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

Step-by-step 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

Answer 2

Step-by-step 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 =
`(m_(1) - m_(2))/(m_(1) + m_(2)) * g`

=
`(17.7 - 11.1)/(17.7 + 11.1) * 9.8`

=
`(6.6)/(28.8) * 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 * 9.8 - T = 17.7 * 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.