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Ber [7]
4 years ago
12

70 kg man walks out on a 10 kg beam that rests on, but is not attached to, two supports. When the beam just starts to tip, what

is the force exerted on the beam by the right support
Physics
1 answer:
Serggg [28]4 years ago
5 0

Answer:

The force is  F  =  784 \ N

Explanation:

From the question we are told that

      The mass of the man is  m = 70 \ kg

      The mass of the beam is m_b  =  10 \ kg

     

Now from the question we can deduce that when this beam start to tip that both the force exerted by the weight of the man and that of the beam is been supported by the  right support so

 The force exerted on the right support is mathematically evaluated as

           F  =  (m +  m_b) * g

substituting values

         F  =  (70 +  10 ) * 9.8

         F  =  784 \ N

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Consider a light, single-engine airplane such as the Piper Super Cub. If the maximum gross weight of the airplane is 7780 N, the
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Solution

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W=(1/2)S_{area}(V_{Stalling-speed} )^{2} (P_{Air-density} )(C_{Lmax} )\\ so\\V_{Stalling-speed}=\sqrt{\frac{2W}{S_{area}*(P_{Air-density} )(C_{Lmax} )} }\\V_{Stalling-speed}=\sqrt{\frac{2*7780}{16.6*2.1*1.225} }\\  V_{Stalling-speed}=19.08 m/s

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