# What is the effective weight of the bar when I'm doing a landmine squat?

When I'm doing a landmine squat the bar weighs 20 kg. As it is pivoted in the corner of the room the effective weight of the bar must be less than its 20 kg but it must contribute something to the weight of the lift - but how much. Its obviously somewhere in between 0.1 kg and 19.9 kg but how much?

Many Thanks

• While this question obviously concerns fitness, you might get a more thorough response at Physics. May 6 at 13:33

At the very bottom (barbell on the ground) the effective weight of the bar is half. The bar is supported at one end by you and at the other end by the landmine contraption and therefore the lifter is twice the distance from the fulcrum as the centre of mass. Throughout the range of motion the effective weight of the barbell changes with respect to the angle. At 60° the force required to lift the bar is 25% of the weight and at 90° it would be 0%.

This is accurate for only the bar since the weight is assumed to be a uniform distribution. Once you add a load to the end it changes a bit since the lifter and the load are the same distance from the fulcrum. On the floor, the effective weight would be 100% of the load + half the bar. At 60° the effective weight would be 50% of the load + 25% of the bar weight. Still 0% at 90°.

So, in the landmine squat, say we move from 45° to 60° in a rep the weight is changing from 35% to 25% of the bar weight and 71% to 50% of the loaded weight.

You could calculate it like so: ((bar_weight/2) + loaded_weight)*cos(angle_to_floor).

• Someone's got to double check that for me though.
– C. Lange
May 6 at 4:57
• if you remove the loaded weight, then the formula is total weight * cos(angle) /2 for the bar, which is what you have. the loaded weight is not constant throughout the bar, and is based on a second-order lever. while negligible difference depending on where your applying force. for example, a 45lb plate if your picking up the bar at the end of the sleeve could feel like 37-38 lbs given the formula for a second order lever if solving for |Fe| in -Fe × de = Fr × dr May 6 at 15:46
• @AceCabbie -- right. So, my original assumption here was that the loaded plates and force applied were at the same point for simplicity. You're saying that the plates are at about 80% of the length of the barbell, where force applied is at the end. So it would be (bar_weight/2) + (loaded_weight/1.23)? (then the same angle mult.)
– C. Lange
May 6 at 16:06
• This answer's calculation is correct (although it has simplified the problem to be easier to calculate, so it's not perfectly accurate). Here is a landmine squat calculator based on this answer: desmos.com/calculator/nqznvau2v9 May 6 at 19:55
• For any reader's sake, though, this is purely for curiosity at this point. If you're actively trying to get better at landmine squats just ignore the bar weight and track the plates you put on.
– C. Lange
May 7 at 15:52

The answer to this question can get a little complicated, depending on how accurate you want to be.

Regardless of bar position, the centre of mass of the bar is always half way between the landmine attachment and your grip, meaning that its weight is divided evenly between the two. So if you want to know how much to add to the plates on the bar to get the total weight, 10kg would be a reasonable initial approximation.

Caveat: In a landmine squat, you have some leverage due to the fact that you're gripping the bar further away from the point of rotation than the plates, which means that the landmine itself will take about 10-15% of the weight of the plates on the bar, leaving you with 85-90% of the weight of the plates, plus 50% . So if you have a 20kg plate and a 20kg bar, the total weight would actually be 10kg from the bar, and maybe only 18kg from the plate, adding up to 28kg.

Now, I should also mention the angle of the bar, since that's been factored in in another answer. I don't think the bar angle alone affects the load placed on the lifter, but rather the combination of the bar angle and the angle of the lifter's body (measured between the centre of their feet and the point where they're holding the bar) does. The body angle doesn't change as much as the bar angle does, since the lifter's feet are fixed in position, which results in the lifter never really pushing at 90° to the bar, and instead leaning more into the bar at the top of the lift. The effective load decreases as the bar angle increases (becomes more upright) and increases as the angle at which the lifter pushes on the bar moved away from perpendicular. I estimate that this will mean the lifter needs to exert force equivalent to 82% of the previously calculated weight at the top of the lift, and 95% of the previously calculated weight at the bottom of the lift. If desired, I can add in my calculations, but I really think they're unnecessarily complicated as an answer to this question, and I'd recommend just ignoring the angles of the bar and lifter.

As an aside, I think the linked article vastly overstates the benefit of landmine squats, and unfairly criticises back squats. I can't think of a single situation in which I'd recommend a landmine squat over any kind of free-weight squat.

• Thanks for the answer. TBH I just grabbed the first article with a pic of a landmine squat. I like them as it's good for me to practice a decent squat, My squat is pretty shaky May 6 at 8:56