Forces when lifting a weight and punching

[sgstudent]

When a person pushes up a weight like in this picture http://www.freefitnessguru.com/Anatomy/images/military press 5 combined.jpg the weight is exerted on the wrist. So when the weight is stationary, the downwards force acting on the wrist is equal to the weight of the object. However, when he starts lifting it up, would this downwards force remain the same just that now the upwards force on the wrist>downwards force by the weight of the object?

While what if the wrist cannot handle the weight so it might snap. So in this case the wrist would be the weakest link here. But for example if a weak person loads too much weight, and the bone structure/muscles in the wrists can handle the forces however the triceps and shoulder (primary movers) cannot handle the weight then now they would be the weakest link?

When a person punches a bag, we would use the formula Δmv/t so a person punches at v1=10m/s and stops at 0m/s. But what would be the m? Would it just be the fist? Because the hips are also being used in the punch so I'm not sure which m to use. Would it just be the fist? If not what should it be?

Thanks in advance

 

[Dale]

However, when he starts lifting it up, would this downwards force remain the same just that now the upwards force on the wrist>downwards force by the weight of the object?

While the weight is accelerating upwards the downwards force on the wrist becomes greater than the weight of the object.

While what if the wrist cannot handle the weight so it might snap. So in this case the wrist would be the weakest link here. But for example if a weak person loads too much weight, and the bone structure/muscles in the wrists can handle the forces however the triceps and shoulder (primary movers) cannot handle the weight then now they would be the weakest link?

If there is a mechanical failure then there is some place where the mechanical failure first occurs. If there is no mechanical failure then it is hard to know where a mechanical failure would occur under different conditions without some complicated computer modeling.

When a person punches a bag, we would use the formula Δmv/t so a person punches at v1=10m/s and stops at 0m/s. But what would be the m? Would it just be the fist? Because the hips are also being used in the punch so I'm not sure which m to use. Would it just be the fist? If not what should it be?

A better approach would be to measure the change in momentum of the bag, not the person. A still better approach would be to just measure the force directly.

 

[sgstudent]

While the weight is accelerating upwards the downwards force on the wrist becomes greater than the weight of the object.

If there is a mechanical failure then there is some place where the mechanical failure first occurs. If there is no mechanical failure then it is hard to know where a mechanical failure would occur under different conditions without some complicated computer modeling.

A better approach would be to measure the change in momentum of the bag, not the person. A still better approach would be to just measure the force directly.

oh that's right. Because now the upwards force on the weight is greater so the Newton's Third Law reaction force on the hand would also be greater. So the wrist would have to accelerate at the same speed as the weight. So this would mean the force acting on the wrist to counteract the downwards force must be greater than the downwards force?

I think measuring the bag would also be better. But if we wanted to measure the person punching it instead, how should we go about doing that?

Thanks so much for the help :)

 

[Lsos]

I think it's difficult to calculate the momentum behind the punch. The mass of the fist certainly comes in play, but as you said so does the mass of the forearm, arm, hips, shoulders, etc. However, they are all moving at different speeds, and not in line with the fist...all dependent on the skill of the puncher. It's probably best to measure this.

 

[Dale]

oh that's right. Because now the upwards force on the weight is greater so the Newton's Third Law reaction force on the hand would also be greater. So the wrist would have to accelerate at the same speed as the weight. So this would mean the force acting on the wrist to counteract the downwards force must be greater than the downwards force?

Yes, by a small amount equal to the weight of the hand times its acceleration.

I think measuring the bag would also be better. But if we wanted to measure the person punching it instead, how should we go about doing that?

I would recommend some force plates installed in the ground, one under each foot, as well as several cameras with fiducial markers mounted on the subject to track the motion.