By symmetry

$C_V = B_V = 400 ~ \text{lb}$

$\Sigma M_A = 0$

$8B_H = 6B_V + 8(500)$

$8B_H = 6(400) + 8(500)$

$B_H = 800 ~ \text{lb}$

Moment at *B*

$M_B = 500(2) = 1000 ~ \text{lb}\cdot\text{ft}$

Axial Force at *B*

$P_a = 400 + B_V \cos \alpha + B_H \cos \theta$

$P_a = 400 + 400(4/5) + 800(3/5)$

$P_a = 1200 ~ \text{lb compression}$

Maximum Compressive Stress Will Occur at Point *B*

$\sigma_c = \dfrac{P_a}{bd} + \dfrac{6M_B}{bd^2}$

$\sigma_c = \dfrac{1200}{4(4)} + \dfrac{6(1000)(12)}{4(4^2)}$

$\sigma_c = 1200 ~ \text{psi}$ *answer*

**Axial, Shear, and Moment Diagrams**

Not actually necessary but just in case you need it.

$P_1 = 800 \cos \theta = 800(3/5) = 480 ~ \text{lb}$
$P_2 = 900 \sin \theta = 900(4/5) = 720 ~ \text{lb}$

$P_3 = 800 \cos \theta = 800(3/5) = 480 ~ \text{lb}$

$P_4 = 400 \sin \theta = 400(4/5) = 320 ~ \text{lb}$

$P_5 = 500 \sin \theta = 500(4/5) = 400 ~ \text{lb}$

$V_1 = 800 \sin \theta = 800(4/5) = 640 ~ \text{lb}$

$V_2 = 900 \cos \theta = 900(3/5) = 540 ~ \text{lb}$

$V_3 = 800 \sin \theta = 800(4/5) = 640 ~ \text{lb}$

$V_4 = 400 \cos \theta = 400(3/5) = 240 ~ \text{lb}$

$V_5 = 500 \cos \theta = 500(3/5) = 300 ~ \text{lb}$

## Comments

## a block weighting 1000N is

a block weighting 1000N is kept on a rough plane inclines at 40 digress to the horizontal the coefficient of friction between the block and the plane is 0.6.Determine the smallest force inclines at 15 digress to the plane required just to move the block up the plane.