The velocity is a vectorial quantity, whereas speed is a scalar quantity, meaning it depends on the direction!
As such, the velocity is changing because the direction is changing.
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
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20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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Answer:
hope this answer helps this is what I understand
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)
Please reply to this with your question I would like to help.