Tag: barbell

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Static Analysis of an Unbalanced Barbell

Alright, this is one that people have asked me to explain to them at the gym using actual numbers on more than one equation. It’s easy to understand that a barbell with weights only loaded on one side is able to balance, but people have asked some variant of “how stable is a bar that has weights on one side?” Well, let’s get into it!

Quick Physics Primer

In order to answer this one, we’ll take advantage of a branch of classical mechanics that people typically call statics; that is, analysis of forces on a body that result in no motion. Two concepts that you will have to understand are the notions of balance of forces, and balance of moments. A balance of forces is easy enough to imagine, a simple example is if you were to push horizontally on a wall, the wall pushes back on you with the exact same amount of force in the opposite direction, thereby causing the forces acting on the person-wall system to balance to 0. Balance of moments is exactly the same, but using moments, that is, forces that cause rotational motion.; most people will call this a torque, but physics uses the term moment. Think of two kids on a seesaw, one weighs 90 pounds and the other 80 pounds, each 4 feet from the center pivot point. The moment created by the 90 pound child we will call M1, and has magnitude 90 lbs x 4 feet = 360 lb-ft. Similarly, the moment created by the second child is 80 lbs x 4 feet = 320 lb-ft. So, if they are at rest and the smaller child is up in the air and the larger child is sitting on the ground, how much force is the ground exerting against the larger child? Easy! We know our forces and moments have to balance because there is no motion, so by looking at our moments we can construct the equation 360 = 320 + ground. Solving for the ground force we are left with 40 lb-ft, which upon dividing by the distance the force is from the center pivot, we obtain 10 pounds, as one would expect. Alright, now on to answering the question we actually care about.

Solution

Since we are dealing with moments, we need to know the actual dimensions of the barbell and weights we are asking about. Let’s assume we are using a Rogue Ohio Bar, and it’s being loaded with Rogue HG 2.0 bumper plates. I’ve chosen these plates because they are rather thick and will create a worst case scenario since they will force the loaded weight to hang further to one side. Now, with this bar and plates the dimensions we are about are as follows:

Unbalanced 45 lb Barbell

With all of this in mind, let’s create a free body diagram to visualize all of the forces and moments.

Unbalanced Barbell FBD

Alright, let’s talk about this picture. There are five main things that I’ve added that we care about: Forces acting on the system from the weight plates F1, and F2, forces acting on the system from the weight of the barbell, F3, forces acting on the system from the reactionary (balancing) forces caused by the squat rack R1, R2, the coordinate system that I’ve chosen to use, describing the X, Y, Z directions and coordinate system origin, O, from which all of my measurements will be relative to, and the distances these forces are from the origin. Since the barbell is axisymmetric we can apply the 45 pound force resulting from the mass of the barbell as a point force acting at its center of mass. Similarly, we can do this with the two plates.
You may notice that I chose to place my coordinate system origin through the line of action of the first reactionary force, this is on purpose. This allows me to neglect the moment created by R1 since it passes through the origin, resulting in a moment of M = 0"\cdot R_{1} = 0.

Working Out The Math

With all of this illustrated, let’s go ahead and construct our force and moment balance equations.

Sum of Forces:
1) \qquad \displaystyle \sum F = -F_{1} - F_{2} + R_{1} - F_{3} + R_{2} = 0

Sum of Moments:
2) \displaystyle \qquad \sum\nolimits M = 9.125" \cdot F_{1} + 5.375" \cdot F_{2} - 23.5" \cdot F_{3} + 47" \cdot R_{2} = 0

If we rearrange these equations we obtain:
1) \displaystyle \qquad F_{1} + F_{2} + F_{3} = R_{1} + R_{2}
2) \displaystyle \qquad 9.125" \cdot F_{1} + 5.375" \cdot F_{2} + 47" \cdot R_{2} = 23.5" \cdot F_{3}

We also know that forces F1, F2, and F3 are all 45 pounds. We can plug these values into equations 1 and 2 and solve for the second reactionary force. Doing this allows us to find out how much force the far side squat rack support arm is pushing back against the bar with. When this value drops to 0 or lower, the bar will lose contact with the rack since the support arm would then have to be pulling that side of the bar down in the negative y direction for forces to balance.

1) \displaystyle \qquad 45lb + 45lb + 45lb = R_{1} + R_{2}
2) \displaystyle \qquad 9.125" \cdot 45lb + 5.375" \cdot 45lb + 47" \cdot R_{2} = 23.5" \cdot 45lb

Simplifying further yields:
1) \displaystyle \qquad 135lb = R_{1} + R_{2}
2) \displaystyle \qquad 410.625 in-lb + 241.875 in-lb + 47" \cdot R_{2} = 1057.5 in-lb

Finally solving for the second reactionary moment:
2) \displaystyle \qquad 47" \cdot R_{2} = 405 in-lb

And dividing by the lever arm distance to obtain the reactionary force:
\displaystyle R_{2} = 8.617 lb

Now that we know R2, we can plug this in to equation 1 to find the first reactionary force.
1) \displaystyle \qquad 135lb = R_{1} + 8.617lb

Solving for R1 we obtain:
\displaystyle R_{1} = 126.383lb

Cool! So we can see that with this setup the barbell is still stable on the squat rack as the second support arm is exerting more than 8.5 lbs upwards to keep the barbell from falling.

Moving On

In a second part to this I will address the worst case scenario of when the bar is pulled all the way to one side, thereby creating longer lever arms for the plates, and a shorter lever arm for the mass of the barbell. We can even go as far as to solve for what kind of force would be required to start the bar tipping off the rack if someone were to bump in to it from the far side. Until then, happy lifting!