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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:

brainly.com/question/28106868

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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

#SPJ1

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77julia77 [94]

Answer:

$V=200cm^3\\\\\rho =0.500g/cm^3$

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Explanation:

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In this case, given the width, length and height, we can compute the volume as follows:

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Moreover, since the density is computed via the division of the mass by the volume:

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3 years ago

A jogger runs 20 mi West and then 6.0 mi North. Find the magnitude and direction of the resultant displacement.

Black_prince [1.1K]

Answer:

The magnitude of the resultant displacement is 21 mi and its direction is 16.7° north of west

Explanation:

Hi there!

Please see the figure for a better understanding of the problem. The total displacement vector will be the sum of both displacements:

The vector for the first displacement is:

First displacement = (20 mi, 0)

The second displacement:

Second displacement = (0, 6.0 mi)

The resultant displacement will be:

R = (20 mi, 0) + (0, 6.0 mi) = (20 mi + 0, 0 + 6.0 mi) = (20 mi, 6.0 mi)

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The magnitude of the vector displacement is 21 mi.

To find the direction of the vector R, we have to apply trigonometry:

In a right triangle the following trigonometric rule applies:

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In this case:

cos θ = 20 mi / magnitude of R

θ = 16.7°

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2 years ago

A spring is hung vertically with a 425g mass attached to it. The mass is at rest. If the mass causes the spring to stretch 0.67

egoroff_w [7]

Answer:

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From Hooke's law we deduce that F=kx where F is the applied force and k is the spring constant while x is the extension or compression of the spring. Making k the subject of the above formula then

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