Question

Two blocks connected by a light string are being pulled across a frictionless horizontal tabletop by a hanging 16.2-N weight (block C). Block A has a mass of 2.20 kg. The mass of block B is only 1.00 kg. The strings remain taut at all times.
Assuming the pulley is massless and frictionless, what are the values of the tensions 1 and 2?

Answer 1

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

       $W_c -T_2 = m_c a$

Block A

X axis  

       $T_2 - T_1 - W_a_x = m_a a$  

Y axis  

       $N_a - W_a_y = 0$  

Block B

X axis

      $T_1 - W_b_x = m_b a$  

Y axis

      $N_b - W_b_y =0$

Let's  use trigonometry to find the components of the weight.

Block A

         cos θ = $\frac{W_a_y}{W_a}$  

         sin θ = $\frac{W_a_x}{W_a}$

         $W_a_y = W_a cos \theta$

         $W_a_x= W_a sin \theta$

Block B

        cos θ = $\frac{W_b_y}{W_b}$

        sin θ =  $\frac{W_b_x}{W_b}$

        $W_b_y = W_b cos \theta \\W_b_x = W_b sin \theta$

Let's write our system of equations.

     $W_c - T_2 = m_c a \\ T_2 - T_1 - W_a_x = m_a a \\T_1 - W_b_x = m_b a$

 

Let's find the acceleration of the bodies, adding the equations.

     $W_c - W_a_x - W_b_x = ( m_a+m_b+m_c) a$  

         

The weight is

    W = mg

Let's  substitute

         $(m_c - m_a -m_b ) g \ sin \theta = ( m_c+m_a+m_b) \ a \\a= \frac{ m_c-m_a-m_b }{ m_a+m_b+m_c} \ g sin \theta$

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

          $a= \frac{1.65 -2.20 -1.00}{1.65+2.20+1.00} \ 9.8 \ sin \theta \\a= -0.3096 sin \theta$

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

           $W_c - T_2 = m_c a \\T_2 = m_c ( g-a)\\T_2 = 1.65 ( 9.8 + 0.226)$

            T₂ = 16.54 N

From the equation of block B

          $T_1 - W_b_x = m_b a\\T_1 = m_b (a + g sin \theta)\\T_1 = 1.00 (-0.226 + 9.8 \ sin 45)$

           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