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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whitneytr12 whitneytr12 · Answer 1 · 2020-04-23T03:22:37+02:00

Answer:

[tex]T_{1}=105.38 N[/tex]

[tex]T_{2}=187.34 N[/tex]

Explanation:

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

[tex]-f_{1f}+T_{1}=m_{A}a[/tex]

[tex]-\mu_{k}N_{1}+T_{1}=m_{A}a[/tex]

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

[tex]-\mu_{k}m_{A}g+T_{1}=m_{A}a[/tex]

Now we can find T₁:

[tex]\mu_{k}m_{A}g+m_{A}a=T_{1}[/tex]

The acceleration is the same for both boxes.

[tex]T_{1}=m_{1}(\mu_{k}g+a)[/tex]

[tex]T_{1}=18*(0.240*9.81+3.5)[/tex]

[tex]T_{1}=105.38 N[/tex]

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

[tex]-f_{2f}-T_{1}+T_{2}=m_{B}a[/tex]

[tex]-\mu_{k}m_{B}g-T_{1}+T_{2}=m_{B}a[/tex]

Let's solve it for T₂.

[tex]T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g[/tex]

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

[tex]T_{2}=14(3.5+0.240*9.81)+105.38[/tex]

[tex]T_{2}=187.34 N[/tex]

I hope it helps you!