Respuesta :
Answer:
a) T = -22796.5 N, b) F = 3000 N
Explanation:
a) For this part we use Newton's second law
Let's set a reference frame with the x-axis in the direction of motion and the y-axis in the vertical direction.
We make a free-body diagram for each body,
the tractor has the bar force (T) and the push force (F) and the friction force (fr₁)
Y axis
N₁ -W₁ = 0
N₁ = M₁ g
X axis
F - T - fr₁ = M₁ a
the Trailer has the bar force (T) and the friction force (fr₂)
Y axis
N₂ - W₂ = 0
N₂ = m₂ g
X axis
T - fr₂ = m₂ a
let's write the system of equations
F - T - fr₁ = M₁ a (1)
T - fr₂ = m₂ a
we solve
F - (fr₁ + fr₂) = (M₁ + m₂) a
indicate that the total friction forces are fr = 3000N
fr = fr₁ + fr₂
F =[tex]\frac{(M_1+m_2) a}{fr}[/tex]
let's calculate
F =[tex]\frac{(2000+1500) \ 3}{3000}[/tex]
F = 3.5 N
The friction force is
fr = μ N
the norm of the system is N = N₁ + N₂
μ = [tex]\frac{fr}{N_1 + N_2}[/tex]
μ = [tex]\frac{3000}{2000+1500}[/tex]
μ = 0.858
with this value we can find the friction force 1 and substitute in equation 1
F - T - μ N₁ = M₁ a
T = F - M₁ (a + μ g)
T = 3.5 - 2000 (3 + 0.858 9.8)
T = -22796.5 N
b) when the system moves with constant velocity the acceleration is zero
F - T - fr₁ = 0
T - fr₂ = 0
we solve
F + (fr₁ + fr₂) = 0
F = fr₁ + fr₂
F = 3000 N
