Answer:
f_D = =3.24 N/m
Explanation:
data given
properties of air
k = 0.0288 W/m.K
WE KNOW THAT
Reynold's number is given as
= 1.941 *10^4
drag coffecient is given as
solving for f_D
Drag coffecient for smooth circular cylinder is 1.1
therefore Drag force is
f_D = =3.24 N/m
Explanation:
First consider that each hand works as a fulcrum: a pivot point where the barbell can rotate.
Now consider only the left hand. If the center of mass of the barbell is between hands (in the middle) it is displaced respect the fulcrum, therefore the weight which is pushing the bar downwards becomes a rotational force. The same thing happens to the other hand. Now, if more weight is added to the left hand the center of mass is displaced towards the left hand and depending how much weight is added, the center of mass will change its position and therefore the torque each hand experiences changes.
If the center of mass is still between hands: The torque remains almost the same changing only the magnitudes but not the direction.
If the center of mass is on the hand: there is no torque for the left hand because there is no leaver.
If the center of mass is to the left: now the torque changes direction and both hands need to stop it in the same direction.
(see diagram below)
