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2 years ago

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:

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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

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<h3>Explain about the Elastic Collision?</h3>

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2 years ago

A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a

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Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

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Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

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