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torisob [31]
3 years ago
13

A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the s

maller piston to obtain a force of 1600 N at the larger piston
Physics
1 answer:
yan [13]3 years ago
8 0

Answer:

The  force is F_1  = 400.8 \  N

Explanation:

From the question we are told that

   The first  diameter is  d_1 =  4.0 \ cm =  0.04 \ m

   The second diameter is  d_2  =  8.0 \ cm  = 0.08 \  m

   

Generally the first area is  

         A_1  =  \pi  * \frac{d^2_1 }{4}

=>      A_1  = 3.142  * \frac{0.04^2}{4}

=>       A_1  =  0.00126 \ m^2

The  second area is  

     A_2 =  \pi  * \frac{d^2_2 }{4}

     A_2  = 3.142  * \frac{0.08^2}{4}

     A_2  =  0.00503 \ m^2

For a hydraulic press the pressure at both end must be equal .

Generally  pressure is mathematically represented as

    P =  \frac{F}{A}

=>  

   \frac{F_1}{A_1 }  =  \frac{F_2}{A_2 }

=>   F_1  =  \frac{1600}{0.00503}  *  0.00126

=>    F_1  = 400.8 \  N

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Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

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substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

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