Lab 5: Weightlifting
Procedure and Analysis:
The lab began with the group making a rough estimate of the force required from each arm to lift the weight up to be about 700N per arm.
We then drew a force diagram for the force acting on the barbell and when the lifter held the barbell over his head.
Were:
B= the angle=35 degrees
Fn= The normal force m= mass g=gravity F= the force exerted by each arm
We then used the following equation W = mg to calculate the weight of the barbell.
Were:
W= weight m=mass g=gravity
W = 135kg ∙ 9.81m/s2 = 1324.35 N
We then estimate the angle of each arm acting on the bar to be about 60 degrees. We then broke down both Vector F into its component vectors. Since the the group assumed that the force needed to lift the barbell is the same in each arm we multiply our component vector sin(b) times 2.
F = 2W (sin(B))
Were:
F= the force in both arms
W= weight
B= the angles that Vector F makes with the xaxis
F = 2(1324.35N)(sin(35)) = 1517.68 N
This calculation is for both arms the force exerted by each arm is about 758.8N
Part B:
The group now proceed to make a force diagram for when the barbell was held at waist level just before the lift. The group came up with the following diagram.
Were:
F= the force in one arm m= mass g= gravity
Fn= The Normal Force
B= an angle
We then procedure to repeat the procedure in part A but realize that their was no need since the Force diagram generated in part B was the exact same diagram as the one in part A the only difference was that in part B everything was below the xaxis but the calculations would still be the same.
We then proceed to make a force diagram for when the barbell as it is lifted past the eyes.
In this diagram the barbell is lifted past the eyes thus the arms are at a 90 degree angle.
When we break the Force vectors into y and x components the x components equal 0 since there is no force in the x direction thus
F=W(sin(C))
The lab began with the group making a rough estimate of the force required from each arm to lift the weight up to be about 700N per arm.
We then drew a force diagram for the force acting on the barbell and when the lifter held the barbell over his head.
Were:
B= the angle=35 degrees
Fn= The normal force m= mass g=gravity F= the force exerted by each arm
We then used the following equation W = mg to calculate the weight of the barbell.
Were:
W= weight m=mass g=gravity
W = 135kg ∙ 9.81m/s2 = 1324.35 N
We then estimate the angle of each arm acting on the bar to be about 60 degrees. We then broke down both Vector F into its component vectors. Since the the group assumed that the force needed to lift the barbell is the same in each arm we multiply our component vector sin(b) times 2.
F = 2W (sin(B))
Were:
F= the force in both arms
W= weight
B= the angles that Vector F makes with the xaxis
F = 2(1324.35N)(sin(35)) = 1517.68 N
This calculation is for both arms the force exerted by each arm is about 758.8N
Part B:
The group now proceed to make a force diagram for when the barbell was held at waist level just before the lift. The group came up with the following diagram.
Were:
F= the force in one arm m= mass g= gravity
Fn= The Normal Force
B= an angle
We then procedure to repeat the procedure in part A but realize that their was no need since the Force diagram generated in part B was the exact same diagram as the one in part A the only difference was that in part B everything was below the xaxis but the calculations would still be the same.
We then proceed to make a force diagram for when the barbell as it is lifted past the eyes.
In this diagram the barbell is lifted past the eyes thus the arms are at a 90 degree angle.
When we break the Force vectors into y and x components the x components equal 0 since there is no force in the x direction thus
F=W(sin(C))