Call the length of string below point B as h, then diagonal of triangle = a - h. angle to vertical of string = theta
T + Tcostheta = W
use pythagoras to find h (a - h)^2 = b^2 + h^2 so h = (a^2 - b^2)/2a
and costheta = h/(a-h) = (a^2 - b^2) /(2a(a - [(a^2 - b^2)/2a] ) which reduces to T (2a^2)/(a^2 + b^2) = W hence the first part.
Then for part 2, P = T sintheta = W(a^2+b^2)/(2a^2) x ( b/[a -(a^2-b^2)/2a] which believe it or not all reduces to W b/a. Just go slowly on the math, easy to make a mistake, but you are going for partial credit so errors are not too critical.
A light string of length a is attached to two points A and B on the same level and distance b apart, where b < a. A smooth ring of weight W is threaded on the string and is pulled by a horizontal force P, so that it rests in equilbrium vertically below B. Show that tension in the strings is W(a^2 + b^2)/2a^2 and find the force P
1 answer
1 answer