Question

QUESTION FOUR [Total: 20 Marks a. An exporter arrived with his goods at the Panama Canal; he was informed by the official manning the canal that the vessel carrying his goods would be towed along the canal. The rope that is used to tow the barge is inclined at 20° at the direction of the canal. The tension in the rope is 200N. i. Sketch the set-up narrated above. ii. Calculate the values of the rectangular components of this tension, one component being in the direction of the canal. AP [7marks] On your way to Kumasi, the driver of the VIP bus that you were in stop at the SOS village for a rest. You sat on a steel beam and remembered the lecture on turning moment. You started to examine this uniform steel beam and observed that the uniform steel beam AB is 6.0 m long and has a mass equivalent to 4.0 kN acting at its centre. The beam rests on two supports at C and B. A point load of 20.0 kN is attached to the beam. The manager of the village gave you the picture of the steel girder as shown in figure 1. Determine the value of the force F that causes the beam to just lift off the support B F 4kN 20KN 3 m 1 m B A 2.5 m RB Rc Fig 1

please I need help

Answer 1

200 sin20 N N 002 לד 200 200 cos20 N

Component of Tension force along the direction of canal = 200 cos20 N $\approx$ 188 N

Component of Tension force in the direction perpendicular to canal = 200 sin20 N $\approx$ 68 N

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Weight of the bar 4 kN and point load 20 kN together make clockwise moment if we consider support point C as Fulcrum.

Clockwise moment of weight of bar i.e. 4 kN and point load 20 kN = 4 $\times$ 0.5 + 20 $\times$ 2.5 = 52 kN-m

If Force F makes counter clockwise moment just above 52 kN-m about support point C , then bar will lift off support B

Magnitude of force F to make counter clockwise moment 52 kN about support point C is calculated as follows

F $\times$ 2.5 = 52 kN-m   or   F = 20.8 kN