|Dec18-12, 09:07 AM||#1|
Force in the inclined bar with the weight on one side.
Hello, I have a confusing question, please help.
Suppose I have a structure like on the figure:
I want to find force acting along the bar, for example to find critical stresses. My brain thinks in two ways:
1. I can distribute P in the application point as Fb=P*sin(90-al);
2. I can find it from the sum of all of the forces on the y-axis, so it is Fb=P/cos(al).
I feel that the second one is more right than the other one, but the second one seems to be more real. Suppose alpha is zero, in this case Fb is equal to P, so when alpha is increasing it should decrease and the force P will to cause the moment.
If I take a pack of milk in my hand and increase the alpha, I feel that while alpha is increasing it becomes more difficult to hold it, but I think that it's because of the moment produced by the force Fm=P*cos(90-al), but not because of force.
Where is the trouble?
|Dec18-12, 09:49 AM||#2|
It is a little unclear what you are trying to do.
To understand what happens start by drawing a proper free body diagram of your bar, showing all forces.
As you have drawn it, the bar is not in equilibrium so you cannot write any equations.
|force in the bar, inclined bar|
|Similar Threads for: Force in the inclined bar with the weight on one side.|
|Pulley Problem: Tension on one side, Weight on another||Introductory Physics Homework||13|
|Why is the tension of a rope not doubled with one weight on each side?||Classical Physics||9|
|Pulley system with 2 inclined planes on either side, a block, and a cylinder||Introductory Physics Homework||0|
|Side-by-Side Blocks with Force pushing it||Introductory Physics Homework||4|
|Inclined plane + Inclined force. Normal reaction force = ?||Introductory Physics Homework||3|