Question

A tractor of mass 2000kg Pulls a trailer of mass 1500kg. The total frictional force is 3000N and the acceleration of the tractor is 3ms^-2. Calculate;
(a) the force exerted on the tractor by the tow-bar when the acceleration is 3ms^-2
(b) the force exerted when the tractor and trailer are moving at a constant speed of 4m/s

Answer 1

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 =$\frac{(M_1+m_2) a}{fr}$

         

let's calculate

        F =$\frac{(2000+1500) \ 3}{3000}$

        F = 3.5 N

The friction force is

       fr = μ N

the norm of the system is N = N₁ + N₂

       μ = $\frac{fr}{N_1 + N_2}$

       μ = $\frac{3000}{2000+1500}$

       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