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Balancing Normal Forces (Statics).

1. Homework Statement
A horizontal uniform bar of mass 260 kg and length L = 3.0 m is placed on two supports, labeled A and B, located as shown in the diagram. A block of mass 60 kg is placed on the right end of the bar.

problems_MIT_rayyan_check_points_Pictures_BK85.png

Find the normal forces FA and FB exerted on the bar by the supports. Enter your answer in Newtons (N). Use g = 9.8 m/s2.

2. Homework Equations
##\Sigma F = m*a##
## \tau = F * D ##


3. The Attempt at a Solution

I had thought to use a system of equations to solve the problem, however I simply cannot find/get to work a second equation.

I believe I have one correct equation: ## F_A + F_B = (260 + 60) * 9.8 ##. This being from Newton's Second Law. I tried to work out another equation with the torques of A and B being equal to zero (based on an equilibrium assumption), using the position of the red square as the starting point I tried: ## -.5*F_A + 1.5*F_B = 0 ##. This however proved fruitless. At this point I am out of ideas.

As always, my thanks in advance for any assistance.
 

SammyS

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1. Homework Statement
A horizontal uniform bar of mass 260 kg and length L = 3.0 m is placed on two supports, labeled A and B, located as shown in the diagram. A block of mass 60 kg is placed on the right end of the bar.

problems_MIT_rayyan_check_points_Pictures_BK85.png

Find the normal forces FA and FB exerted on the bar by the supports. Enter your answer in Newtons (N). Use g = 9.8 m/s2.

2. Homework Equations
##\Sigma F = m*a##
## \tau = F * D ##


3. The Attempt at a Solution

I had thought to use a system of equations to solve the problem, however I simply cannot find/get to work a second equation.

I believe I have one correct equation: ## F_A + F_B = (260 + 60) * 9.8 ##. This being from Newton's Second Law. I tried to work out another equation with the torques of A and B being equal to zero (based on an equilibrium assumption), using the position of the red square as the starting point I tried: ## -.5*F_A + 1.5*F_B = 0 ##. This however proved fruitless. At this point I am out of ideas.

As always, my thanks in advance for any assistance.
You need to find the torque about some particular point. The torque about that point due to ALL of the forces must be zero. Two of those forces are FA and FB. But there are two more, due to gravity acting on the block and on the bar itself. (Equivalently, you can set the sum of clockwise torques to be equal to the sum of counter-clockwise torques.)
 
OK. So if we chose the particular point as B our torque equation would be ##1 * (\frac{260}{3} * 9.8) + 1*(60*9.8) = 1.5*F_B + .5 * F_A ##. Am I on the right track?
 

SteamKing

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OK. So if we chose the particular point as B our torque equation would be ##1 * (\frac{260}{3} * 9.8) + 1*(60*9.8) = 1.5*F_B + .5 * F_A ##. Am I on the right track?
Not quite.

Picking point B is OK. However, how far from B if the normal force F applied?

It looks like you just copied your moment equation from your first attempt at solution without realizing that the selection of a different reference point changes the moments of each normal force. When you pick a particular point as a reference for moment calculations, the moment arms of all forces must be measured from this same point.

Also, you should realize that moments have a certain orientation. A clockwise moment is different from a counter-clockwise moment.
 
Alright, trying to work more carefully now: ## 1∗(\frac{260}{3}∗9.8)+1∗(60∗9.8) = 0*F_B + 1.5*F_A##. The distance from B is zero, so if I use that as the reference point its torque is zero, it is still one meter to the point where the red block is, so the ones stay there. And point A is 1.5 meters from B, thus my logic for the above equation.

(Side note: The deadline has based and I have the answers (but I still want to know how), I get something that is within the acceptable 5% margin of error using ## 1∗(\frac{260}{3}∗9.8)+1∗(60∗9.8) = 0*F_B + 3*F_A##. However this a. doesn't make sense to me and b. I still feel I should be able to hit their answer on the nose seeing as rounding is practically nonexistent in systems of equations solved on CAS.)
 

SteamKing

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If you have divided the mass of the uniform bar by 3 in order to calculate the amount of the bar which is not balanced about point B, you should use the correct distance from its center of mass to point B as the moment arm, in other words, 2 m - 0.5 m = 1.5 m, instead of 1.0 m.

How far is point A from point B? According to your calculations, it is 2.0 m. Check the diagram carefully.
 
But my mass is already based on there being a distance of one meter from B to the block. I say since it is a uniform bar we know that in the one meter of bar there is 260/3 kilograms present.

Isn't it 1.5 meters? which is what I had in my initial equation, I don't recall ever putting in 2.0.
 

SammyS

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But my mass is already based on there being a distance of one meter from B to the block. I say since it is a uniform bar we know that in the one meter of bar there is 260/3 kilograms present.

Isn't it 1.5 meters? which is what I had in my initial equation, I don't recall ever putting in 2.0.
Are you using B as the reference point for torque, as you said in post #5? How far is the center of mass from B ?
 

SteamKing

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But my mass is already based on there being a distance of one meter from B to the block. I say since it is a uniform bar we know that in the one meter of bar there is 260/3 kilograms present.

Isn't it 1.5 meters? which is what I had in my initial equation, I don't recall ever putting in 2.0.
Sorry, 2.0 was a typo on my part. I meant to type 3.0.

If it's 1.5 m from Point A to point B, then why does your moment equation say "3 * FA"?
 

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