problem12_89

University Physics with Modern Physics with Mastering Physics (11th Edition)

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
12.89: From symmetry, the component of the gravitational force parallel to the rod is zero. To find the perpendicular component, divide the rod into segments of length dx and mass , 2 L M dx dm = positioned at a distance x from the center of the rod. The magnitude of the gravitational force from each segment is . 2 2 2 2 2 a x dx L GmM a x dM Gm dF + = + = The component of dF perpendicular to the rod is dF , 2 2 a x a + and so the net gravitational force is - - + = = L L L L a x dx L GmMa dF F . ) ( 2 2 3 2 2 The integral can be found in a table, or found by making the substitution
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . a x tan = Then, , sec ) ( , sec 2 2 2 2 2 a a x d a dx = + = and so + = = = = + , sin 1 cos 1 sec sec ) ( 2 2 2 2 2 3 3 2 2 3 2 2 a x a x a d a a d a a x dx and the definite integral is . 2 2 L a a GmM F + = When a >> L, the term in the square root approaches , and 2 2 a GmM F a as expected....
View Full Document

This document was uploaded on 02/05/2008.

Ask a homework question - tutors are online