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nadezda [96]
2 years ago
15

a uniform beam of length l and mass mb is supported by two pillars located l/3 from either end, as shown in the figure. a duck o

f mass md stands on one end. a scale is placed under each pillar. the entire system is in equilibrium.
Physics
1 answer:
Cloud [144]2 years ago
7 0

When the system is in equilibrium, the sum of the moment about a point is

zero.

Force \ shown  \ by \  the  \ scale \  under \  the \  right  \ pillar \ is \ F = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}

Reasons:

Length of the beam = l

Mass of the beam = m_B

Location \ of \  the \  two \  pillars = \dfrac{l}{3}  \  from  \  either  \  end

Mass of the duck = m_D

Required:

Force shown by the scale under the right pillar.

Solution:

The location of the duck = On the left end of the beam

When the system is in equilibrium, we have; ∑M = 0

Taking moment about the left pillar, we get;

Clockwise moment = m_D \times g \times  \dfrac{l}{3} + F \times \dfrac{l}{3}

Anticlockwise moment = m_B \times g \times  \dfrac{l}{6}

At equilibrium, clockwise moment = Anticlockwise moment

Therefore;

m_D \times g \times  \dfrac{l}{3} + F \times \dfrac{l}{3} = m_B \times g \times  \dfrac{l}{6}

F \times \dfrac{l}{3} = m_B \times g \times  \dfrac{l}{6} - m_D \times g \times  \dfrac{l}{3}

F = \dfrac{m_B \times g \times  \dfrac{l}{6} - m_D \times g \times  \dfrac{l}{3}}{\dfrac{l}{3} }  = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}

Force \ shown  \ by \  the  \ scale \  under \  the \  right  \ pillar, \  F = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}

Learn more here:

brainly.com/question/12227548

brainly.com/question/14778371

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The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

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the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

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Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

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We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

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we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

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        t = √ (10 / 4.9)

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Now let's use the first equation and the last one

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         9.8 t = v₀  sin θ

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we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

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the other angle that gives the same result is

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thisis the correct

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Setler [38]

I think the answer is B.

Hope this helps.

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