ch12-p041 - 41. The diagrams below show the forces on the...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
41. The diagrams below show the forces on the two sides of the ladder, separated. F A and F E are the forces of the floor on the two feet, T is the tension force of the tie rod, W is the force of the man (equal to his weight), F h is the horizontal component of the force exerted by one side of the ladder on the other, and F v is the vertical component of that force. Note that the forces exerted by the floor are normal to the floor since the floor is frictionless. Also note that the force of the left side on the right and the force of the right side on the left are equal in magnitude and opposite in direction. Since the ladder is in equilibrium, the vertical components of the forces on the left side of the ladder must sum to zero: F v + F A W = 0. The horizontal components must sum to zero: T F h = 0. The torques must also sum to zero. We take the origin to be at the hinge and let L be the length of a ladder side. Then F A L cos θ W ( L /4) cos T ( L /2) sin = 0.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Page1 / 2

ch12-p041 - 41. The diagrams below show the forces on the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online