Q12: Three forces A=19 kN, B=18 kN and C=33 kN. What is the angle of the equilibrant Force?

A = 19 kN @ 110o

B = 18 kN @ 73o

C = 33 kN @ 343o

Get X componentsAx = 19*cos(110) = -6.4984 kNBx = 18*cos(73) = 5.26269 kNCx = 33*cos(343) = 31.5581 kNGet Y componentsAy = 19*sin(110) = 17.8542 kNBy = 18*sin(73) = 17.2135 kNCy = 33*sin(343) = -9.6483 kNAdd components

Fx = -6.4984+5.26269+31.5581 = 30.32239 kN

Fy = 17.8542+17.2135+-9.6483 = 25.4194 kNMagnitude of Resultant (Pythagoras)R = (30.32239^2 + 25.4194^2)^0.5 = 39.56758 kNAngle of ResultantAngle = atan(25.4194/30.32239) = 39.97326 o

Angle of Equilibrant!39.97326 + 180 = 219.97326 o

Using AutoCad…

Q12 Equilibrant ForceMonday, 24 February 20145:58 PM

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Q17: Angle A=74°, and angle B=46°. The man is 84 kg. (a) Find the force in Rope A.

A = ? kN @ 74o

B = ? kN @ 226W = 84*9.81 = 824.04 N

Using AutoCad… Draw known force(s) W=824.04@270o

Draw the unknown lines at their angles (74 and 226)Trim these lines together

Q17 AbseilingMonday, 24 February 20146:25 PM

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Example: A Lifting Eye

Two ropes are attached to this lifting eye. Force A is 1200N at 75o, and Force B is 1600N at 60o from horizontal.What is the force in the bolt?

Adding forces and replacing witha single force = RESULTANTAdding forces and making it go back to zero = EQULIBRANT

Resultant is always 180o apart from Equilibrant

Lifting EyeThursday, 23 February 201210:22 AM

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W = 9.81*5000 /1000 = 49.05 kN

Draw the unknown forcesat the angles of 135 and 30 degrees.This give intersection point.Now get lengths of the lines.

45

60

180-45-60 = 75

49.05

F1?

F2?

Sine rule; opposite / sin(angle) = constant

49.05 / sin(75) = 50.783F1 / sin(60) = 50.783 so F1 = 50.783*sin(60) = 43.979 kNF2 / sin(45) = 50.783 so F2 = 50.783*sin(45) = 35.909 kN

Two cranesThursday, 24 February 20119:49 AM

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Fdy = 17*sin(45)=12.021 kNFby = 17*sin(45)=12.021 kNTotal Fy = 0Fdy + Fby + Fcy = 0Fcy = - Fdy - Fby = -12.021 - 12.021 = -24.042 kN

Ex 5.3

10m0.2m

Ang = A?

Wooden trussMonday, 7 March 20117:29 PM

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Convert to N: 15 * 9.81 = 147.15 NFind angle: sin(A) = 2 / 10So A = asin(2/10) = 11.537 degrees

From CAD, force in cable = 367.865N

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Now add forces (Force Polygon)

Q23: InsulatorThursday, 23 February 201211:01 AM

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Now add these forces (in a force polygon), starting from the first known force of the 39240N (39.24kN) weight force (vertical).Properties check on one of the sling forces gives 24.898kN.

Convert 4 tonnes to N;4000*9.81 = 39240 N

Q25: Generator Lifting SlingsThursday, 23 February 201211:01 AM

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(The other sling force also gives the same value, of course)

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Q26: Two slings (as shown) are lifting this 3.1 tonne generator. If the maximum allowable cable tension is 26.6 kN, what is minimum angle A?

F1 = 26.6 kN @ ?o

F2 = 26.6 kN @ ?W = 3100*9.81 = 30.411 kN

Angle is 214.864 degreesBut according to question…

Here we know the Newtons (Length) of the sling forces but not the angle, so we use intersection of arcs to find the angle. (Usually we know the angle and not the length, which we normally find by trimming lines)

Q26: Find sling angleMonday, 24 February 20146:36 PM

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But according to question…214.864-180 = 34.864 o

This is the minimum angle for this sling - (Note: This low angle has almost halved the load rating of the sling! This is why slings are typically rated up to 60 degrees only)

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First of all, divide the weight between two sets of slings, so the slings on each side take 12 tonnes.

Now find the angle of the lower chain slings using CAD.

From above, angle is 29.06o

Now add forces (for one side of container) in a force polygon; (again using CAD)Start with the known force (the weight force) of 12 Tonnes.12 Tonnes = 12000*9.81 = 117720 N (117.72 kN)

Q28: Container lower slingsWednesday, May 24, 201711:13 AM

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This time the concurrent point is at the hook. Here, we have the full weight applied (27 tonnes)

We need to determine the angle of the upper chain slings (using CAD) - assume the spreader bar is 2400 wide.

From above, angle is 43.78o

Now add forces in a force polygon; (again using CAD)Start with the known force (the weight force) of 27 Tonnes.27 Tonnes = 27000*9.81 = 264870 (264.87 kN)

So the upper slings have a force of 191.41 kN each.

Q29: Container upper slingsWednesday, May 24, 201711:13 AM

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