Answer:
a) a = 17.1 m / s², b) F = 3.04 N
Explanation:
This is an exercise of Newton's second law, in this case the selection of the reference system is very important, we have two possibilities
* a reference system with the horizontal x axis, for this selection the normal and the friction force have x and y components
* a reference system with the x axis parallel to the plane, in this case the weight and the applied force have x and y components
We are going to select the second possibility, since it is the most used in inclined plane problems, let's analyze the angle of the applied force (F) it has an angle 10º with respect to the x axis, if we rotate this axis 30º the new angle is
θ = 10 -30 = -20º
The negative sign indicates that it is in the fourth quadrant. Let's use trigonometry to find the components of the applied force
sin (-20) = F_{y} / F
cos (-20) = Fₓ / F
F_{y} = F sin (-20)
Fₓ = F cos (-20)
F_y = 18 sin (-20) = -6.16 N
Fₓ = 18 cos (-20) = 16.9 N
The decomposition of the weight is the customary
sin 30 = Wₓ / W
cos 30 = W_y / W
Wₓ = W sin 30 = mg sin 30
W_y = W cos 30 = m g cos 30
Wₓ = 0.8 9.8 sin 30 = 3.92 N
W_y = 0.8 9.8 cos 30 = 6.79 N
Notice that in the case the angle is measured with respect to the axis y perpendicular to the plane
Now we can write Newton's second law for each axis
X axis
Fₓ - fr = m a
Y Axis
N - [tex]F_{y}[/tex] - Wy = 0
N =F_{y} + Wy
N = 6.16 + 6.79
They both go to the negative side of the axis and
N = 12.95 N
The friction force has the formula
fr = μ N
we substitute
Fₓ - μ N = m a
a = (Fₓ - μ N) / m
we calculate
a = (16.9 - 0.25 12.95) / 0.8
a = 17.1 m / s²
b) now the block slides down with constant speed, therefore the acceleration is zero
ask for the value of the applied force, we will suppose that with the same angle, that is, only its modulus was reduced
Newton's law for the x axis
Fₓ -fr = 0
Fₓ = fr
F cos 20 = μ N
F = μ N / cos 20
we calculate
F = 0.25 12.95 / cos 20
F = 3.04 N
this is the force applied at an angle of 10º to the horizontal
