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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
R2 = R1 + WB
R2 = 422.43 + 700
R2 =1122.43N ...(iv)
P = µR1 + µR2
P = 0.4(422.43 + 1122.43)
P = 617.9N .......ANS
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