A uniform 300nt. scaffold is 8 meters long. it hangs from two ropes, one at ech end. A 700nt painter stands two meters from one end. What are the tensions in the two ropes?
Let the tension in the two ropes are T1(near) T2(far) Newtons, =>T1+T2 = 300 + 700 = 1000 -----(i) The unit length weight of scaffold = 300/8 = 37.5 N By taking the torque at painter, =>T1 x 2 + [37.5 x 6] x 3 = T2 x 6 + [37.5 x 2] x 1 =>2T1 + 675 = 6T2 + 75 =>3T2 - T1 = 300 ------(ii) by (i) + (ii):- =>4T2 = 1300 =>T2 = 1300/4 = 325 N =>T1 = 675 N
Let's mark left end as point A and right end as point B Let tensions in left and right rope are SA and SB respectively. Let G is weight of painter and Q is weight of scaffold. Let center of weight of painter is a=2m from A. Length of scaffold is L=8m Center of weight of scaffold is at its middle point b=L/2=8/2=4m from point A Two conditions of equilibrium: 1° there is no rotation ---> the sum of moments (torques) in any point of scaffold is zero. For point A: ΣMA = 0 (let's consider clockwise direction as positive) G*a + Q*b - SB*L = 0 ....(1) 2° there is no translation ---> the sum of all vertical forces including forces of tension in two ropes is zero ΣFy = 0 (let's consider downward direction as positive) G + Q - SA - SB = 0 .....(2) From (1) SB = (G*a + Q*b) / L SB = (700*2 + 300*4) / 8 = 325 N and from (2) SA = G + Q - SB SA = 700 + 300 - 325 = 675 N
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