A 84.3-N backpack is hung from the middle of an aluminum wire, as the drawing sh

Question

A 84.3-N backpack is hung from the middle of an aluminum wire, as the drawing shows. The temperature of the wire then drops by 15.9 C

Explanation / Answer

Consider the tensions in the part left to the backpack and to its right separately. Due to symmetry of the arrangement the tensions are identical. From a free body sketch you could see that the vertical part of this two tensional forces is equals the weight of the backpack. i.e. 2·T·sin(f) = m·g => T = m·g / (2·sin(f)) f is the angle between the wire and the horizontal as indicated in the drawing. This angle changes with length of the wire. Let d be the (constant horizontal distance backpack to the support and l the length of wire from support tu backpack. Basic triangular geometry tells that cos(f) = d/l d = l·cos(f) we know the angle of thew "hot" wire as f0 = 3° Let l0 be it length at the hot temperature (its unknown but it cancels out later) d = l0·cos(f0) Due to temperature drop the wire contracts by ?l = l0·a·?T with a = 23×10?6K?¹ thermal expansion coefficient of aluminum So the length changes to l1 = l0 - ?l = l0·(1 - a·?T) Since distance d does not change, the angle decreases such that: d = l1·cos(f1) = l0·(1 - a·?T)·cos(f1) Equate two expression for d and solve for this angle: l0·cos(f0) = l0·(1 - a·?T)·cos(f1) cos(f0) = (1 - a·?T)·cos(f1) f1 = cos?¹( cos(f0) / (1 - a·?T) ) = cos?¹( cos(3°) / (1 - 23×10?6K?¹ · 20K) ) = 2.4457° Hence: T = 80N / (2·sin(2.4457°)) = 937.4N

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