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Tanzania [10]
4 years ago
15

You hold a barbell in a horizontal position. The barbell’s center of mass is exactly mid-way between your two hands. A friend wa

lks up and attaches a weight on the end near your left hand. How does the torque on the bar exerted by your left hand compare to the torque on the bar from your right hand if the bar remains horizontal and at rest?

Physics
1 answer:
Harrizon [31]4 years ago
6 0

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)

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As you know a cube with each side 4 m in length has a volume of 64m3. Each side of the cube is now doubled in length. What is th
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The ratio of the new volume to the old volume is 8 to 1.

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Let's first list what we know:

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Since the new cube has sides double the length of the sides of the old cube, and 4 doubled is 8, the length of the sides of the new cube is 8.

The equation for the volume of a cube is V = s^3, where "V" is the volume and "s" is the lengths of the sides.

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The volume of the new cube is 512 m^3.

The ratio of the new volume to the old volume is 512 : 64.

Let's simplify the ratio:

512 : 64

8 : 1

The ratio of the new volume to the old volume is 8 to 1.

P.S. This question should be in the mathematics subject, not the physics subject. (I pretty much only do math problems, so yes, it does matter. I don't know about the others though.)

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