Posted by Diane on Sunday, December 10, 2006 at 1:32pm.
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.
Taking torque around AC string
block P + Board force down= BD tension
(0.3m * 96N)+ (.6m * 16N)= BD (1.2m)
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:
cos (theta) = 3m/5m
theta= 53.13 degrees
Ff (friction foce on bottom) - FN (normal force on top) = 0
Torque taken from top point of ladder
(800N X 2.5X cos(53.1))= FN(5)
Answer This Question
More Related Questions
- Physics questions ( I posted all my work) - A uniform rod AB is 1.2 m long and ...
- Physics - A 1.34‒kg ball is attached to a rigid vertical rod by means of ...
- phy - 1. A Uniform rod AB 3m long weighing 3 Kg is held horizontally by two ...
- Grade 12 - a 4.0kg block is attached to a vertical rod by two strings.when the ...
- Physics - A uniform rod 8m long weighing 5kg is supported horizontally by two ...
- Physics - a uniform rod of 3 kg and length 1 m is suspended from a fixed point ...
- Physics - The sketch shows a mobile in equilibrium. Each of the rods is 0.16 m ...
- physics - A doubly tethered model airplane of mass 1.0 kg is attached to a ...
- physics - The 4.00-kg block in the figure is attached to a vertical rod by means...
- Physics - A 0.79 kg ball is connected by means of two ideal strings to a ...