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 frict

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.

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Gia Bảo 6 months 2021-07-26T22:18:21+00:00 1 Answers 18 views 0

Answers ( )

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    2021-07-26T22:20:18+00:00

    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

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