Question

A tractor is being used to pull two large logs across a field. A chain connects the logs to eachther; the front log is connected to the tractor by a separate chain. The mass of the front log is 180 kg. The mass of the back log is 220 kg. The coefficient of friction between the logs and the field is approximately 0.45. The tension in the chain connecting the tractor to the front log is 1850 N. Determine the tension in the chain that connects the two logs.

Answer 1

Answer:

$T = 1017.5 N$

Explanation:

As we know that the friction force on the wooden log is given as

$F_f = \mu F_n$

so we have

$F_f = \mu mg$

For front log we have

$F_f = (0.45)\times (180)(9.81)$

$F_f_1 = 794.6 N$

For back log we have

$F_f = (0.45)\times (220)(9.81)$

$F_f_2 = 971.2 N$

Now by force equation we have

$F - F_f_1 - F_f_2 = (m_1 + m_2)a$

$1850 - 971.2 - 794.6 = (220 + 180) a$

so we have

$a = 0.21 m/s^2$

now for the tension in the string between two logs we have

$T - F_f_2 = m_2 a$

$T - 971.2 = 220(0.21)$

$T = 1017.5 N$