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A balanced lever has two weights on it, the first with mass #14 kg # and the second with mass #9 kg#. If the first weight is # 2 m# from the fulcrum, how far is the second weight from the fulcrum?

Answer 1

Nathan Ashcroft

#3.1m#

A moment is the product of force and the distance over which the force acts.

#M = Fd#

In a balanced lever, both moments are equal to each other, so

#M_1 = M_2#

#F_1d_1 = F_2d_2#

We can work out the force either side by #F=ma#, where #a~~10ms^-2#, so

#F_1 = 14 xx 10 = 140N#

#F_2 = 9 xx 10 = 90N#

and put these values, along with #d_1 = 2m# (from the question) into the equation:

#2m xx 140N = d xx 90N#

#d = 280/90 = 3.1m#

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Answer 2

Elena Albarran

Using the principle of torque balance for a balanced lever, the torque exerted by the first weight (14 kg) equals the torque exerted by the second weight (9 kg).

Torque = force × distance from fulcrum

For the first weight:
Torque1 = (14 kg) × (9.8 m/s^2) × (2 m)

For the second weight:
Torque2 = (9 kg) × (9.8 m/s^2) × (x m)

Setting Torque1 equal to Torque2:
(14 kg) × (9.8 m/s^2) × (2 m) = (9 kg) × (9.8 m/s^2) × (x m)

Solving for x:
(14 kg) × (2 m) = (9 kg) × (x m)
28 kgm = 9 kg * x
x = (28 kgm) / (9 kg)
x ≈ 3.11 m

Therefore, the second weight is approximately 3.11 meters from the fulcrum.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.