My physics knowledge is pretty basic, somebody suggested that I'll get the definitive answer of below question here.

Imagine a barbell of mass m (kg) which is placed on the ground initially,And then it's lifted by a lifter at some angle from the ground. The question is, how much weight is actually lifted by the lifter. So, if the barbell's weight is 20kg and I add one 20kg weight plate to it, then, am I lifting 40kg when I lift it high enough that the bar makes a 30" angle? or, it's less or more.

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To be precise, the barbell is lifted like below image. for back workout :).

enter image description here

the end of the barbell which doesn't have plates can be considered touching the ground for simplicity.

  • $\begingroup$ Are you sure you're not a gym nerd? Maybe edit your question and remove the slurs. "That's our word!" $\endgroup$
    – BillOnne
    Oct 11, 2022 at 15:55
  • $\begingroup$ Your question needs clarification. Are you pulling the barbell up a ramp? $\endgroup$ Oct 11, 2022 at 16:08
  • $\begingroup$ @David - No, the barbell is placed on the ground in like a straight line. $\endgroup$
    – shashank
    Oct 11, 2022 at 16:18
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Oct 11, 2022 at 16:33
  • $\begingroup$ Welcome @shshank. Is all the weight (20kg or 40kg) at the raised end? How exactly is the lifter holding the barbell: Both hands on the end with the weights or somewhere along the bar? Is the lower end fixed on the ground so it can't move but only rotate? $\endgroup$ Oct 11, 2022 at 17:06

1 Answer 1


Aah, you are asking about landmine exercises.

If the lift is optimally applied, i.e. perpendicular to the bar, as you lift the plates the bar will take more and more of the weight until it takes all of the weight when the bar is vertical. The precise answer for the effective weight as a function of angle is given in @C-Lange's answer to the Physical Fitness Stack Exchange question What is the effective weight of the bar when I'm doing a landmine squat?:

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

Stick in 30° for angle_to_floor, 40 kg for loaded_weight, and include the weight of bar, and you'll have your answer. (You wrote 30" for the angle, but since 30 arcseconds corresponds to lifting the end a fraction of a millimetre, I assume you meant 30°.)

If you want to understand the physics better, you might want to look at What is the work done to lift an object standing on a pivot from flat to angle θ or How much force is required to lift the back of a 2,480 lb car in the air from the bumper?.

Note, however, that if the direction of the lift is not optimal, the effective weight could actually be more than than the weight of the plates. For example, in the extreme case of pulling along the bar instead of perpendicular to it, I think it is intuitive that you couldn't get the plates off the ground. If I haven't made any mistake in my mental trigonometry, the vertical component of a force $F$ parallel to the bar is $F\sin{\theta}$, where $\theta$ is the angle of the bar with the floor, so the force needed to lift the weight is $W/\sin{\theta}$ which goes to infinity for $\theta=0$ and is never less than $W$ at any angle. I doubt any landmine exercise has people pulling directly along the bar, but if your pull goes past 90° from the floor (in the vertical plane through the bar) then you are both lifting the weight and trying to pull the bar out of its socket. If the bar was loose in the socket, it should pull out (which could be dangerous), but if the bar is fixed in the socket you'll be exerting a force greater than the weight.


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