Question

1. In the figure below there are 6 boxes of different masses. The coefficient of kinetic friction between each box and the surface is 0.12.

box 1: 3.0kg

box 2: 6.0kg

box 3: 1.0kg

box 4: 2.0kg

box 5: 3.0kg

box 6: 4.0kg

FA=96N[R]

a) Find the acceleration of the boxes caused by the applied force

b) Find the tension in the string joining the 6.0 kg and 1.0 kg boxes.

Answer 1

a) The acceleration can be found by Newton's second law:

$$F_\text{net}=F-f_\Sigma=F-\mu g·\Sigma m,\\\space\\ a=\frac{F_\text{net}}{\Sigma m}=\frac{F-\mu g·\Sigma m}{\Sigma m},\\\space\\ a=\frac{96-0.12·9.8(3+6+1+2+3+4)}{3+6+1+2+3+4}=3.88\text{ m/s}^2.$$

b) Forces acting on box 3: tension from box 2, friction of all 4 other boxes:

$$m_3a=T_2-\mu g(m_3+m_4+m_5+m_6),\\ T_2=m_3a+\mu g(m_3+m_4+m_5+m_6)=15.64\text{ N}.$$