
#1
Feb307, 09:50 PM

P: 80

1. The problem statement, all variables and given/known data
An athlete whose mass m is performing weight lifting excersies. Starting from rest positions, he lifts, with constant acceleration, a barbell that weighs w. He lifts the barbell a distance of x in time of t. Use Newton's laws to find the total force his feet exert on the ground as he lifts the barbell. 2. Relevant equations F=ma 3. The attempt at a solution I drew the free body diagram already, with the force his feet exert on the ground downward, the normal force upward, and the gravity force downward. Is this correct? Is there any other force I'm missing? 



#2
Feb307, 10:15 PM

HW Helper
P: 858

interesting problem... things to note, when the barbell is being lifted its weight is different from when it is stationary (or not under acceleration). The normal upward force on the barbell is the force provided by the lifting. So, total foce on ground would I guess come from the body weight and the modified barbell weight




#3
Feb307, 10:40 PM

P: 80

The barbell weight changes? I don't get this part...




#4
Feb307, 10:58 PM

Emeritus
Sci Advisor
PF Gold
P: 5,198

Newton's Laws  Weight LiftingMa = F  w ==> wa/g + w = F ==> w(a/g + 1) = F Now, if he pushes up on the barbell with this force, then presumably the barbell pushes down on him with the same force (Newton's third law, this is why the problem says to use Newton's lawS (plural)). Which means that the floor, in addition to supporting his weight mg as it normally does, must also support force F. That's the best I can come up with for this problem. Somebody let me know if I'm totally on crack... 



#5
Feb307, 11:16 PM

HW Helper
P: 858

cepheid
what you have done have actually demonstrated that the (apparent) weight of the barbell has changed since F = w (a/g+1) which is different from its original weight w. It is like inside a lift when the lift accelerate up you feel that your feet is pushing down the floor of the lift harder. The term "weight" can sometimes be confusing. 



#7
Feb1107, 05:37 PM

P: 8

So... the normal force is N = mg + w(2x/(t^2)+1) ? Something like that?



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