Question

Blocks 1 and 2 of masses m1 and m2, respectively are connected by a light string, as show above. These blocks are further connected to a block of Mass M by another light string that passes over a pulley of negligible mass and friction. Blocks 1 and 2 move with a constant velocity v down the inclined plane, which makes an angle θ with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f.

1. Determine the coefficient of kinetic friction between the inclined plane and block 1
2. Determine the value of the suspended mass M that allows the inclined plane and block 1.
3. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane.

Answer 1

Answer:

1)    μ₁ = (1 + m₂/m₁ )/3  tan θ - m3/3m₁   scs θ , 2)   W₃ = g (m₁ + m₂) sin θ + 3f , 3) a₁ = g sin θ - f / m₁

Explanation:

For this problem we must use Newton's second law, as the speed of the blocks is constant zero acceleration

Let's look at the attachment for the names of the forces. Let's write the equations for each body

Body 3

     T- W₃ = 0

     T = W₃

Body 2

Y Axis

     N- Wy₂ = 0

     N = $W_{y}$₂

X axis

     Wₓ₂- T₃ + T₁ -fr₂ = 0

Let's use trigonometry to find the weight components

     sin θ = Wₓ₂ / W₂

    cos θ = $W_{y}$₂ / W₂

    Wₓ₂ = W₂ sin θ

    $W_{y}$₂ = W₂ cos θ

They indicate that the friction force of block 2 is

    fr₂ = 2f

We replace

    W₂ sin θ - T₃ + T₁ - 2f = 0

Body 1

Y Axis

     N₁ - $W_{y}$₁ = 0

X axis

     Wₓ₁ -T₁- fr₁ = 0

If we use the same trigonometry equation as for body 2

    Wₓ₁ = W₁ sin θ

    $W_{y}$₁ = W₁ cos θ

Give the rubbing force of block 1

   fr₁ = f

We replace

   W₁ sin θ - T₁ - f = 0

Let's write our system of equations

   T₃ - W₃ = 0

   W₂ sintea - T₃ + T₁ - 2f = 0

   W₁ sintea - T₁ - f = 0

Add the three equations

    W₂ sin θ + W₁ sin θ -W₃ - 3f = 0

    g sin θ (m₂ + m₁) - m₃ g = 3f

    f = g [(m₁ + m₂) sin θ - m₃] / 3

   The formula for friction force is

    fr₁ = μ N₁

    fr₁ = f = μ W₁ cos θ

    μ₁ = f / m₁ g cos θ

    μ₁ = g [(m₁ + m₂) sin θ - m₃] / 3 m₁ g cos θ

    μ₁ = (1 + m₂/m₁ )/3  tan θ - m3/3m₁   scs θ

2)  add 2 and 3 equation

    W₂ sin θ + W₁ sin θ - T₃ - 3f = 0

    T₃ = g (m₁ + m₂) sin θ + 3f

    W₃ = g (m₁ + m₂) sin θ + 3f

3) when cutting the rope T1 = 0

   W₁ sin θ - f = m₁ a₁

   a₁ = g sin θ - f / m₁