Answer:
the acceleration of the wedge ![\mathbf{a_A = 5.0702 \ ft/s^2}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba_A%20%3D%205.0702%20%20%5C%20ft%2Fs%5E2%7D)
the acceleration of the block relative to the wedge is ![\mathbf{a_{B/A} \approx 20.50 \ ft/s^2}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba_%7BB%2FA%7D%20%5Capprox%2020.50%20%5C%20ft%2Fs%5E2%7D)
Explanation:
Let assume that the angle at which the block starts to slide is 30°
Given that ;
the weight of block B =
= 12-lb
the weight of wedge 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)](https://tex.z-dn.net/?f=%5Csum%20f_x%20%3D%20m_Aa_A%20%5C%5C%20%5C%5C%20N_1%20sin%2030%5E0%20%3D%20m_Aa_A%20%5C%5C%20%5C%5C%20%200.5%20N_1%20%3D%20%28%20%5Cdfrac%7BW_A%7D%7Bg%7D%29a_A%20%20---%20%281%29)
![\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)](https://tex.z-dn.net/?f=%5Csum%20F_x%20%20%3D%20m_Ba_x%20%5C%5C%20%5C%5C%20%5Csum%20F_x%20%20%3D%20m_B%28a_x%20cos%20%5C%2030%5E0%20-%20a_%7BB%2FA%7D%29%20%5C%5C%20%5C%5C%20-W_b%20sin%2030%5E0%20%3D%20%28%20%5Cdfrac%7BW_B%7D%7Bg%7D%29%28a_A%20cos%20%5C%2030%5E0%20-%20a_%7BB%2FA%7D%29%20%5C%5C%20%5C%5C%20a_%7BB%2FA%7D%20%3D%20a_A%20%5C%20cos%20%5C%2030%5E0%20%2B%20g%20%5C%20sin%2030%5E0%20%20%20%20-----%20%282%29)
![\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)](https://tex.z-dn.net/?f=%5Csum%20Fy%20%3D%20m_B%20ay%20%20%5C%5C%20%5C%5C%20%20%5Csum%20Fy%20%3D%20m_B%20%28-a_A%20%5C%20si%2030%5E0%29%20%5C%5C%20%5C%5C%20N_1%20-%20W_B%20cos%2030%5E0%20%3D%20%20%20-%20%28%5Cdfrac%7BW_B%7D%7Bg%7D%29%20a_A%20%20%5C%20sin30%5E0%20%20%20%20-----%20%283%29)
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](https://tex.z-dn.net/?f=0.5%20N_1%20%3D%20%28%20%5Cdfrac%7BW_A%7D%7Bg%7D%29a_A%20%20%20%5C%5C%20%5C%5C%20%20%5Cdfrac%7B1%7D%7B2%7D%20N_1%20%3D%20%28%20%5Cdfrac%7BW_A%7D%7Bg%7D%29a_A%20%5C%5C%20%5C%5C%20%20N_1%20%3D%202%20%20%28%20%5Cdfrac%7BW_A%7D%7Bg%7D%29a_A)
From equation (3) ; we have:
![N_1 - W_B cos 30^0 = - (\dfrac{W_B}{g}) a_A \ sin30^0](https://tex.z-dn.net/?f=N_1%20-%20W_B%20cos%2030%5E0%20%3D%20%20%20-%20%28%5Cdfrac%7BW_B%7D%7Bg%7D%29%20a_A%20%20%5C%20sin30%5E0)
replacing
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](https://tex.z-dn.net/?f=2%20%20%28%20%5Cdfrac%7BW_A%7D%7Bg%7D%29a_A%20-%20W_B%20cos%20%5C%2030%5E0%20%3D%20-%20%28%5Cdfrac%7BW_B%7D%7Bg%7D%29%20a_A%20%5C%20sin%20%5C%2030%5E0)
Making
the subject of the formula; we have :
