Question

An Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. Assume that the rope and pulley are massless and that there is no friction in the pulley. If the masses have the values m 1 = 17.7 kg m1=17.7 kg and m 2 = 11.1 kg, m2=11.1 kg, find the magnitude of their acceleration a a and the tension T T in the rope. Use g = 9.81 m/s 2 .

Answer 1

Explanation:

According to Newton's second law of motion,

            $m_{1}g - T = m_{1}a$ ......... (1)

and,    $T - m_{2}g = m_{2}a$ ......... (2)

When we add both equations, (1) and (2) then the expression obtained for "a" is as follows.

             a = $\frac{m_{1} - m_{2}}{m_{1} + m_{2}} \times g$

                = $\frac{17.7 - 11.1}{17.7 + 11.1} \times 9.8$

                = $\frac{6.6}{28.8} \times 9.8$

                = 2.24 $m/s^{2}$

Now, putting the value of "a" in equation (1) then we will calculate the tension as follows.

              $m_{1}g - T = m_{1}a$

              $17.7 \times 9.8 - T = 17.7 \times 2.24$

                   173.46 - T = 39.648

                       T = 133.812 N

Thus, we can conclude that the magnitude of their acceleration is 2.24 $m/s^{2}$ and the tension T is 133.812 N in the rope.

Answer 2

Answer:

Explanation:

m1 = 17.7 kg

m2 = 11.1 kg

Let a be the acceleration and T be the tension in the string.

use Newton's second law

m1 g - T = m1 x a ....(1)

T - m2 g = m2 x a ..... (2)

Adding both the equations

(m1 - m2) g = ( m1 + m2 ) x a

(17.7 - 11.1 ) x 9.8 = (17.7 + 11.1) x a

64.68 = 28.8 a

a = 2.25 m/s²

Put the value of a in equation (1)

17.7 x 9.8 - T = 17.7 x 2.25

173.46 - T = 39.825

T = 133.64 N