Two blocks A and B have a weight of 11 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of static friction are μA = 0.15 and μB = 0.26.

Determine the angle θ which will cause motion of one of the blocks.

What is the friction force under each of the blocks when this occurs? The spring has a stiffness of k = 2 lb/ft and is originally unstretched.

the block that starts moving first is block A ,    fr = 1.625 N ,  fr = 1.5 N

Explanation:

For this exercise we use Newton's second law, for which we take a reference system with the x axis parallel to the plane and the y axis perpendicular to the plane

X axis

       fr- Wₓ = 0

       fr = Wₓ

Axis y

      N- = 0

      N = W_{y}

Let's use trigonometry to find the components of the weight

     sin θ = Wₓ / W

     Cos θ = W_{y} / W

     Wₓ = W sin θ

     W_{y} = W cos θ

     Wₓ = 11 sin θ

     W_{y} = 11 cos θ

The equation for friction force is

      fr = μ N

We substitute

      μ (W cos θ) = W sin θ

      μ = tan θ

We can see that the system began to move the angle.

         θ = tan⁻¹ μ

So the angles are

Block A      θ = tan⁻¹ 0.15

           θ = 8.5º

Block B      θ = tan⁻¹ 0.26

             θ = 14.6º

So the block that starts moving first is block A

The friction force is

Block A

         fr = Wx = W sin θ

         fr = 11 sin 8.5

         fr = 1.625 N

Block B

         fr = 6 sin 14.6

         fr = 1.5 N

       F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

        V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

        V_{water} = 1.09 10⁻³ m³

the amount of water is

       ρ = m / V

       m = ρ V

       m = 1000 1.09 10⁻³

       m = 1.09 kg

so the weight of the liquid in the container for this time is

       W = mg

       W = 1.09 9.8

       W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

        P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2  ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

         P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

         v² = v₀² - 2 g (y-y₀)

         v =

let's calculate

         v = √(2 9.8 40.5)

         v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

         P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

         P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

         P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

         P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

         P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

         P₁ = 1.28 10⁵ Pa

The definition of Pressure is

         P = F / A

         F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

          m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

            F = P A

            F = 1.28 10⁵  7.46 10⁻⁵

            F = 9.5 N

with these data let's use the equilibrium equation

           F_ scale -W - F = 0

           F_scale = W + F

           F_scale = 10.682 + 9.5

           F_scale = 20.18 N