A 10.00 gram superbug walks out along a 1.000 meter long uniform horizontal flag
ID: 2070857 • Letter: A
Question
A 10.00 gram superbug walks out along a 1.000 meter long uniform horizontal flagpole (mass = 100.0 grams) which is connected to the side of a building with a light weight cable and a low friction bearing. The cable makes an angle of 30.00o with the pole. The maximum tension that the cable can support is 1.1107 N.a. How far along the pole can the fearless bug walk before the cable snaps? [Ans. 66.7 cm]
b. Just before the cable snapped, what was the magnitude of the total force in the bearing?
[Ans. 1.095N]
Explanation / Answer
(a) Tension T in the cable can be resolved in T Cos 30 acting horizontally towards the building and T Sin 30 acting vertically upwards. Weight of the flagpole will act downwards from its centre. Let bug can go to x distance out along the flagpole. Taking moments of forces about the point where flagpole is attached to the building, (1.1107* Sin 30)*1 = (10/1000)*9.8*x + (100/1000)*9.8*(1/2) So, x = 0.667 m = 66.7 cm (b) Magnitude of horizontal force in the bearing = T Cos 30 = 1.1107*Cos 30 = 0.9619 N Magnitude of vertical force in bearing = (10/1000)*9.8 + (100/1000)*9.8 - (1.1107* Sin30) = 0.52265 N Net force = (0.9619^2 + 0.52265^2)^0.5 = 1.095 N
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