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Nataly [62]
3 years ago
7

An woman whose weight is 804 N stands on a long horizontal plank of wood 1.55 m from one end. The plank is uniform and is suppor

ted at each end by vertical posts. It has a mass of 27 kg and is 5.5 m long. With what force does the post that is closest to the woman push upward on the board
Physics
1 answer:
meriva3 years ago
6 0

Answer:

Explanation:

There will be reaction force by  each vertical post on horizontal plank . Let it be R₁ and R₂ . R₁ is reaction force by the post nearer to woman

Taking torque of all forces about the end far away from the woman

Torque by reaction force = R₁ x 5.5

= 5.5 R₁ upwards

Torque by weight of woman in opposite direction , downwards

= - 804 x ( 5.5 - 1.55 )

= - 3175.8

Torque by weight of the plank in opposite direction , downwards .

= - 27 x 5.5 / 2

= - 74.25

Torque by R₂ will be zero as it passes through the point about which torque is being taken .

Total torque

= 5.5 R₁ - - 3175.8 - - 74.25  = 0 ( For equilibrium )

5.5 R₁ = 3250

R₁ = 590.9 N .

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a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

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a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

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E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

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b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

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Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
2 years ago
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