Now try some more complex combinations. With 15kg at the 1.5m radius, balance with a 10kg and 5kg mass on the other side. Notice that the 10kg and 5kg masses cannot be positioned at the same radius. Balance the 20kg mass at 1.75m with just the 5kg and 10kg mass. Do the same with the 20kg mass at 0.75m. You must use both the 5kg mass and the 10kg mass. Neatly record your data. Again show how your data supports your explanation of torque. If it does not support your explanation exactly as stated, then improve your explanation?

Question

Physics

Guest

27 June, 03:30

Now try some more complex combinations. With 15kg at the 1.5m radius, balance with a 10kg and 5kg mass on the other side. Notice that the 10kg and 5kg masses cannot be positioned at the same radius. Balance the 20kg mass at 1.75m with just the 5kg and 10kg mass. Do the same with the 20kg mass at 0.75m. You must use both the 5kg mass and the 10kg mass. Neatly record your data. Again show how your data supports your explanation of torque. If it does not support your explanation exactly as stated, then improve your explanation?

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Answers (1)

Know the Answer?

Zane Mata · Answer 1

a) x = 2.33 m

b) x = 1 m

Explanation:

a)

We have 20 kg mass at one side, at a distance of 1.75 m. W need to balance it with 5 kg and 15 kg masses on the other side. To balance both sides, there torque must be equal. So from conservation of torque:

(20 kg) (1.75 m) = (5 kg + 10 kg) (x)

where,

x = the distance of 5 kg and 10 kg masses from the center.

Therefore,

(20 kg) (1.75 m) = (15 kg) (x)

x = 35 kg m/15 kg

x = 2.33 m

b)

Now, we need to balance the same 20 kg mass with the 5 kg and 10 kg masses. But this time the distance of 20 kg mass from center is 0.75 m. Therefore,

(20 kg) (0.75 m) = (5 kg + 10 kg) (x)

x = 15 kg m/15 kg

x = 1 m