![a_A = \dfrac{gW_B \ cos \ 30^0}{2W_A + W_B \ sin \ 30^0 }](https://tex.z-dn.net/?f=a_A%20%3D%20%5Cdfrac%7BgW_B%20%5C%20cos%20%5C%2030%5E0%7D%7B2W_A%20%2B%20W_B%20%5C%20sin%20%5C%2030%5E0%20%7D)
where ; g = ![32.2 \ ft/s^2](https://tex.z-dn.net/?f=32.2%20%5C%20ft%2Fs%5E2)
the weight of block B =
= 12-lb
the weight of wedge A =
= 30-lb
Solving for the acceleration of wedge A
; we have;
![a_A = \dfrac{(32.2 \ ft/s^2 )(12 \ lb) \ cos \ 30^0}{2( 30 \ lb )+ 12 \ lb \ sin \ 30^0 }](https://tex.z-dn.net/?f=a_A%20%3D%20%5Cdfrac%7B%2832.2%20%5C%20ft%2Fs%5E2%20%29%2812%20%20%5C%20lb%29%20%5C%20cos%20%5C%2030%5E0%7D%7B2%28%2030%20%5C%20lb%20%29%2B%2012%20%5C%20lb%20%20%5C%20sin%20%5C%2030%5E0%20%7D)
![a_A = \dfrac{(334.63 \ ft/s^2 )}{66 \ }](https://tex.z-dn.net/?f=a_A%20%3D%20%5Cdfrac%7B%28334.63%20%5C%20ft%2Fs%5E2%20%29%7D%7B66%20%5C%20%20%7D)
![\mathbf{a_A = 5.0702 \ ft./s^2}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba_A%20%3D%205.0702%20%20%5C%20ft.%2Fs%5E2%7D)
Thus; the acceleration of the wedge ![\mathbf{a_A = 5.0702 \ ft/s^2}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba_A%20%3D%205.0702%20%20%5C%20ft%2Fs%5E2%7D)
To determine the acceleration of the block relative to the wedge
; Let consider equation 2
From equation 2;
![a_{B/A} = a_A \ cos \ 30^0 + g \ sin 30^0](https://tex.z-dn.net/?f=a_%7BB%2FA%7D%20%3D%20a_A%20%5C%20cos%20%5C%2030%5E0%20%2B%20g%20%5C%20sin%2030%5E0)
we know that:
![\mathbf{a_A = 5.0702 \ ft/s^2}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba_A%20%3D%205.0702%20%20%5C%20ft%2Fs%5E2%7D)
g = 32.2 ft/s²
Thus;
![a_{B/A} = 5.0702 * \ cos \ 30^0 + 32.2 ft/s^2 * \ sin 30^0](https://tex.z-dn.net/?f=a_%7BB%2FA%7D%20%3D%205.0702%20%2A%20%20%5C%20cos%20%5C%2030%5E0%20%2B%2032.2%20ft%2Fs%5E2%20%2A%20%5C%20sin%2030%5E0)
![a_{B/A} = 5.0702 ft/s^2 * \ 0.8660 + 32.2 ft/s^2 * 0.5](https://tex.z-dn.net/?f=a_%7BB%2FA%7D%20%3D%205.0702%20ft%2Fs%5E2%20%2A%20%20%5C%200.8660%20%2B%2032.2%20ft%2Fs%5E2%20%2A%200.5)
![a_{B/A} =4.3907932 \ ft/s^2 + 16.1 \ ft/s^2](https://tex.z-dn.net/?f=a_%7BB%2FA%7D%20%3D4.3907932%20%5C%20%20ft%2Fs%5E2%20%2B%2016.1%20%20%5C%20ft%2Fs%5E2)
![a_{B/A} = 20.4907932 \ ft/s^2](https://tex.z-dn.net/?f=a_%7BB%2FA%7D%20%3D%2020.4907932%20%5C%20ft%2Fs%5E2)
![\mathbf{a_{B/A} \approx 20.50 \ ft/s^2}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba_%7BB%2FA%7D%20%5Capprox%2020.50%20%5C%20ft%2Fs%5E2%7D)
Thus; the acceleration of the block relative to the wedge is ![\mathbf{a_{B/A} \approx 20.50 \ ft/s^2}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba_%7BB%2FA%7D%20%5Capprox%2020.50%20%5C%20ft%2Fs%5E2%7D)