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

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}`

Step-by-step 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 = ( (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 = ( (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 = - ((W_B)/(g)) a_A \ sin30^0 ----- (3)`

From equation (1)

`0.5 N_1 = ( (W_A)/(g))a_A \\ \\ (1)/(2) N_1 = ( (W_A)/(g))a_A \\ \\ N_1 = 2 ( (W_A)/(g))a_A`

From equation (3) ; we have:

`N_1 - W_B cos 30^0 = - ((W_B)/(g)) a_A \ sin30^0`

replacing
`N_1 = 2 ( (W_A)/(g))a_A` in above equation (3); we have :

`2 ( (W_A)/(g))a_A - W_B cos \ 30^0 = - ((W_B)/(g)) a_A \ sin \ 30^0`

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

`a_A = (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 = ((32.2 \ ft/s^2 )(12 \ lb) \ cos \ 30^0)/(2( 30 \ lb )+ 12 \ lb \ sin \ 30^0 )`

`a_A = ((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}`