Question #49734

A hydraulic jack system is created in the configuration below. A person pushes down with force F on one end of the lever arm of total length l= 1.2m. The lever arm is attached to a fixed pivot on the opposite end and in turn pushes straight down on the input piston, located halfway down the lever arm, with Fi. The ratio of the output pistons area to the input pistons area is Ao/Ai=55. You want the output force to sustain the weight of a W = 6000N object.
a) Algebraically find the input pistons force Fi in terms of the persons force F using a torque argument.
b) Find the required force the person must exert in order to support the object by relating the expression for Fi to Fo.

Expert's answer

Answer on Question#49734 - Physics - Molecular Physics

A hydraulic jack system is created in the configuration below. A person pushes down with force FF on one end of the lever arm of total length l=1.2ml = 1.2\mathrm{m} . The lever arm is attached to a fixed pivot on the opposite end and in turn pushes straight down on the input piston, located halfway down the lever arm, with FiF_{i} . The ratio of the output pistons area to the input pistons area is AoAi=55\frac{A_{o}}{A_{i}} = 55 . You want the output force to sustain the weight of a W=6000NW = 6000\mathrm{N} object.

a) Algebraically find the input pistons force FiF_{i} in terms of the persons force FF using a torque argument.

b) Find the required force the person must exert in order to support the object by relating the expression for FiF_{i} to FoF_{o} .

Solution:


Applying the principle of moments to the lever we obtain


l2Fi=lF\frac {l}{2} F _ {i} = l \cdot F


or


Fi=2FF _ {i} = 2 F


Since the pressures under input and output pistons are equal, we can write the following relation for FiF_{i} and FoF_{o}

FiAi=FoAo\frac {F _ {i}}{A _ {i}} = \frac {F _ {o}}{A _ {o}}


We can easily find FoF_{o} from this expression


Fo=AoAiFi=2FAoAiF _ {o} = \frac {A _ {o}}{A _ {i}} F _ {i} = 2 F \frac {A _ {o}}{A _ {i}}


Since the magnitude of the output force has to be equal to the weight of the object, we obtain


W=2FAoAiW = 2 F \frac {A _ {o}}{A _ {i}}


or


F=W2AiAo=6000N2155=54.5NF = \frac {W}{2} \frac {A _ {i}}{A _ {o}} = \frac {6000 \mathrm{N}}{2} \frac {1}{55} = 54.5 \mathrm{N}


Answer:

a) Fi=2FF_{i} = 2F

b) F=W2AiAo=54.5NF = \frac{W}{2} \frac{A_i}{A_o} = 54.5\mathrm{N}

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