Question

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

Answer 1

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

         $F = (70 + 10 ) * 9.8$

         $F = 784 \ N$

Answer 2

The force exerted on the beam by the right support is 784 Newton.

Given the data in the question;

• Mass of the man; $m_m = 70kg$

• Mass of the beam; $m_b = 10kg$

Force exerted on the beam by the right support; $F = W = \ ?$

When the beam just starts to tip, the right support holds up the combined mass of the man and the beam.

Hence;

$M_{net} = m_m + m_b\\\\M_{net} = 70kg + 10kg\\\\M_{net} = 80kg$

Now, To determine the force exerted on the beam by the right support, we use the general formula for weight or equation of force of gravity which is expressed as:

$F = W = m * g$

Where m is mass and g represents the acceleration due to gravity( $9.8m/s^2$ )

We substitute our values into the equation

$F = 80kg * 9.8m/s^2\\\\F = 784kg.m/s^2\\\\F = 784N$

Therefore, the force exerted on the beam by the right support is 784 Newton.

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