Homework for Quiz 2 Picture Tools Philip McDonald- File Home Insert Draw Design

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Homework for Quiz 2 Picture Tools Philip McDonald- File Home Insert Draw Design Layout References Mailings Review View Help Grammarly Format Tell me what you want to do | Share (6) Newton's aws (Feb 5 1. A spider pulls down on a strand of web with a weight W -wj is suspended by a web witll two strands which neet at the point c = ( d/2.-h ), onie attached to the ceiling at the point d = di. Te tension in the strand that connects to the ceiling is W/2 and the tension in the strand that connects to the wall is 2 W 3. Find the vector position of the connection point of the web strand that connects to the d/2 wall Llimt The forces fron the teisiois ii the straids have to point along the directio of the web strands (and therefore have the sa vectors). Questions to think about: (a) What are the unit vectors pointing along the three strands? (b) Draw free body diagrams for the spider and for the entire web. For each: a. Identify lhe foroes. b. Identify the reaction forces (c) The normal force is always perpendicular to the surface it comes from. Which direction is the normal force on the strands: into the wall or out of it. How do the forces on the individual strands balance? Page 1 of b 1564 words + 100% 447 PM Type here to search 2/15/201 22

Explanation / Answer

1. given

weight of spider = -Wj

location of the point where the web threads meet c = (d/2, -h)

also, d = di

tension in the right strand = W/2

tension in the left strand = 2W/3

we have to find x

hence

from force balance in the x direction

W/2 * d/2*sqrt((d/2)^2 + h^2) = (2W/3) * d/2*sqrt((d/2)^2 + (h - x)^2)

9((d/2)^2 + (h - x)^2) = 16((d/2)^2 + h^2)

9d^2/4 + 9h^2 + 9x^2 - 18hx = 16*d^2/4 + 16h^2

9x^2 - 18hx = 7*d^2/4 + 7h^2

also, form force balance in y direction

W/2 * h/*sqrt((d/2)^2 + h^2) + (2W/3) * (h - x)/*sqrt((d/2)^2 + (h - x)^2) = W

3h/sqrt((d/2)^2 + h^2) + 4(h - x)/sqrt((d/2)^2 + (h - x)^2) = 6

hence

solving we het

x = (18h +- sqroot((18h)^2 + 252(d^2/4 + h^2)))/18

so vector location of point is

x = -xj

a. unit vectors along the three strands are

right strand -> (di/2 + hj)/sqrt((d/2)^2 + h^2)

left stran -> (-di/2 + (h - x)j)/sqrt((d/2)^2 + (h - x)^2)

bottom strand -> -j

b. forces are

tension in the right string

tension in the left string

tensionin the bottom string

c. normal reacviton in the wall where the strands are attached is into the wall as the the strings are pulling out of the wall

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