Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of static fric

Question

Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of static friction are μA = 0.16 and μB = 0.23. Determine the incline angle θ for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 2.1 lb/ft .

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Verity 4 years 2021-08-10T14:25:15+00:00 1 Answers 15 views 0

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  1. RuslanHeatt
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    2021-08-10T14:26:37+00:00
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    Answer:

    \theta=10.20^{\circ}  

    \Delta l=0.10 ft    

    Explanation:

    First of all, we analyze the system of blocks before starting to move.

    \Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0  

    \Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0

    11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0

    11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0

    11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0  

    16sin(\theta)-2.91cos(\theta)=0  

    tan(\theta)=0.18  

    \theta=arctan(0.18)  

    \theta=10.20^{\circ}  

    Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

    Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

    P_{A}sin(\theta)-F_{fA}-F_{spring}=0

    Where:

    F_{spring} = k\Delta l=2.1\Delta l

    P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0

    \Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}

    \Delta l=0.10 ft    

    Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

    I hope it helps you!

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