Problem 540 | Friction on Wedges

From the FBD of 20-kN block
$\Sigma F_H = 0$

$R_1 \sin 15^\circ = R_2 \cos 30^\circ$

$R_1 = 3.346R_2$
 

$\Sigma F_V = 0$

$R_1 \cos 15^\circ = R_2 \sin 30^\circ + 20$

$(3.346R_2) \cos 15^\circ = R_2 \sin 30^\circ + 20$

$2.732R_2 = 20$

$R_2 = 7.32 \, \text{ kN}$
 

 

From the FBD of the upper block
$\Sigma F_V = 0$

$P = 2R_2 \sin 30^\circ$

$P = 2(7.32) \sin 30^\circ$

$P = 7.32 \, \text{ kN}$           answer
 

When one block weigh 50 kN and the other is 20 kN, the first to impend when movement starts is the 20 kN block. Thus the reaction R₂ = 7.32 kN, similar to the above value of R₂. Thus, the answer which is P = 7.32 kN will not change. See the free body diagram below and note that the friction reaction f₃ is not equal to the maximum available friction under the 50 kN block.