Solving a System of Forces Problem: Analytical Approach

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To solve the system of forces acting on the steel bar, first identify a suitable pivot point to simplify torque calculations. Calculate the torques produced by each force, ensuring that their sum equals zero for the bar to be balanced and not experience angular acceleration. Additionally, resolve the forces in both x and y directions to confirm that their resultant is also zero, indicating that the center of mass is not accelerating. A body is in equilibrium when both the resultant forces and moments are zero. Understanding these principles will help in analyzing the forces and achieving equilibrium.
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On the steel bar of the figure acts a system of forces. Check the balance.
http://img140.imageshack.us/img140/3714/dibujo2ch.jpg
F1(2,-1.5) = (1000N,120º)
M1 = 200Nm
F2(2,1.5) = (1000N,240º)
M2 = 400Nm
F3(0,2.5) = (1000N,0º)
M3 = 1900Nm

¿How do I solve it? ¿Can anyone teach me to do it?
Thanks very much in advance :)
 
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I'm sorry, luv. Your diagram makes no sense to me at all. I can't tell if you're dealing with an S-shaped bar or just what.

Some generic advice: Balance is all about torques. Pick a pivot point on the bar - exactly where doesn't matter mathematically, but there are good places and bad places to put it. Generally, putting the pivot point where one force acts causes that force to cancel out of the torques and makes life easier.

Once you have the pivot point selected, calculate the torque applied from each force involved. Add the torques - if they sum to 0, then the bar is not experiencing torque, is therefore experiencing no angular acceleration, and is therefore balanced. If there's a torque, then no.

Does that help?
 
Diane_ said:
I'm sorry, luv. Your diagram makes no sense to me at all. I can't tell if you're dealing with an S-shaped bar or just what.

Some generic advice: Balance is all about torques. Pick a pivot point on the bar - exactly where doesn't matter mathematically, but there are good places and bad places to put it. Generally, putting the pivot point where one force acts causes that force to cancel out of the torques and makes life easier.

Once you have the pivot point selected, calculate the torque applied from each force involved. Add the torques - if they sum to 0, then the bar is not experiencing torque, is therefore experiencing no angular acceleration, and is therefore balanced. If there's a torque, then no.

Does that help?

Sorry for the diagram, but I used some words and letters in spanish, perhaps this can clear the problem:
The bar has a shape like this: |_
........_|
.......|
I used "M" to represent the torques, and I specified the directions of them.

Can you now help me? Thanks :smile:
 
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Resolve the forces in x and y direction to see that the resultant force in x and y directions are zero. If the resultant force is zero then the center of mass is not accelerating.
Find the resultant momentum about any point. If it is zero then angular acceleration is zero.

A body is said to be in equilibrium if resultant of all the forces acting on it is zero and resultant moment is zero.

(In your fig. all the moments are anticlockwise ! )
 
mukundpa said:
Resolve the forces in x and y direction to see that the resultant force in x and y directions are zero. If the resultant force is zero then the center of mass is not accelerating.
Find the resultant momentum about any point. If it is zero then angular acceleration is zero.

A body is said to be in equilibrium if resultant of all the forces acting on it is zero and resultant moment is zero.

(In your fig. all the moments are anticlockwise ! )

That's the point! I don't know how to solve it, that's why I'm asking for anyone to teach me and show analitically how to do it... Thanks...
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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