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Verizon [17]
3 years ago
5

a vertical solid steel post 29cm in diameter and 2.0m long is required to support a load of 8200kg, ignore the weight of the pos

t. determine the stress in the post​
Physics
1 answer:
erik [133]3 years ago
7 0

Answer:

The stress is  \sigma  =  1.218*10^{6} \  N/m^2

Explanation:

From the question we are told that

   The diameter of the post is  d =  29 \ cm  =  0.29 \  m

   The length is L  =  2.0 \  m

    The weight of the loading mass

Generally  the  radius of the post is mathematically represented as

     r =  \frac{0.29}{2}

=>   r = 0.145 \  m

Generally the area of the post is  

       A =  \pi r^2

=>     A =  3.14 *  0.145 ^2

=>     A =  0.066 \ m^2

Generally the weight exerted by the load is mathematically represented as

        F =  m  *  g

=>      F =  8200  *  9.8

=>      F =  80360 \  N

Generally the stress is mathematically represented as

         \sigma  =  \frac{F}{A}

=>      \sigma  =  \frac{80360 }{0.066}

=>      \sigma  =  1.218*10^{6} \  N/m^2

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The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25
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From the question,

Work done by the frictional force = Kinetic energy of the object

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Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.

Make d the subject of the equation.

d = 1/2m(v²-u²)/F.................. Equation 2

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