A 1110 N uniform boom is supported by a cable perpendicular to the boom, as seen

Question

A 1110 N uniform boom is supported by a cable perpendicular to the boom, as seen in the figure below.

The boom is hinged at the bottom, and an m = 2090 N weight hangs from its top. Assume the angles to be a = 60.6 deg and theta = 90.0 deg - a .Find the tension in the supporting cable.

Find the x-components of the reaction force exerted on the boom by the hinge (choose to the right as positive).

Find the y-components of the reaction force exerted on the boom by the hinge (choose upwards as positive).

Explanation / Answer

If I read your figure corrrectly, the cable is attached at a distance of ' 1 ' (it could be ' l ' however) from the pivot. Unfortunately, you did NOT give the length of the boom. I will call it ' L .' The figure is not very well drawn, since the combination of the angles 25 and 65 degrees (25 + 65 = 90) means that the cable is PERPENDICULAR to the boom. (It certainly DOESN'T look it!) Equilibrium means that the moments of the forces around the pivot must balance. [Taking moments about the PIVOT is ALWAYS advantageous as it ELIMINATES the unknown force operating at the pivot itself.] The weight of the uniform boom acts through its midpoint, L/2 along it, of course. Let T be the tension in the cable. Then: T*1 = (1200N*L/2 + 2883.40N*L)*cos 65 = 3483.40N*L*cos 65 = 1472.15...*LN. So T*1 (or T*l?) = 1472.15...*LN. ### I must say that this seems to be a fairly pointless problem as far as the particular inputs are concerned. While the PRINCIPLE of balancing the moments is fine, the actual inputs seem to have NO merit in terms of promoting a particularly pleasing or aesthetic looking result. Live long and prosper. ### Ah, DIVIDING by 9.81, this gives almost exactly 150! That means that this is a tension of 150 kg weight. Nothing so simple comes from dividing 2883.40 by either 9,81 or 9.80. I simply DO NOT see the point of providing that particular figure to such a ludicrous number of digits and therefore implied accuracy! Could you enlighten me about the setting of this problem? P.S. You just posted some CRUCIAL "additional details": "To clarify the picture, the lenghts are not given exact numbers. They are just L and .75L" Well, why the HELL didn't you say so, to start with?! Did you possibly imagine that the ratio of these lengths was irrelevant?! Boy, do you have a lot to learn! Furthermore, what the hell was the symbol ' 1' or ' l ' doing near the middle of the shorter pair of arrows?! Returning to my earlier form of the solution, it should be replaced by: T*0.75*L = 1472.15...*LN, so that T = (4/3)*1472.15...*LN, that is T = 1962.87... N, equivalent to a weight of ~ 200kg. I now suspect that the original setter of this equation originally made the tension the equivalent of that due to supporting a 200kg mass under gravity, and then worked out the very odd value for the weight hanging at the end of the boom.

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