SKEB 1313 STATICS AND DYNAMICS
Tutorial 5 (Rigid Bodies – Moment in 3D)
Q1
Based on Figure Q1, determine the moment at point 𝐵, given the tension in the cable EA is
𝑇𝐸𝐴 = 250 N, exerted at point E.
Figure Q1
Q2
Based on Figure Q2, calculate the moment at point 𝑂, given the tension in the cable CD is 𝑇 =
2 kN, exerted at point C.
Figure Q2
Q3
The 14-m boom 𝐴𝐵 is attached at 𝐴. A steel cable is stretched from the free end 𝐵 of the boom
to a point 𝐶 located on the vertical wall, as shown in Figure Q3.
i.
ii.
If the tension in the cable 𝐵𝐶 is 3750 N exerted at B, determine the moment about point
A.
Calculate the moment about the 𝑦-axis due to the tension of cable 𝐵𝐶
Figure Q3
(ii)
Q4
A door is held open by a force 𝐹𝐶𝐷 = 360 N of the chain 𝐶𝐷 as shown in Figure Q4. Determine
the moment of the force 𝐹𝐶𝐷 about the door hinges at point A.
Figure Q4
𝐹𝐶𝐷 = 360 𝑁
𝐶(−1 − 1.2 cos 30 , 1.2 sin 30 , 0)
𝐷(0,0, −1.5)
𝑑𝑥 = 2.0392 𝑚; 𝑑𝑦 = −0.6 𝑚; 𝑑𝑧 = −1.5 𝑚; 𝑑 = 2.6016 𝑚 [1 mark]
2.0392
0.6
1.5
𝐹𝐶𝐷 = 360 𝑁 [2.6016 𝑖 − 2.6016 𝑗 − 2.6016 𝑘] [1 mark]
Option 1: use 𝑟𝐴𝐷
𝑟𝐴𝐷
𝐴(−1,0, −2)
𝐷(0,0, −1.5)
𝑟𝑥 = 1 𝑚; 𝑟𝑦 = 0; 𝑟𝑧 = 0.5 𝑚 [1 mark]
𝑟𝐴𝐷 = 1 𝑖 + 0.5 𝑘 [1 mark]
𝑀𝐴 = 𝑟𝐴𝐷 × 𝐹𝐶𝐷
𝑀𝐴 = (1 𝑖 + 0.5𝑘) × (282.1771 𝑁𝑖 − 83.0258 𝑁𝑗 − 207.5646 𝑁𝑘) [1 mark]
i
j
k
i
j
𝑟𝐴𝐷
1
0
0.5
1
0
𝐹𝐶𝐷
282.1771
-83.0258
-207.5646
282.1771
-83.0258
[3 marks]
𝑀𝐴 = 0𝑖 + 141.0886𝑗 − 83.0258𝑘 − 0𝑘 − (−41.5264𝑖) − (−207.5646𝑗) [1 mark]
𝑀𝐴 = 41.5264 𝑁𝑚 𝑖 + 348.6532 𝑁𝑚 𝑗 − 83.0258 𝑁𝑚 𝑘 [1 mark]
Option 2: use 𝑟𝐴𝐶
𝑟𝐴𝐶
𝐴(−1,0, −2)
𝐶(−2.0392,0.6,0)
𝑟𝑥 = −1.0392 𝑚; 𝑟𝑦 = 0.6; 𝑟𝑧 = 2 𝑚 [1 mark]
𝑟𝐴𝐶 = −1.0392 𝑖 + 0.6 𝑗 + 2 𝑘 [1 mark]
𝑀𝐴 = 𝑟𝐴𝐶 × 𝐹𝐶𝐷
𝑀𝐴 = (−1.0392 𝑖 + 0.6 𝑗 + 2 𝑘) × (282.1771 𝑁𝑖 − 83.0258 𝑁𝑗 − 207.5646 𝑁𝑘) [1 mark]
i
j
k
i
j
𝑟𝐴𝐷
-1.0392
0.6
2
-1.0392
0.6
𝐹𝐶𝐷
282.1771
-83.0258
-207.5646
282.1771
-83.0258
[3 marks]
𝑀𝐴 = −124.5388 𝑖 + 564.3542 𝑗 + 86.2804 𝑘 − 169.3063 𝑘 + 166.0516 𝑖 − 215.7011 𝑗
[1 mark]
𝑀𝐴 = 41.5128 𝑖 + 348.6532 𝑗 − 83.0259 𝑘 [1 mark]
Q5
The friction at sleeve 𝐴 can provide a maximum resisting moment of 250 𝑁𝑚 about the 𝑧-axis.
Determine the largest magnitude of force 𝐅 that can be applied to the bracket so that the bracket
will not turn.
Figure Q5
Answer:
𝐹⃗𝐵 = (𝐹 cos 60 𝑖 + 𝐹 cos 45 𝑗 − 𝐹 cos 60 𝑘)𝑁 [1 mark]
𝑟𝐴𝐵 = 0.3𝑚 𝑖 + 0.1𝑚 𝑗 − 0.15𝑚 𝑘 [1 mark]
⃗⃗⃗⃗⃗⃗
𝑢𝑧 = 𝑘 [1 mark]
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
𝑀𝐴 = ⃗⃗⃗⃗⃗⃗
𝑟𝐴𝐵 × 𝐹⃗𝐵 [1 mark]
⃗⃗⃗⃗⃗⃗
𝑀
𝐴 = (0.3𝑚 𝑖 + 0.1𝑚 𝑗 − 0.15𝑚 𝑘) × (𝐹 cos 60 𝑖 + 𝐹 cos 45 𝑗 − 𝐹 cos 60 𝑘) [1 mark]
⃗⃗⃗⃗⃗⃗
𝑀
𝐴 = (−0.1𝐹 cos 60 + 0.15𝐹 cos 45)𝑖 + (−0.15𝐹 cos 60 + 0.3𝐹 cos 60)𝑗 +
(0.3𝐹 cos 45 − 0.1𝐹 cos 60) 𝑘 [1 mark]
𝑀𝑧 = ⃗⃗⃗⃗⃗
𝑢𝑧 ∙ ⃗⃗⃗⃗⃗⃗
𝑀𝐴 [1 mark]
𝑀𝑧 = 𝑘 ∙ [(−0.1𝐹 cos 60 + 0.15𝐹 cos 45)𝑖 + (−0.15𝐹 cos 60 + 0.3𝐹 cos 60)𝑗 +
(0.3𝐹 cos 45 − 0.1𝐹 cos 60) 𝑘] [1 mark]
250 = 0.3𝐹 cos 45 − 0.1𝐹 cos 60 [1 mark]
𝐹=
250
0.3 cos 45 − 0.1 cos 60
𝐹 = 1541.95 𝑁 [1 mark]