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
1 answer
for checking my work. Your work is correct.
0 answers