Answer to Question #177936 in Mechanics | Relativity for tasfia tarannum

Question

2. Box 1, a wooden box, has a mass of 8.60 kg  

and a coefficient of kinetic friction with the  

inclined plane of 0.35. Box 2, a cardboard box,  

sits on top of box 1. It has a mass of 1.30 kg.  

The coefficient of kinetic friction between the  

two boxes is 0.45. The two boxes are linked by  

a rope which passes over a pulley at the top of  

the incline, as shown in the diagram. The inclined plane is at an angle of 38.0° with respect to the  horizontal.  

(a) What is the acceleration of each box?

(b) Now consider all surfaces are frictionless. Then calculate the amount of force with direction  to prevent the sliding of the boxes.

Answer 1

The diagram for this situation looks like this,

$m_1 = 8.90kg$

$m_2 = 1.30kg$

$\mu_A = 0.35$

$\mu_B = 0.45$

The next step is to draw the free body diagram

For box 2 - Start by assuming the forces in y direction where there is no acceleration,

$\sum F_y = F_{NB}-m_2gcos38^\circ = m_2a_y = 0$

This can be solved to give the Normal force $F_{NB}$

$F_{NB} = m_2gcos38^\circ = (1.30)(9.80)cos38^\circ = 10.04N$

Now we can find out the net force in x- direction where there is an acceleration up the slope

$\sum F_x=T-f_B-m_2gsin38^\circ = m_2a_x$

$T= m_2a_x+f_B+m_2gsin38^\circ$

$= m_2a_x+\mu_BF_{NB}+m_2gsin38^\circ$

$= m_2a_x+(0.45)(10.04)+(1.30)(9.80)sin38^\circ$

$= m_2a_x+12.36N$

Now move on to bo x 1. Going back to the free-body diagram and summing forces in the y-direction gives:

$\sum F_y=F_{NA}-m_1gcos38^\circ-F_{NB} = m_1a_y = 0$

This equation can be solved to give the normal force associated with the interaction between the inclined plane and box 1.

$F_{NA} = m_1gcos38^\circ+f_{NB} = (8.60)(9.80)cos38^\circ +10.04 = 76.45N$

Things are a little more complicated in the x- direction, but adding up the forces gives:

$\sum F_x = m_1gsin38^\circ-m_2a_x-12.36\mu_BF_{NB}-\mu_AF_{NA}$

Moving the acceleration terms to the left side gives:

$m_1a_x + m_2a_x = m_1gsin38^\circ - m_2a_x - 12.36- \mu_B F_{NB} - \mu_A F_{NA}$

Moving the acceleration terms to the left side gives

$m_1a_x + m_2a_x m_1gsin38^\circ - 12.36 - \mu_BF_{NB} - \mu_AF_{NA}$

Solving for the acceleration gives

$a_x = \dfrac{(m_1gsin38^\circ - 12.36-\mu_BF_{NB}-\mu_AF_{NA})}{(m_1+m_2)}$

$a_x = \dfrac{8.252}{9.90}$

$a_x = 0.834 ms^{-2}$