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 ?

Answer 1

`T_(1)=105.38 N`

`T_(2)=187.34 N`

Step-by-step 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!