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cricket20 [7]
2 years ago
11

A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

Physics
2 answers:
Keith_Richards [23]2 years ago
8 0

The hi.nge will apply a force of 142.10N on the beam in the vertical direction.

We must learn more about the tension in order to find the solution.

<h3>How can I determine the vertical component of the force the hi.nge has on the beam?</h3>
  • Let's create the system's free body diagram.
  • We must balance the total vertical force to zero in order to get the vertical component of the force applied to the beam by the height.

                           F_V=mg-Tsin\alpha

  • We must identify the tension in order to find the solution.

                            Tlsin\alpha =mg\frac{l}{2}sin\beta  \\T=\frac{mgsin90}{2sin57} =169.43N

  • Consequently, the force that the height exerts on the beam will have a vertical component that is,

                     F_v=(29*9.8)-(169.43*sin57)=142.10N

This leads us to the conclusion that the vertical component of the force the hi.nge exerts on the beam will be 142.10N.

Learn more about the tension here:

brainly.com/question/28106868

#SPJ1

castortr0y [4]2 years ago
5 0

The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

<h3>How to find the vertical component of the force exerted by the hi.nge on the beam?</h3>
  • Let's draw the free body diagram of the system.
  • To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.

                      F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\

  • To find the answer, we have to find the tension,

                     Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N

  • Thus, the vertical component of the force exerted by the hi.nge on the beam will be,

                F_V=(29*9.8)-(169.43*sin57)=142.10N

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

Learn more about the tension here:

brainly.com/question/28106868

#SPJ1

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Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

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Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to F:

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m.a_{C}=G\frac{Mm}{r^{2}} (4)

Simplifying:

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