Respuesta :
Newton's second law allows us to find the results for the string tensions are:
- T₁ = 6.7 N
- T₂ = 16.54 N
Newton's second law gives a relationship between force, mass and acceleration of bodies
∑ F = ma
Where the bold letters indicate vectors, F is the force, m the mass and the acceleration.
Free-body diagrams are representations of the forces applied to bodies without the details of them.
The reference system is a coordinate system with respect to which the forces decompose, in this case the x-axis is parallel to the plane and the positive direction in the direction of movement, the y-axis is perpendicular to the plane.
In the attachment we see a free-body diagram of the three-block system.
Let's apply Newton's second law to each body.
Block C
Y-axis
[tex]W_c -T_2 = m_c a[/tex]
Block A
X axis
[tex]T_2 - T_1 - W_a_x = m_a a[/tex]
Y axis
[tex]N_a - W_a_y = 0[/tex]
Block B
X axis
[tex]T_1 - W_b_x = m_b a[/tex]
Y axis
[tex]N_b - W_b_y =0[/tex]
Let's use trigonometry to find the components of the weight.
Block A
cos θ = [tex]\frac{W_a_y}{W_a}[/tex]
sin θ = [tex]\frac{W_a_x}{W_a}[/tex]
[tex]W_a_y = W_a cos \theta[/tex]
[tex]W_a_x= W_a sin \theta[/tex]
Block B
cos θ = [tex]\frac{W_b_y}{W_b}[/tex]
sin θ = [tex]\frac{W_b_x}{W_b}[/tex]
[tex]W_b_y = W_b cos \theta \\W_b_x = W_b sin \theta[/tex]
Let's write our system of equations.
[tex]W_c - T_2 = m_c a \\ T_2 - T_1 - W_a_x = m_a a \\T_1 - W_b_x = m_b a[/tex]
Let's find the acceleration of the bodies, adding the equations.
[tex]W_c - W_a_x - W_b_x = ( m_a+m_b+m_c) a\\[/tex]
The weight is
W = mg
Let's substitute
[tex](m_c - m_a -m_b ) g \ sin \theta = ( m_c+m_a+m_b) \ a \\a= \frac{ m_c-m_a-m_b }{ m_a+m_b+m_c} \ g sin \theta[/tex]
Indicate ma mass of the block a ma = 1.00 kg, the mass of the block b mb = 2.2 kg and the weight of the block c Wc = 16.2 N, let's find the mass of block c.
m_c = Wc / g
m_c = 16.2 / 9.8
m_c = 1.65 kg
we substitute the values
[tex]a= \frac{1.65 -2.20 -1.00}{1.65+2.20+1.00} \ 9.8 \ sin \theta \\a= -0.3096 sin \theta[/tex]
The negative sign indicates that the system is descending, to be able to give a specified value an angle is needed, they assume that the angle of the ramp is 45º
a = - 0.3196 sin 45
a = -0.226 m / s
Taking the acceleration we are going to look for the tensions.
From the equation of block C
[tex]W_c - T_2 = m_c a \\T_2 = m_c ( g-a)\\T_2 = 1.65 ( 9.8 + 0.226)[/tex]
T₂ = 16.54 N
From the equation of block B
[tex]T_1 - W_b_x = m_b a\\T_1 = m_b (a + g sin \theta)\\T_1 = 1.00 (-0.226 + 9.8 \ sin 45)[/tex]
T₁ = 6.7 N
In conclusion using Newton's second law we can find the results for the string tensions are:
- T₁ = 6.7 N
- T₂ = 16.54 N
Learn more here: brainly.com/question/20575355
