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Snowcat [4.5K]
3 years ago
7

A 15kg beam that is 10m long is placed on a fulcrum that is 3m from the end an 80kg person sits at the end closer to the fulcrum

what does the mass of an object plaved at the opposite end need to be in order to balance the beam?
Physics
1 answer:
Andrei [34K]3 years ago
7 0

Answer:

 m₃ = 30 kg

Explanation:

This is a problem of rotational equilibrium, let's write Newton's law for rotational equilibrium.

Let's fix our reference system in the support, the positive torques are those that create an anti-clockwise turn

Let's look for the distances to the point of support

The distance of man

      x₁ = -3 m

The distance of the bar is

      x₂ = L / 2 -3

     x₂ = 10/2 -3

     x₂ = 2 m

Remote object at the end

      x₃ = L-3

     x₃ = 10-3

     x₃ = 7 m

They give us the mass of man (m1) and the mass of the bar (m2)

Let's write the torques

      W₁ x₁ - W₂ x₂ - w₃ x₃ = 0

       m₁ g 3 - m₂ g 2 - m₃ g 7 = 0

       m₃ = (3m₁ -2m₂) / 7

Let's calculate

      m₃ = (3 80 -2 15) / 7

     m₃ = 30 kg

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a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical
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Answer:

Vx =  10.9 m/s ,  Vy = 15.6 m/s

Explanation:

Given velocity V= 19 m/s

the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°

θ = 55°

to Find Vx = ? and Vy= ?

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As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

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