Question

The 12-lb block B starts from rest and slides on the 30-lb wedge A, which is supported by a horizontal surface. Neglecting friction, determine (a) the acceleration of the wedge, and (b) the acceleration of the block relative to the wedge.

Answer 1

Answer:

the acceleration of the wedge $\mathbf{a_A = 5.0702 \ ft/s^2}$

the acceleration of the block relative to the wedge is $\mathbf{a_{B/A} \approx 20.50 \ ft/s^2}$

Explanation:

Let assume that the angle at which the block starts to slide is 30°

Given that ;

the weight of block B = $W_B$ =  12-lb

the weight of wedge A = $W_A$ = 30-lb

The 12-lb block B starts from rest and slides on the 30-lb wedge A, which is supported by a horizontal surface.

We can resolve the equation of motion  for the wedge and the block into the vertical component and horizontal components as follows:

$\sum f_x = m_Aa_A \\ \\ N_1 sin 30^0 = m_Aa_A \\ \\ 0.5 N_1 = ( \dfrac{W_A}{g})a_A --- (1)$

$\sum F_x = m_Ba_x \\ \\ \sum F_x = m_B(a_x cos \ 30^0 - a_{B/A}) \\ \\ -W_b sin 30^0 = ( \dfrac{W_B}{g})(a_A cos \ 30^0 - a_{B/A}) \\ \\ a_{B/A} = a_A \ cos \ 30^0 + g \ sin 30^0 ----- (2)$

$\sum Fy = m_B ay \\ \\ \sum Fy = m_B (-a_A \ si 30^0) \\ \\ N_1 - W_B cos 30^0 = - (\dfrac{W_B}{g}) a_A \ sin30^0 ----- (3)$

From equation (1)

$0.5 N_1 = ( \dfrac{W_A}{g})a_A \\ \\ \dfrac{1}{2} N_1 = ( \dfrac{W_A}{g})a_A \\ \\ N_1 = 2 ( \dfrac{W_A}{g})a_A$

From equation (3) ; we have:

$N_1 - W_B cos 30^0 = - (\dfrac{W_B}{g}) a_A \ sin30^0$

replacing  $N_1 = 2 ( \dfrac{W_A}{g})a_A$  in  above equation (3); we have :

$2 ( \dfrac{W_A}{g})a_A - W_B cos \ 30^0 = - (\dfrac{W_B}{g}) a_A \ sin \ 30^0$

Making $a_A$ the subject of the formula; we have :

$a_A = \dfrac{gW_B \ cos \ 30^0}{2W_A + W_B \ sin \ 30^0 }$

where ; g = $32.2 \ ft/s^2$

the weight of block B = $W_B$ =  12-lb

the weight of wedge A = $W_A$ = 30-lb

Solving for the acceleration of wedge A  $a_A$; we have;

$a_A = \dfrac{(32.2 \ ft/s^2 )(12 \ lb) \ cos \ 30^0}{2( 30 \ lb )+ 12 \ lb \ sin \ 30^0 }$

$a_A = \dfrac{(334.63 \ ft/s^2 )}{66 \ }$

$\mathbf{a_A = 5.0702 \ ft./s^2}$

Thus; the acceleration of the wedge $\mathbf{a_A = 5.0702 \ ft/s^2}$

To determine  the acceleration of the block relative to the wedge $a_{B/A}$; Let consider equation 2

From equation 2;

$a_{B/A} = a_A \ cos \ 30^0 + g \ sin 30^0$

we know that:

$\mathbf{a_A = 5.0702 \ ft/s^2}$

g = 32.2 ft/s²

Thus;

$a_{B/A} = 5.0702 * \ cos \ 30^0 + 32.2 ft/s^2 * \ sin 30^0$

$a_{B/A} = 5.0702 ft/s^2 * \ 0.8660 + 32.2 ft/s^2 * 0.5$

$a_{B/A} =4.3907932 \ ft/s^2 + 16.1 \ ft/s^2$

$a_{B/A} = 20.4907932 \ ft/s^2$

$\mathbf{a_{B/A} \approx 20.50 \ ft/s^2}$

Thus; the acceleration of the block relative to the wedge is $\mathbf{a_{B/A} \approx 20.50 \ ft/s^2}$