Question:

A block of mass 2 kg is placed on a frictionless inclined plane with an angle of 30 degrees above the horizontal. The block is connected to a pulley by a string passing over it. On the other end of the pulley, a second block of mass 4 kg is hanging freely. The system is in equilibrium.

a) Draw a free-body diagram for each block, clearly showing all the forces acting on them.

b) Determine the coefficient of static friction between the block on the inclined plane and the surface.

c) Determine the tension in the string.

Assume that g = 9.8 m/s².

Answer:

a)

Free-body diagram for the 2 kg block on the inclined plane:

     Fg →  |
            |               ࿓ N →
            |               |
            | ࿓ fs →        |
            | _______       |
            |/\/\/\/\|      | ࿓ fN →
            |__mg___|-------|
                       ࿓ fK →

Fg : Weight of the block (mg)

N : Normal force exerted on the block by the inclined plane

fs : Static friction force exerted on the block by the inclined plane (opposing its motion)

fN : Component of the normal force acting perpendicular to the inclined plane

fK : Component of the normal force acting parallel to the inclined plane (opposing its motion)

Free-body diagram for the 4 kg hanging block:

  Fg →  |
         |
         |
         |
         |
         |
         ࿓ T →

Fg : Weight of the block (mg)

T : Tension in the string

b) To determine the coefficient of static friction between the block on the inclined plane and the surface, we need to analyze the forces acting on the block in the horizontal direction.

The net force acting on the block in the horizontal direction is given by:

Net force = fs - fK

Since the system is in equilibrium, the net force in the horizontal direction is zero. This implies fs = fK.

We can express the static friction force as:

fs = μs * N

Where μs is the coefficient of static friction and N is the normal force.

The component of the normal force acting perpendicular to the inclined plane can be calculated as:

fN = N * sinθ

Where θ is the angle of the inclined plane.

Setting fs = fK, we have:

μs * N = N * sinθ

μs = sinθ

Plugging in the given angle of 30 degrees, we find:

μs = sin(30°) = 0.5

Thus, the coefficient of static friction between the block on the inclined plane and the surface is 0.5.

c) To determine the tension in the string, we need to analyze the forces acting on the hanging block.

The net force acting on the hanging block in the vertical direction is given by:

Net force = T - Fg

Since the system is in equilibrium, the net force in the vertical direction is zero. This implies T = Fg.

The weight of the hanging block can be calculated as:

Fg = m * g

Where m is the mass of the hanging block and g is the acceleration due to gravity.

Plugging in the given mass of 4 kg and g = 9.8 m/s², we find:

Fg = 4 kg * 9.8 m/s² = 39.2 N

Therefore, the tension in the string is 39.2 N.