting on the 31 kg mass? Answer in units of N.
(part 2 of 3)
What is the largest angle which the incline can have so that the mass does not slide down the incline? Answer in units of degrees.
(part 3 of 3)
What is the acceleration of the block down the incline if the angle of the incline is 33 ± ? Answer in units of m/s2.

Answer 1

For all three parts use the force from gravity as
sin(angle)*m*g
and the Normal force as
Cos(angle)*m*g

Part 1 of 3

F=sin(23)*31*9.8
=118.7N

Part 2 of 3
Max is the normal force times the static coefficient
when that is equal to the force pulling the block down the slope
sin(angle)*m*g=
cos(angle)*m*g*u(static)
tan(angle)=u(static)
angle=ATan(u)
=ATan(.49)
=26.1 degrees

Part 3 of 3 is to determine the magnitude of the resultant force down the incline and use
F=m*a to find a
m*a=m*g*(sin(33)-cos(33)*u(kinetic))
a=sin(33)-cos(33)*.42

=.192 m/s^2
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