Newton's second law we can find the answers for the block system with a pulley are
a) The acceleration is 0.39 m / s² descending the inclined plane
b) The tension of the rope is 3.9 N
Newton's Second Law gives the relationship between the net external force is proportional to the mass and the acceleration of the body
∑ F = m a
Where bold letters indicate vectors, F is force, m is mass, and a is the acceleration.
Reference systems are coordinate systems with respect to which to perform vector decomposition measurements. In this case we take a system with the x axis parallel to the plane and the positive direction the direction of movement, the ej y is perpendicular to the plane,
The free body diagram is a diagram of the forces in the system without the physical body details, in the attached we have a free body diagram. Let's write the equation for each axis
y-axis
N - [tex]W_y[/tex] = 0
N = [tex]W_y[/tex]
x-axis
Wₓ -fr -T = m a
Let's use trigonometric to break down the weight
cos 37 =[tex]\frac{W_y}{W}[/tex]
sin 37 = [tex]\fracW_{x}{W}[/tex]
W_y = W cos 37
Wₓ = W sin 37
The friction force is the macroscopic manifestation of the interaction between the two bodies, it has the formula
fr =μ N
fr = μ W sin 37
We substitute
W sin 37 - μ W cos 37 - T = m a
Now let's use Newton's second law for rotational motion in the pulley
Σ τ = I α
Where τ is the torque, I the moment of inertia and α is the angular acceleration of the pulley.
Torque is the product of the force times the perpendicular distance to the axis of rotation, let's take the clockwise direction as positive
The weight forces in the y axis and normal are on the axis, their distance is zero, so they do not create torque
T R = I α
Let's use the relationship between linear and angular variables
a = α R
α = [tex]\frac{a}{R}[/tex]
TR = I a / R
Let's write our equations
W sin 37 - μ W cos 37 - T = m a
T = I [tex]\frac{a}{R^2 }[/tex]
We have two equation es and two unknowns, let's solve the system
m g sin 37 - μ mg cos 37 - I a / Rr² = m a
m (g sin 37 - μ g cos 37) =m a ( 1 + [tex]\frac{I}{mR^2}[/tex])
a = g [tex]g \ \frac{sin 37 - \mu \ cos 37}{ 1 + \frac{I }{mR^2 } }[/tex]
Let's calculate
a = [tex]9.8 \ \frac{sin 37 - 0.2 cos 37 }{ 1 + \frac{0.4}{1.0 \ 0.2^2 } }[/tex]
a = 0.39 m / s²
We look for the tension of the rope
T = I [tex]\frac{a}{R^2 }[/tex]
T = 0.4 0.39 / 0.2²
T = 3.9 N
In conclusion using Newton's second law we can find the answers for the block system with a pulley are
a) The acceleration is 0.39 m / s² descending the inclined plane
b) The tension of the rope is 3.9 N
Learn more here: brainly.com/question/19860811
