Two boxes, mA = 18 kg and mB = 14 kg, are attached by a string under tension T1 . The rightmost box is being pulled horizontally across the

Question

Two boxes, mA = 18 kg and mB = 14 kg, are attached by a string under tension T1 . The rightmost box is being pulled horizontally across the floor by a different string under tension T2 . The coefficient of kinetic friction between the boxes and the floor is µK = 0.240. If the boxes are accelerating at 3.5 m/s2 to the right, what is the tensions T1 and T2 ?

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Sigridomena 6 months 2021-08-02T06:24:35+00:00 1 Answers 2 views 0

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    2021-08-02T06:26:14+00:00

    Answer:

    T_{1}=105.38 N

    T_{2}=187.34 N

    Explanation:

    Applying the second Newton’s law for the first box we have.

    -f_{1f}+T_{1}=m_{A}a

    -\mu_{k}N_{1}+T_{1}=m_{A}a

    We know that the normal force is the product between the weight and the kinetic friction, so we have:

    -\mu_{k}m_{A}g+T_{1}=m_{A}a

    Now we can find T₁:

    \mu_{k}m_{A}g+m_{A}a=T_{1}

    The acceleration is the same for both boxes.

    T_{1}=m_{1}(\mu_{k}g+a)

    T_{1}=18*(0.240*9.81+3.5)

    T_{1}=105.38 N

    Now let’s analyze the forces of the second box.

    -f_{2f}-T_{1}+T_{2}=m_{B}a

    -\mu_{k}m_{B}g-T_{1}+T_{2}=m_{B}a

    Let’s solve it for T₂.

    T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g

    T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g

    T_{2}=m_{B}(a+\mu_{k}g)+T_{1}

    T_{2}=14(3.5+0.240*9.81)+105.38

    T_{2}=187.34 N

    I hope it helps you!

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