Answer :
Answer:
a) a= 8.33 m/s², T = 12.495 N , b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²
a) The maximum acceleration of the trailer and the corresponding tension in the cable is [tex]8.33 m/s^2[/tex] and 12.495 N
b) The maximum deceleration of the trailer is [tex]2.45 m / s^2[/tex]
- The calculation is as follows;
Here we apply Newton's second law to the lower charge
[tex]fr_1 + fr_2 = ma[/tex]
The equation for the force of friction is
[tex]fr = \mu N[/tex]
Y Axis
[tex]N - W_1 -W_2 = 0\\\\N = W_1 + W_2\\\\N = (m_1 + m_2) g[/tex]
Because the beams are the same, it has the similar mass
N = 2 m g
Now
[tex]\mu_1 2mg + \mu_2 mg = m a\\\\a = (2\mu_1 + \mu_2) g\\\\a = (2 0.30 + 0.25) 9.8\\\\ a= 8.33 m/s^2[/tex]
Now cable tension with beam 2
[tex]T = m_2 a[/tex]
T = 1500 8.33
T = 12.495 N
b)
the maximum deceleration of the trailer is
[tex]fr = m_1 a_2\\\\N- w_2 = 0\\\\N = W_2 = mg\\\\ \mu_2 mg = m a_2\\\\a = \mu_2 g\\\\a = 0.25 (9.)8\\\\ a = 2.45 m / s^2\\\\[/tex]
Find out more information about the cable here: https://brainly.com/question/13818952?referrer=searchResults