A balanced lever has two weights on it, the first with mass #8 kg # and the second with mass #3 kg#. If the first weight is # 2 m# from the fulcrum, how far is the second weight from the fulcrum?

Answer 1

#5.333m#

We apply the second condition for static equilibrium which states that the resultant torque (moments) must be zero, ie. #sum vectau = 0#, where #vec tau= vecr xx vecF# is the cross product of the position vector #vecr# from the axis of rotation (fulcrum) to the applied force and the applied force #vecF#.

So application in this case yields :

#r_1F_1-r_2F_2=0#
#therefore (2xx8g)-(r_2xx3g)=0#
#therefore r_2=(2xx8xx9.8)/(3xx9.8)=5.333m#.
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Answer 2

Using the principle of moments, the product of the mass and the distance from the fulcrum is equal on both sides. Thus, ( 8 , \text{kg} \times 2 , \text{m} = 3 , \text{kg} \times \text{distance} ). Solving for the distance of the second weight from the fulcrum, we get: ( \text{distance} = \frac{8 , \text{kg} \times 2 , \text{m}}{3 , \text{kg}} = 16/3 , \text{m} ), or approximately ( 5.33 , \text{m} ).

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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.

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.

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.

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.

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