Block A of weight 520N rest on the horizontal top of block B having weight 700N as shown in
the figure-1 Block A is tied by string C at horizontally at an angle 30°. The coefficient of friction
b/w all surfaces are 0.4. Determine the minimum force of P just to move the block B. How much
the tension in the cable then?
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Answer:Consider First FBD of block A Figa∑H = 0µR1 = T cos 30°0.4R1 = 0.866TR1 = 2.165T ........i∑V = 0W = R1 + Tsin 30°520 = 2.165 T + 0.5 T520 = 2.665 TT = 195.12 N .......iiPutting in i we get R1 = 422.43 N ......iii Consider First FBD of block A Fig 9.30∑V = 0R2 = R1 + WBR2 = 422.43 + 700R2 =1122.43N∑H = 0 .........ivP = µR1 + µR2P = 0.4422.43 + 1122.43P = 617.9N
Explanation:
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