Question

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.

Answer 1

Answer:

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- $W_{y}$ = 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