Free-body Diagram (FBD)

Mass-acceleration Dynamics (MAD)

Begin with a free-body diagram (FBD), which shows the forces acting on the crate, and a mass-acceleration diagram (MAD), which shows the inertia vector of the crate.

Since there is no force in the z direction, the motion is confined to the x-y plane.

Using Newton's Second Law, sum the forces in both directions. Note that there is no acceleration in the y direction.

     ΣF_y = ma_y
     N - mg = 0
     N = mg

In the x direction, the sum of the force are

     ΣF_x = -f = ma_x

The friction force depends on the static coefficient of friction (no motion) to start moving,

     f = μ_s N = μ_s mg

The equations can be combined to give an expression for a_x in terms of μ_s,

     -μ_s mg = ma_x

Horizontal Box Stops

Horizontal Box Slides

Divide both sides by m,

     a_x = dv/dt = -μ_s g

Then Integrate a_x to find an expression for v,

     
     v(t) = dx/dt = -μ_sgt + v_o

An expression for x(t) can be found by integrating v(t),

     
     x(t) = -1/2 μ_sgt² + v_ot

Knowing that v_o = 10 ft/s, the expression for v(t) can be used to find the time required for the belt to come to a stop,

     v(t) = 0 = -μ_sgt + 10
     0.5(32.2) t = 10
     t = 0.6211 s

Substituting the time into the x(t) function gives the minimum distance to stop,

     x = 10 (0.6211) -1/2 (0.5)(32.2)(0.6211)²
       = 3.106 ft

Mass-acceleration Dynamics (MAD)

As with all dynamic problems, it is best to start with a free-body and a mass-acceleration diagrams. These are shown at the left for this problem. First, rotate the x-y axis to be parallel and perpendicular to the conveyor belt. This will simplify calculations.

Using Newton's Second Law, sum the forces in both directions. Noting that there is no acceleration in the y direction, these equations become

     ΣF_y = ma_y
     N - mg cosθ = 0
     N = mg cosθ

In the x direction,

     ΣF_x = ma_x
      -f + mg sinθ = ma_x

The friction force in terms of μ_s and g is

     f = μ_s N = μ_s mg cosθ

Combining equations, and dividing through by m gives an expression for a_x in terms of μ_s.

     a_x = dv/dt = g sinθ - μ_s g cosθ>

Inclined Belt Box Stops

Inclined Belt Box Slides

Integrating a_x gives an expression for v,

     
      v(t) = gt (sinθ - μ_s cosθ) + v_o

Now the next step can be taken by integrating the velocity function to find the position function.

     v(t) = dx/dt

Integrating v(t) gives x(t) as

     
     x(t) = 1/2 gt² (sinθ - μ_s cosθ) + v_o t

Knowing that v_o = 10 ft/s, the time required for the belt to come to a stop is

     t (32.2) (sin15 - 0.5 cos15) = -10
     t = 1.386 s

Substitute the time into the expression for x(t) gives the minimum distance to stop,

   x = 1/2 (32.2)(1.386)² (sin15 - 0.5cos15) + 10(1.386)
      = 6.928 ft