A pulley P is attached to the ceiling at O by a piece of metal that can swing (b) freely. One end of a rope is attached to the ceiling at A. The rope is passed through the pulley P and a weight is attached to the other end of the rope at M, as shown in the diagram. A The distance OA is 1 m, the length of the rope is 2 m, and the length of the piece of metal OP=r metres, where 0

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
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Chapter2: Equations And Inequalities
Section2.7: More On Inequalities
Problem 44E
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THIS QUESTION IS A MATHS QUESTION, DO NOT MISINTERPRET IT AS A PHYSICS QUESTION: THERE WAS NO MISTAKE WITH MY CATEGORISATION INTO "ADVANCED MATH"

A pulley P is attached to the ceiling at O by a piece of metal that can swing
(b)
freely. One end of a rope is attached to the ceiling at A. The rope is passed
through the pulley P and a weight is attached to the other end of the rope at M,
as shown in the diagram.
A
The distance OA is 1 m, the length of the rope is 2 m, and the length of the piece
of metal OP=r metres, where 0<r<1. Let X be the point where the line MP
produced meets OA. Let OX=x metres and XM= l metres.
(i)
By considering triangles OXP and AXP, show that
l = 2+ vr2 – x² – v1–2x+r² .
(1? -x²)- x²(1–2x + r²)
vt -x?VI-2x+r°{NP=x²+ xvI-2r+r*)
dl
(ii) Show that
dx
(iii) You are given the factorisation
(* -x*)-x'(1-2x +r*)=(x – 1)(2x² -²x -r*).
(Do NOT prove this.)
Find the value of x for which M is closest to the floor. Justify your
answer.
Transcribed Image Text:A pulley P is attached to the ceiling at O by a piece of metal that can swing (b) freely. One end of a rope is attached to the ceiling at A. The rope is passed through the pulley P and a weight is attached to the other end of the rope at M, as shown in the diagram. A The distance OA is 1 m, the length of the rope is 2 m, and the length of the piece of metal OP=r metres, where 0<r<1. Let X be the point where the line MP produced meets OA. Let OX=x metres and XM= l metres. (i) By considering triangles OXP and AXP, show that l = 2+ vr2 – x² – v1–2x+r² . (1? -x²)- x²(1–2x + r²) vt -x?VI-2x+r°{NP=x²+ xvI-2r+r*) dl (ii) Show that dx (iii) You are given the factorisation (* -x*)-x'(1-2x +r*)=(x – 1)(2x² -²x -r*). (Do NOT prove this.) Find the value of x for which M is closest to the floor. Justify your answer.
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