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3 years ago

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Suppose we wish to use a 6.0 m iron bar to lift a heavy object by using it as a lever. If we place the pivot point at a distance

of 0.8 m from the end of the bar that is in contact with the load and we can exert a downward force of 527 N on the other end of the bar, find the maximum load that this person is able to lift (pry) using this arrangement (neglect the mass of the bar in this problem).

1 answer:

Ber [7]3 years ago

8 0

Answer:

The maximum load that this person is able to lift is 34.3 N

Explanation:

Applying the balancing torque, the expression is equal:

F₁L₁ = F₂L₂

$mgL_{1} =F_{2} L_{2}$

Where

g = 9.8 m/s² = gravity

L₁ = 0.8 m

F₂ = 527 N

L₂ = 6 - 0.8 = 5.2 m

Replacing and clearing the mass m:

$m=\frac{F_{2}L_{2} }{gL_{1} } =\frac{527*5.2}{9.8*0.8} =3.5kg$

The maximum load that this person is able to lift is:

F = m * g = 3.5 * 9.8 = 34.3 N

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