Question

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).

Answer 1

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