Compute minimum value of the horizontal force:

Block A of weight 520N rest on horizontal top of block B having weight of 700N as shown in the figure given below. Block A is tied to support C by cable at 300 horizontally. The coefficient of friction is 0.4 for all the contact surfaces. Compute minimum value of the horizontal force P only to move the block B. How much is the tension in cable then. 

Sol.: Consider First free body diagram of block A

∑H = 0

µR₁ = Tcos30°

0.4R₁ = 0.866T

R₁ = 2.165T                                                                                                                           ...(i)

∑V = 0

W = R₁ + Tsin30°

520 = 2.165T + 0.5T

520 = 2.665T

T = 195.12N                                                                                                                       ...(ii)

Putting in equation (i) we get

R₁ = 422.43N                                                                                                                      ...(iii)

Consider First free body diagram of block A

∑V = 0

R₂ = R₁  + WB

R₂ = 422.43 + 700

R₂ =1122.43N                                                                                                                     ...(iv)

∑H = 0

P = µR₁ + µR₂

P = 0.4(422.43 + 1122.43)

P = 617.9N                                                  .......ANS