Posted by
**Diane** on
.

A uniform rod AB is 1.2 m long and weighs 16N. It is suspended by strings AC and BD. A block P weighing 96N is attached at E, 0.30N from A. The tension force in the string BD has a magnitude.

My work:

Taking torque around AC string

clockwise=counterclockwise

block P + Board force down= BD tension

(0.3m * 96N)+ (.6m * 16N)= BD (1.2m)

38.4= BD(1.2m)

BD= 32N

An 800N man stands halfway up a 5.0m long ladder of negligible weight. The base of the ladder is 3.0m from the wall. Assuming that the wall-ladder contact is frictionless, the wall pushes against the ladder with a force magnitude of:

my work:

cos (theta) = 3m/5m

theta= 53.13 degrees

horizontal components

Ff (friction foce on bottom) - FN (normal force on top) = 0

Ff=FN

vertical components:

Torque taken from top point of ladder

clockwise= counterclockwise

(800N X 2.5X cos(53.1))= FN(5)

FN=240N

Thank you