Question:
A long uniform beam is pivoted at one end. A force of 300N is applied to hold the beam horizontally.
What is the weight of the beam?
Attempt:
Need Help... Don't know how to start
I can't really help you exactly.. though doesn't it have something to see with the strenght of gravity? You need an amount of strenght equal to the gravity to hold the beam in air.
This is just a gamble... so you best wait 'till someone proves me wrong (which 'll probably be the case but anyway...)
The strenght of a gravity field:
As they apply 300N to hold the beam in the air...
I could be terribly wrong though...
Edit: and 'topsquark' has proven me wrong so ignore me ^^
I don't have time to fully analyze this, but the CM of the beam will be 1.25 m from the pivot, so that's where the weight acts from. If we take the sum of the torques using the pivot as our axis of rotation we will get an equation for w. (The net torque is, of course, 0 Nm.)
-Dan
Since the beam is uniform the center of gravity C is 1.25 m from the pivot.
The weight W of the beam creates a right-turning momentum
while the force F = 300 N creates a left-turning momentum of
Both momenti must be equal that means the beam doesn't turn in any direction:
I want to quickly mention why this doesn't work. There is a reaction force at the pivot point and we don't know what size that force is. The reason this isn't a difficulty using the torque method is that by taking the axis of revolution as the pivot the torque due to this reaction force is 0 Nm because it has a 0 moment-arm.
-Dan
Torque is the rotational version of force. . A "pivot" is a point around which a beam (or other object) would rotate about since it is attached to that point. (This differs slightly from "axis of rotation" because an axis of rotation could be anywhere on the object. A pivot is a point that is attached to the object and forces the object to rotate around it.)
-Dan