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A log is lifted by a crane from a barge onto the quayside using a harness. Only the log of weight 6000N and the straps A and B, the log is 4 metres again. Strap A is 0.5m from end and strap B is 1m from the other end.

a) write down the value of the the tension in N in the crane attached to the harness

b) by taking moments about the bottom of strap B, calculate the upward force in N exerted by strap A

c) calculate the upward force N expected by strap B
Original post by Custardcream000
A log is lifted by a crane from a barge onto the quayside using a harness. Only the log of weight 6000N and the straps A and B, the log is 4 metres again. Strap A is 0.5m from end and strap B is 1m from the other end.

a) write down the value of the the tension in N in the crane attached to the harness

b) by taking moments about the bottom of strap B, calculate the upward force in N exerted by strap A

c) calculate the upward force N expected by strap B


Hello there,

I'm sorry that your post has not received any replies yet. It some seem that the information in the questions is somewhat unclear; it is possibly mistyped. I am assuming a set-up like the attached images.

a) If the log is held in equilibrium, then the total tension T T must equal the weight of 6000N 6000 N .

b) Considering moments about B,

[br]Mclockwise=FA(2.5)[br][br]Mclockwise=6000(1)=6000[br][br]M_{clockwise} = F_A (2.5)[br][br]M_{clockwise} = 6000 (1) = 6000[br]

If the log is held in equilibrium . . .

[br]Mclockwise=Manticlockwise[br][br]FA(2.5)=6000[br][br]FA=2400N[br][br]M_{clockwise} = M_{anticlockwise}[br][br]F_A (2.5) = 6000[br][br]F_A = 2400 N[br]

. . . A force of 2400N 2400 N is applied at point A.

c) As T T accounts for both FA F_A and FB F_B . . .

[br][br]FA+FB=T[br][br](2400)+FB=(6000)[br][br]FB=3600N[br][br][br]F_A + F_B = T[br][br](2400) + F_B = (6000)[br][br]F_B = 3600 N[br]

. . . a force of 3600N 3600 N is exerted at B.

I hope that this has been helpful.
(edited 8 years ago)
d.PNG1.8KB
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